Rezolvarea completă a modelului oficial pentru Bacalaureat 2026 la Matematică M_pedagogic . Pentru fiecare cerință găsești enunțul pe scurt, rezolvarea pas cu pas, răspunsul final și cum se punctează conform baremului oficial. Timp de lucru la examen: 3 ore; se acordă 10 puncte din oficiu.
Cum să folosești rezolvarea: lucrează întâi singur subiectul (deschide PDF-ul de mai sus), apoi compară fiecare pas. Dacă ai ajuns la alt drum corect, primești tot punctajul — baremul acceptă orice soluție corectă.
Subiectul I (30 de puncte)
1. Arătați că 64 − 4 2 + 2 2 ⋅ ( 2 − 2 ) = 4 \sqrt{64} - 4\sqrt2 + 2\sqrt2\cdot\left(2 - \sqrt2\right) = 4 64 − 4 2 + 2 2 ⋅ ( 2 − 2 ) = 4
64 = 8 \sqrt{64} = 8 64 = 8 2 2 ⋅ 2 − 2 2 ⋅ 2 = 4 2 − 2 ⋅ 2 = 4 2 − 4 2\sqrt2\cdot 2 - 2\sqrt2\cdot\sqrt2 = 4\sqrt2 - 2\cdot 2 = 4\sqrt2 - 4 2 2 ⋅ 2 − 2 2 ⋅ 2 = 4 2 − 2 ⋅ 2 = 4 2 − 4
8 − 4 2 + 4 2 − 4 = 8 − 4 = 4. 8 - 4\sqrt2 + 4\sqrt2 - 4 = 8 - 4 = 4. 8 − 4 2 + 4 2 − 4 = 8 − 4 = 4.
Termenii cu 2 \sqrt2 2 Barem: 3p calculul termenilor, 2p finalizarea.
2. f ( x ) = 7 x − 4 f(x) = 7x - 4 f ( x ) = 7 x − 4 a a a f ( a ) = 5 a f(a) = 5a f ( a ) = 5 a
f ( a ) = 7 a − 4 f(a) = 7a - 4 f ( a ) = 7 a − 4 7 a − 4 = 5 a ⟺ 2 a = 4 ⟺ a = 2 7a - 4 = 5a \iff 2a = 4 \iff a = 2 7 a − 4 = 5 a ⟺ 2 a = 4 ⟺ a = 2
Răspuns: a = 2 a = 2 a = 2 Barem: 2p f ( a ) f(a) f ( a )
3. Ecuația 9 ⋅ 3 x = 3 5 9\cdot 3^x = 3^5 9 ⋅ 3 x = 3 5
Scriem 9 = 3 2 9 = 3^2 9 = 3 2 3 2 ⋅ 3 x = 3 2 + x 3^2\cdot 3^x = 3^{2+x} 3 2 ⋅ 3 x = 3 2 + x 3 2 + x = 3 5 3^{2+x} = 3^5 3 2 + x = 3 5 2 + x = 5 2 + x = 5 2 + x = 5 x = 3 x = 3 x = 3
Verificare: 9 ⋅ 27 = 243 = 3 5 9\cdot 27 = 243 = 3^5 9 ⋅ 27 = 243 = 3 5
Răspuns: x = 3 x = 3 x = 3 Barem: 3p aducerea la aceeași bază, 2p soluția.
4. O ieftinire cu 15% micșorează prețul cu 75 de lei; prețul inițial
Dacă x x x 15 % 15\% 15% x x x
15 100 ⋅ x = 75 ⇒ x = 75 ⋅ 100 15 = 500 lei . \frac{15}{100}\cdot x = 75 \Rightarrow x = \frac{75\cdot 100}{15} = 500 \text{ lei}. 100 15 ⋅ x = 75 ⇒ x = 15 75 ⋅ 100 = 500 lei .
Atenție: 75 de lei este scăderea , nu prețul nou (prețul nou ar fi 425 de lei).
Răspuns: 500 de lei. Barem: 3p ecuația, 2p rezultatul.
5. A ( 0 , 4 ) A(0,4) A ( 0 , 4 ) B ( 2 , 0 ) B(2,0) B ( 2 , 0 ) C ( 6 , 6 ) C(6,6) C ( 6 , 6 ) M M M O C OC O C M A = M B MA = MB M A = M B
M ( 0 + 6 2 , 0 + 6 2 ) = M ( 3 , 3 ) M\left(\dfrac{0+6}{2}, \dfrac{0+6}{2}\right) = M(3,3) M ( 2 0 + 6 , 2 0 + 6 ) = M ( 3 , 3 )
M A = ( 3 − 0 ) 2 + ( 3 − 4 ) 2 = 9 + 1 = 10 , M B = ( 3 − 2 ) 2 + ( 3 − 0 ) 2 = 1 + 9 = 10 . MA = \sqrt{(3-0)^2 + (3-4)^2} = \sqrt{9 + 1} = \sqrt{10}, \qquad MB = \sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1 + 9} = \sqrt{10}. M A = ( 3 − 0 ) 2 + ( 3 − 4 ) 2 = 9 + 1 = 10 , M B = ( 3 − 2 ) 2 + ( 3 − 0 ) 2 = 1 + 9 = 10 .
Deci M A = M B MA = MB M A = M B Barem: 2p coordonatele lui M M M
6. Arătați că 4 ( sin 60 ∘ ) 2 − ( sin 45 ∘ ) 2 − 2 ( sin 30 ∘ ) 2 = 2 4(\sin60^\circ)^2 - (\sin45^\circ)^2 - 2(\sin30^\circ)^2 = 2 4 ( sin 6 0 ∘ ) 2 − ( sin 4 5 ∘ ) 2 − 2 ( sin 3 0 ∘ ) 2 = 2
Valori: sin 60 ∘ = 3 2 \sin60^\circ = \dfrac{\sqrt3}{2} sin 6 0 ∘ = 2 3 sin 45 ∘ = 2 2 \sin45^\circ = \dfrac{\sqrt2}{2} sin 4 5 ∘ = 2 2 sin 30 ∘ = 1 2 \sin30^\circ = \dfrac12 sin 3 0 ∘ = 2 1 3 4 \dfrac34 4 3 2 4 = 1 2 \dfrac24 = \dfrac12 4 2 = 2 1 1 4 \dfrac14 4 1
4 ⋅ 3 4 − 1 2 − 2 ⋅ 1 4 = 3 − 1 2 − 1 2 = 2. 4\cdot\frac34 - \frac12 - 2\cdot\frac14 = 3 - \frac12 - \frac12 = 2. 4 ⋅ 4 3 − 2 1 − 2 ⋅ 4 1 = 3 − 2 1 − 2 1 = 2.
Barem: 3p valorile, 2p calculul.
Subiectul al II-lea (30 de puncte)
Pe R \mathbb R R x ∗ y = x + y 3 − 4 x * y = \dfrac{x + y}{3} - 4 x ∗ y = 3 x + y − 4
1. Arătați că 7 ∗ 8 = 1 7 * 8 = 1 7 ∗ 8 = 1
7 ∗ 8 = 7 + 8 3 − 4 = 15 3 − 4 = 5 − 4 = 1. 7 * 8 = \frac{7 + 8}{3} - 4 = \frac{15}{3} - 4 = 5 - 4 = 1. 7 ∗ 8 = 3 7 + 8 − 4 = 3 15 − 4 = 5 − 4 = 1.
Barem: 3p înlocuirea, 2p calculul.
2. Determinați x x x 4 ∗ x = x 4 * x = x 4 ∗ x = x
4 ∗ x = 4 + x 3 − 4 4 * x = \dfrac{4 + x}{3} - 4 4 ∗ x = 3 4 + x − 4 4 + x 3 − 4 = x \dfrac{4 + x}{3} - 4 = x 3 4 + x − 4 = x 4 + x − 12 = 3 x ⟺ − 8 = 2 x ⟺ x = − 4 4 + x - 12 = 3x \iff -8 = 2x \iff x = -4 4 + x − 12 = 3 x ⟺ − 8 = 2 x ⟺ x = − 4
Verificare: 4 ∗ ( − 4 ) = 0 − 4 = − 4 4 * (-4) = 0 - 4 = -4 4 ∗ ( − 4 ) = 0 − 4 = − 4
Răspuns: x = − 4 x = -4 x = − 4 Barem: 2p expresia, 3p soluția.
3. Determinați x x x x ∗ x 2 = 0 x * x^2 = 0 x ∗ x 2 = 0
x ∗ x 2 = x + x 2 3 − 4 = x 2 + x − 12 3 . x * x^2 = \frac{x + x^2}{3} - 4 = \frac{x^2 + x - 12}{3}. x ∗ x 2 = 3 x + x 2 − 4 = 3 x 2 + x − 12 .
Ecuația: x 2 + x − 12 = 0 x^2 + x - 12 = 0 x 2 + x − 12 = 0 Δ = 1 + 48 = 49 \Delta = 1 + 48 = 49 Δ = 1 + 48 = 49 x = − 1 ± 7 2 x = \dfrac{-1 \pm 7}{2} x = 2 − 1 ± 7 x 1 = − 4 x_1 = -4 x 1 = − 4 x 2 = 3 x_2 = 3 x 2 = 3
Răspuns: x ∈ { − 4 , 3 } x \in \{-4, 3\} x ∈ { − 4 , 3 } Barem: 3p expresia, 2p soluțiile.
4. Arătați că ( 2 x ) ∗ ( 2 y ) = 2 ⋅ ( x ∗ ( y + 6 ) ) (2x) * (2y) = 2\cdot\left(x * (y + 6)\right) ( 2 x ) ∗ ( 2 y ) = 2 ⋅ ( x ∗ ( y + 6 ) )
Membrul stâng: ( 2 x ) ∗ ( 2 y ) = 2 x + 2 y 3 − 4 (2x) * (2y) = \dfrac{2x + 2y}{3} - 4 ( 2 x ) ∗ ( 2 y ) = 3 2 x + 2 y − 4
Membrul drept: x ∗ ( y + 6 ) = x + y + 6 3 − 4 = x + y 3 + 2 − 4 = x + y 3 − 2 x * (y + 6) = \dfrac{x + y + 6}{3} - 4 = \dfrac{x + y}{3} + 2 - 4 = \dfrac{x + y}{3} - 2 x ∗ ( y + 6 ) = 3 x + y + 6 − 4 = 3 x + y + 2 − 4 = 3 x + y − 2
2 ⋅ ( x ∗ ( y + 6 ) ) = 2 x + 2 y 3 − 4. 2\cdot\left(x * (y + 6)\right) = \frac{2x + 2y}{3} - 4. 2 ⋅ ( x ∗ ( y + 6 ) ) = 3 2 x + 2 y − 4.
Cei doi membri sunt egali pentru orice x , y x, y x , y Barem: 2p membrul stâng, 3p membrul drept și concluzia.
5. Numerele naturale n n n ( 3 n ) ∗ ( ( 2 n ) ∗ n ) ≤ − n (3n) * \left((2n) * n\right) \le -n ( 3 n ) ∗ ( ( 2 n ) ∗ n ) ≤ − n
Lucrăm din interior spre exterior:
( 2 n ) ∗ n = 2 n + n 3 − 4 = n − 4 , (2n) * n = \frac{2n + n}{3} - 4 = n - 4, ( 2 n ) ∗ n = 3 2 n + n − 4 = n − 4 ,
( 3 n ) ∗ ( n − 4 ) = 3 n + n − 4 3 − 4 = 4 n − 4 − 12 3 = 4 n − 16 3 . (3n) * (n - 4) = \frac{3n + n - 4}{3} - 4 = \frac{4n - 4 - 12}{3} = \frac{4n - 16}{3}. ( 3 n ) ∗ ( n − 4 ) = 3 3 n + n − 4 − 4 = 3 4 n − 4 − 12 = 3 4 n − 16 .
Inecuația: 4 n − 16 3 ≤ − n ⟺ 4 n − 16 ≤ − 3 n ⟺ 7 n ≤ 16 ⟺ n ≤ 16 7 ≈ 2,28 \dfrac{4n - 16}{3} \le -n \iff 4n - 16 \le -3n \iff 7n \le 16 \iff n \le \dfrac{16}{7} \approx 2{,}28 3 4 n − 16 ≤ − n ⟺ 4 n − 16 ≤ − 3 n ⟺ 7 n ≤ 16 ⟺ n ≤ 7 16 ≈ 2 , 28
Cum n n n n ∈ { 0 , 1 , 2 } n \in \{0, 1, 2\} n ∈ { 0 , 1 , 2 }
Răspuns: n ∈ { 0 , 1 , 2 } n \in \{0, 1, 2\} n ∈ { 0 , 1 , 2 } n = 0 n = 0 n = 0 Barem: 2p expresia, 3p rezolvarea inecuației.
6. Determinați x ∈ ( 0 , + ∞ ) x \in (0, +\infty) x ∈ ( 0 , + ∞ ) lg x ∗ lg x = ( − 3 ) ∗ lg 1 x \lg x * \lg x = (-3) * \lg\dfrac1x lg x ∗ lg x = ( − 3 ) ∗ lg x 1
Notăm t = lg x t = \lg x t = lg x x > 0 x > 0 x > 0 lg 1 x = − lg x = − t \lg\dfrac1x = -\lg x = -t lg x 1 = − lg x = − t
lg x ∗ lg x = 2 t 3 − 4 , ( − 3 ) ∗ ( − t ) = − 3 − t 3 − 4. \lg x * \lg x = \frac{2t}{3} - 4, \qquad (-3) * (-t) = \frac{-3 - t}{3} - 4. lg x ∗ lg x = 3 2 t − 4 , ( − 3 ) ∗ ( − t ) = 3 − 3 − t − 4.
Egalăm și simplificăm cu − 4 -4 − 4 2 t = − 3 − t ⟺ 3 t = − 3 ⟺ t = − 1 2t = -3 - t \iff 3t = -3 \iff t = -1 2 t = − 3 − t ⟺ 3 t = − 3 ⟺ t = − 1
lg x = − 1 ⟺ x = 10 − 1 = 1 10 \lg x = -1 \iff x = 10^{-1} = \dfrac{1}{10} lg x = − 1 ⟺ x = 1 0 − 1 = 10 1 ( 0 , + ∞ ) (0, +\infty) ( 0 , + ∞ )
Răspuns: x = 1 10 x = \dfrac1{10} x = 10 1 Barem: 2p cele două expresii, 3p soluția.
Subiectul al III-lea (30 de puncte)
Se consideră I 2 = ( 1 0 0 1 ) I_2 = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} I 2 = ( 1 0 0 1 ) A = ( 1 6 0 − 2 ) A = \begin{pmatrix}1 & 6\\ 0 & -2\end{pmatrix} A = ( 1 0 6 − 2 ) B = ( 1 2 0 0 ) B = \begin{pmatrix}1 & 2\\ 0 & 0\end{pmatrix} B = ( 1 0 2 0 ) M ( x , y ) = x A + y B M(x,y) = xA + yB M ( x , y ) = x A + y B
1. Arătați că det A = − 2 \det A = -2 det A = − 2
det A = ∣ 1 6 0 − 2 ∣ = 1 ⋅ ( − 2 ) − 6 ⋅ 0 = − 2. \det A = \begin{vmatrix}1 & 6\\ 0 & -2\end{vmatrix} = 1\cdot(-2) - 6\cdot 0 = -2. det A = 1 0 6 − 2 = 1 ⋅ ( − 2 ) − 6 ⋅ 0 = − 2.
Barem: 3p formula, 2p rezultatul.
2. Numărul a a a A + 2 I 2 = a B A + 2I_2 = aB A + 2 I 2 = a B
A + 2 I 2 = ( 1 6 0 − 2 ) + ( 2 0 0 2 ) = ( 3 6 0 0 ) = 3 ( 1 2 0 0 ) = 3 B . A + 2I_2 = \begin{pmatrix}1 & 6\\ 0 & -2\end{pmatrix} + \begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix} = \begin{pmatrix}3 & 6\\ 0 & 0\end{pmatrix} = 3\begin{pmatrix}1 & 2\\ 0 & 0\end{pmatrix} = 3B. A + 2 I 2 = ( 1 0 6 − 2 ) + ( 2 0 0 2 ) = ( 3 0 6 0 ) = 3 ( 1 0 2 0 ) = 3 B .
Din 3 B = a B 3B = aB 3 B = a B ( 1 , 1 ) (1,1) ( 1 , 1 ) 3 = a 3 = a 3 = a a = 3 a = 3 a = 3
Răspuns: a = 3 a = 3 a = 3 Barem: 3p calculul, 2p valoarea lui a a a
3. Arătați că A ⋅ A + A = 2 I 2 A\cdot A + A = 2I_2 A ⋅ A + A = 2 I 2
A ⋅ A = ( 1 ⋅ 1 + 6 ⋅ 0 1 ⋅ 6 + 6 ⋅ ( − 2 ) 0 ( − 2 ) ⋅ ( − 2 ) ) = ( 1 − 6 0 4 ) . A\cdot A = \begin{pmatrix}1\cdot1 + 6\cdot0 & 1\cdot6 + 6\cdot(-2)\\ 0 & (-2)\cdot(-2)\end{pmatrix} = \begin{pmatrix}1 & -6\\ 0 & 4\end{pmatrix}. A ⋅ A = ( 1 ⋅ 1 + 6 ⋅ 0 0 1 ⋅ 6 + 6 ⋅ ( − 2 ) ( − 2 ) ⋅ ( − 2 ) ) = ( 1 0 − 6 4 ) .
A ⋅ A + A = ( 2 0 0 2 ) = 2 I 2 . A\cdot A + A = \begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix} = 2I_2. A ⋅ A + A = ( 2 0 0 2 ) = 2 I 2 .
Barem: 3p A ⋅ A A\cdot A A ⋅ A
4. Numerele x x x y y y M ( x , y ) ⋅ A = B M(x,y)\cdot A = B M ( x , y ) ⋅ A = B
M ( x , y ) = ( x + y 6 x + 2 y 0 − 2 x ) , M(x,y) = \begin{pmatrix}x + y & 6x + 2y\\ 0 & -2x\end{pmatrix}, M ( x , y ) = ( x + y 0 6 x + 2 y − 2 x ) ,
M ( x , y ) ⋅ A = ( x + y 6 ( x + y ) − 2 ( 6 x + 2 y ) 0 4 x ) = ( x + y − 6 x + 2 y 0 4 x ) . M(x,y)\cdot A = \begin{pmatrix}x + y & 6(x+y) - 2(6x + 2y)\\ 0 & 4x\end{pmatrix} = \begin{pmatrix}x + y & -6x + 2y\\ 0 & 4x\end{pmatrix}. M ( x , y ) ⋅ A = ( x + y 0 6 ( x + y ) − 2 ( 6 x + 2 y ) 4 x ) = ( x + y 0 − 6 x + 2 y 4 x ) .
Egalăm cu B B B 4 x = 0 ⇒ x = 0 4x = 0 \Rightarrow x = 0 4 x = 0 ⇒ x = 0 x + y = 1 ⇒ y = 1 x + y = 1 \Rightarrow y = 1 x + y = 1 ⇒ y = 1 − 6 ⋅ 0 + 2 ⋅ 1 = 2 -6\cdot 0 + 2\cdot 1 = 2 − 6 ⋅ 0 + 2 ⋅ 1 = 2
Răspuns: x = 0 x = 0 x = 0 y = 1 y = 1 y = 1 Barem: 3p produsul, 2p valorile.
5. Dacă x ≠ y x \ne y x = y det ( M ( x , y ) ) = det ( M ( y , x ) ) \det(M(x,y)) = \det(M(y,x)) det ( M ( x , y )) = det ( M ( y , x )) x + y = 0 x + y = 0 x + y = 0
Matricea M ( x , y ) M(x,y) M ( x , y )
det ( M ( x , y ) ) = ( x + y ) ( − 2 x ) = − 2 x ( x + y ) , det ( M ( y , x ) ) = − 2 y ( x + y ) . \det(M(x,y)) = (x + y)(-2x) = -2x(x + y), \qquad \det(M(y,x)) = -2y(x + y). det ( M ( x , y )) = ( x + y ) ( − 2 x ) = − 2 x ( x + y ) , det ( M ( y , x )) = − 2 y ( x + y ) .
Egalitatea devine − 2 x ( x + y ) = − 2 y ( x + y ) ⟺ − 2 ( x + y ) ( x − y ) = 0 -2x(x+y) = -2y(x+y) \iff -2(x+y)(x - y) = 0 − 2 x ( x + y ) = − 2 y ( x + y ) ⟺ − 2 ( x + y ) ( x − y ) = 0
Cum x ≠ y x \ne y x = y x − y ≠ 0 x - y \ne 0 x − y = 0 x + y = 0 x + y = 0 x + y = 0 ■ \blacksquare ■
Barem: 2p cei doi determinanți, 3p raționamentul.
6. Numerele x x x M ( x , x ) ⋅ M ( x , − x ) = M ( − 2 , 2 ) M(x,x)\cdot M(x,-x) = M(-2,2) M ( x , x ) ⋅ M ( x , − x ) = M ( − 2 , 2 )
M ( x , x ) = ( 2 x 8 x 0 − 2 x ) , M ( x , − x ) = ( 0 4 x 0 − 2 x ) . M(x,x) = \begin{pmatrix}2x & 8x\\ 0 & -2x\end{pmatrix}, \qquad M(x,-x) = \begin{pmatrix}0 & 4x\\ 0 & -2x\end{pmatrix}. M ( x , x ) = ( 2 x 0 8 x − 2 x ) , M ( x , − x ) = ( 0 0 4 x − 2 x ) .
M ( x , x ) ⋅ M ( x , − x ) = ( 0 8 x 2 − 16 x 2 0 4 x 2 ) = ( 0 − 8 x 2 0 4 x 2 ) = x 2 ( 0 − 8 0 4 ) . M(x,x)\cdot M(x,-x) = \begin{pmatrix}0 & 8x^2 - 16x^2\\ 0 & 4x^2\end{pmatrix} = \begin{pmatrix}0 & -8x^2\\ 0 & 4x^2\end{pmatrix} = x^2\begin{pmatrix}0 & -8\\ 0 & 4\end{pmatrix}. M ( x , x ) ⋅ M ( x , − x ) = ( 0 0 8 x 2 − 16 x 2 4 x 2 ) = ( 0 0 − 8 x 2 4 x 2 ) = x 2 ( 0 0 − 8 4 ) .
Iar M ( − 2 , 2 ) = − 2 A + 2 B = ( 0 − 8 0 4 ) M(-2,2) = -2A + 2B = \begin{pmatrix}0 & -8\\ 0 & 4\end{pmatrix} M ( − 2 , 2 ) = − 2 A + 2 B = ( 0 0 − 8 4 ) x 2 M ( − 2 , 2 ) = M ( − 2 , 2 ) x^2 M(-2,2) = M(-2,2) x 2 M ( − 2 , 2 ) = M ( − 2 , 2 ) x 2 = 1 x^2 = 1 x 2 = 1
Răspuns: x ∈ { − 1 , 1 } x \in \{-1, 1\} x ∈ { − 1 , 1 } Barem: 3p produsul, 2p soluțiile.
Rezumatul răspunsurilor
Cerință Răspuns I.2 a = 2 a = 2 a = 2 I.3 x = 3 x = 3 x = 3 I.4 500 de lei II.2 x = − 4 x = -4 x = − 4 II.3 x ∈ { − 4 , 3 } x \in \{-4, 3\} x ∈ { − 4 , 3 } II.5 n ∈ { 0 , 1 , 2 } n \in \{0, 1, 2\} n ∈ { 0 , 1 , 2 } II.6 x = 1 10 x = \frac1{10} x = 10 1 III.2 a = 3 a = 3 a = 3 III.4 x = 0 x = 0 x = 0 y = 1 y = 1 y = 1 III.6 x ∈ { − 1 , 1 } x \in \{-1, 1\} x ∈ { − 1 , 1 }
Ce merită reținut din model
Compuneri imbricate de forma a ∗ ( b ∗ c ) a * (b * c) a ∗ ( b ∗ c ) Ecuații cu logaritmi în lege : notează t = lg x t = \lg x t = lg x lg 1 x = − lg x \lg\frac1x = -\lg x lg x 1 = − lg x Matrice triunghiulare : determinantul este produsul diagonalei — economisești timp la cerința 5.