Valorile parametrilor α , β ∈ R \alpha, \beta \in \mathbb{R} α , β ∈ R x , y , z ∈ R x, y, z \in \mathbb{R} x , y , z ∈ R { 2 x − y + 3 z = 0 3 x + y + β z = 0 x + α y + ( α + 1 ) z = 0 și { 2 x − y + 3 z = 0 2 x + y + z = 0 3 x + y + β z = 0 \begin{cases} 2x - y + 3z = 0 \\ 3x + y + \beta z = 0 \\ x + \alpha y + (\alpha+1) z = 0 \end{cases} \quad \text{și} \quad \begin{cases} 2x - y + 3z = 0 \\ 2x + y + z = 0 \\ 3x + y + \beta z = 0 \end{cases} ⎩ ⎨ ⎧ 2 x − y + 3 z = 0 3 x + y + β z = 0 x + α y + ( α + 1 ) z = 0 ș i ⎩ ⎨ ⎧ 2 x − y + 3 z = 0 2 x + y + z = 0 3 x + y + β z = 0 sunt:
β = 2 \beta = 2 β = 2 α = 0 \alpha = 0 α = 0 β = 0 \beta = 0 β = 0 α = 0 \alpha = 0 α = 0 β ∈ R ∖ { 2 } \beta \in \mathbb{R} \setminus \{2\} β ∈ R ∖ { 2 } α ∈ R ∖ { 0 } \alpha \in \mathbb{R} \setminus \{0\} α ∈ R ∖ { 0 } β = 2 \beta = 2 β = 2 α ∈ R ∖ { 0 } \alpha \in \mathbb{R} \setminus \{0\} α ∈ R ∖ { 0 } β ∈ R ∖ { 2 } \beta \in \mathbb{R} \setminus \{2\} β ∈ R ∖ { 2 } α = 0 \alpha = 0 α = 0 Combinând ecuațiile celor două sisteme se obține 4 β − 8 = 0 4\beta - 8 = 0 4 β − 8 = 0 β = 2 \beta = 2 β = 2 10 α = 0 10\alpha = 0 10 α = 0 α = 0 \alpha = 0 α = 0 Dacă A = ( 2 x 1 x 2 x 1 + x 2 x 1 + x 2 1 ) A = \begin{pmatrix} 2x_1x_2 & x_1+x_2 \\ x_1+x_2 & 1 \end{pmatrix} A = ( 2 x 1 x 2 x 1 + x 2 x 1 + x 2 1 ) unde x 1 , x 2 x_1, x_2 x 1 , x 2 x 2 − 3 x + 2 = 0 x^2 - 3x + 2 = 0 x 2 − 3 x + 2 = 0
A − 1 = ( − 1 5 3 5 3 5 − 2 5 ) A^{-1} = \begin{pmatrix} -\dfrac{1}{5} & \dfrac{3}{5} \\ \dfrac{3}{5} & -\dfrac{2}{5} \end{pmatrix} A − 1 = − 5 1 5 3 5 3 − 5 2 A − 1 = ( 1 5 2 5 2 5 2 5 ) A^{-1} = \begin{pmatrix} \dfrac{1}{5} & \dfrac{2}{5} \\ \dfrac{2}{5} & \dfrac{2}{5} \end{pmatrix} A − 1 = 5 1 5 2 5 2 5 2 A − 1 = ( − 1 x 1 2 + x 2 2 x 1 + x 2 x 1 2 + x 2 2 x 1 + x 2 x 1 2 + x 2 2 − 2 x 1 x 2 x 1 2 + x 2 2 ) A^{-1} = \begin{pmatrix} -\dfrac{1}{x_1^2+x_2^2} & \dfrac{x_1+x_2}{x_1^2+x_2^2} \\ \dfrac{x_1+x_2}{x_1^2+x_2^2} & -\dfrac{2x_1x_2}{x_1^2+x_2^2} \end{pmatrix} A − 1 = − x 1 2 + x 2 2 1 x 1 2 + x 2 2 x 1 + x 2 x 1 2 + x 2 2 x 1 + x 2 − x 1 2 + x 2 2 2 x 1 x 2 A − 1 = ( − 1 5 2 5 2 5 − 4 5 ) A^{-1} = \begin{pmatrix} -\dfrac{1}{5} & \dfrac{2}{5} \\ \dfrac{2}{5} & -\dfrac{4}{5} \end{pmatrix} A − 1 = − 5 1 5 2 5 2 − 5 4 A − 1 = ( − 1 5 2 5 2 5 − 2 5 ) A^{-1} = \begin{pmatrix} -\dfrac{1}{5} & \dfrac{2}{5} \\ \dfrac{2}{5} & -\dfrac{2}{5} \end{pmatrix} A − 1 = − 5 1 5 2 5 2 − 5 2 Cu x 1 + x 2 = 3 x_1+x_2=3 x 1 + x 2 = 3 x 1 x 2 = 2 x_1x_2=2 x 1 x 2 = 2 det A = 2 x 1 x 2 − ( x 1 + x 2 ) 2 = − ( x 1 2 + x 2 2 ) \det A = 2x_1x_2 - (x_1+x_2)^2 = -(x_1^2+x_2^2) det A = 2 x 1 x 2 − ( x 1 + x 2 ) 2 = − ( x 1 2 + x 2 2 ) Dacă { a n } n ∈ N ∗ \{a_n\}_{n\in\mathbb{N}^*} { a n } n ∈ N ∗ a n = ∑ k = 0 n C n k ( − 1 ) k + n n + 1 , n ∈ N ∗ a_n = \dfrac{\displaystyle\sum_{k=0}^{n} C_n^k(-1)^k + n}{n+1}, \quad n \in \mathbb{N}^* a n = n + 1 k = 0 ∑ n C n k ( − 1 ) k + n , n ∈ N ∗ atunci:
lim n → ∞ a n < 1 \lim_{n\to\infty} a_n < 1 lim n → ∞ a n < 1 ∄ lim n → ∞ a n \nexists \lim_{n\to\infty} a_n ∄ lim n → ∞ a n lim n → ∞ a n = 1,5 \lim_{n\to\infty} a_n = 1{,}5 lim n → ∞ a n = 1 , 5 lim n → ∞ a n = 1 \lim_{n\to\infty} a_n = 1 lim n → ∞ a n = 1 lim n → ∞ a n > 2 \lim_{n\to\infty} a_n > 2 lim n → ∞ a n > 2 ∑ k = 0 n C n k ( − 1 ) k = ( 1 − 1 ) n = 0 \sum_{k=0}^n C_n^k(-1)^k = (1-1)^n = 0 ∑ k = 0 n C n k ( − 1 ) k = ( 1 − 1 ) n = 0 n ≥ 1 n\ge 1 n ≥ 1 a n = n n + 1 → 1 a_n = \dfrac{n}{n+1} \to 1 a n = n + 1 n → 1 Dacă { S n } n ∈ N \{S_n\}_{n\in\mathbb{N}} { S n } n ∈ N S n = 1 5 + 4 5 2 + ⋯ + ( − 2 ) 2 n 5 n + 1 , n ∈ N S_n = \dfrac{1}{5} + \dfrac{4}{5^2} + \cdots + \dfrac{(-2)^{2n}}{5^{n+1}}, \quad n \in \mathbb{N} S n = 5 1 + 5 2 4 + ⋯ + 5 n + 1 ( − 2 ) 2 n , n ∈ N atunci:
lim n → ∞ S n = 2 \lim_{n\to\infty} S_n = 2 lim n → ∞ S n = 2 lim n → ∞ S n = 0 \lim_{n\to\infty} S_n = 0 lim n → ∞ S n = 0 lim n → ∞ S n = − 1 \lim_{n\to\infty} S_n = -1 lim n → ∞ S n = − 1 lim n → ∞ S n = 1 \lim_{n\to\infty} S_n = 1 lim n → ∞ S n = 1 lim n → ∞ S n = 1 9 \lim_{n\to\infty} S_n = \dfrac{1}{9} lim n → ∞ S n = 9 1 Termenii formează o progresie geometrică cu rația 4 / 5 4/5 4/5 S n = 1 − ( 4 / 5 ) n + 1 → 1 S_n = 1-(4/5)^{n+1} \to 1 S n = 1 − ( 4/5 ) n + 1 → 1 Dacă F , f , g : R → R F, f, g: \mathbb{R} \to \mathbb{R} F , f , g : R → R f ( x ) = e x 2 ( x + 1 ) 2 , g ( x ) = x 2 , F ( x ) = ( f ∘ g ) ( x ) f(x) = e^{x^2}(x+1)^2, \quad g(x) = x^2, \quad F(x) = (f \circ g)(x) f ( x ) = e x 2 ( x + 1 ) 2 , g ( x ) = x 2 , F ( x ) = ( f ∘ g ) ( x ) atunci
F ′ ( x ) = 4 x e x 4 ( x 2 − 1 ) ( x + x 2 + 1 ) ( − x + x 2 + 1 ) F'(x) = 4xe^{x^4}(x^2-1)(x+x^2+1)(-x+x^2+1) F ′ ( x ) = 4 x e x 4 ( x 2 − 1 ) ( x + x 2 + 1 ) ( − x + x 2 + 1 ) F ′ ( x ) = 2 x [ 2 x 2 e x 4 ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) e x 4 ] F'(x) = 2x\left[2x^2e^{x^4}(x^2+1)^2+2(x^2+1)e^{x^4}\right] F ′ ( x ) = 2 x [ 2 x 2 e x 4 ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) e x 4 ] F ′ ( x ) = x e x 4 ( x 2 − 1 ) ( x + x 2 + 1 ) ( − x + x 2 + 1 ) F'(x) = xe^{x^4}(x^2-1)(x+x^2+1)(-x+x^2+1) F ′ ( x ) = x e x 4 ( x 2 − 1 ) ( x + x 2 + 1 ) ( − x + x 2 + 1 ) F ′ ( x ) = 4 x e x 4 ( x 2 + 1 ) ( x + x 2 − 1 ) ( − x + x 2 + 1 ) F'(x) = 4xe^{x^4}(x^2+1)(x+x^2-1)(-x+x^2+1) F ′ ( x ) = 4 x e x 4 ( x 2 + 1 ) ( x + x 2 − 1 ) ( − x + x 2 + 1 ) F ′ ( x ) = 4 x e x 4 ( x 2 + 1 ) ( x + x 2 + 1 ) ( − x + x 2 − 1 ) F'(x) = 4xe^{x^4}(x^2+1)(x+x^2+1)(-x+x^2-1) F ′ ( x ) = 4 x e x 4 ( x 2 + 1 ) ( x + x 2 + 1 ) ( − x + x 2 − 1 ) F ( x ) = e x 4 ( x 2 + 1 ) 2 F(x)=e^{x^4}(x^2+1)^2 F ( x ) = e x 4 ( x 2 + 1 ) 2 F ′ ( x ) = 4 x e x 4 ( x 2 + 1 ) ( x 2 ( x 2 + 1 ) + 1 ) = 2 x [ 2 x 2 e x 4 ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) e x 4 ] F'(x)=4xe^{x^4}(x^2+1)(x^2(x^2+1)+1)=2x\left[2x^2e^{x^4}(x^2+1)^2+2(x^2+1)e^{x^4}\right] F ′ ( x ) = 4 x e x 4 ( x 2 + 1 ) ( x 2 ( x 2 + 1 ) + 1 ) = 2 x [ 2 x 2 e x 4 ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) e x 4 ]