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a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)
= x³ - 125 - x² + 4 + x³ + x² + 4x
= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)
= 2x³ + 4x - 121
b) Tại x = -2 ta có:
A = 2.(-2)³ + 4.(-2) - 121
= 2.(-8) - 8 - 121
= -16 - 129
= -145
c) x² - 1 = 0
x² = 1
x = -1; x = 1
*) Tại x = -1 ta có:
A = 2.(-1)³ + 4.(-1) - 121
= 2.(-1) - 4 - 121
= -2 - 125
= -127
*) Tại x = 1 ta có:
A = 2.1³ + 4.1 - 121
= 2.1 + 4 - 121
= 2 - 117
= -115
Bạn nên viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo).
\(A=\left(\dfrac{x}{x-2}+\dfrac{12}{x^2-4}-\dfrac{x}{x+2}\right):\dfrac{4}{x-2}\left(x\ne2;x\ne-2\right)\)
\(a,A=\left(\dfrac{x}{x-2}+\dfrac{12}{x^2-4}-\dfrac{x}{x+2}\right):\dfrac{4}{x-2}\)
\(=\left[\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x-2}\)
\(=\left[\dfrac{x^2+2x+12-x^2+2x}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x-2}\)
\(=\dfrac{4x+12}{\left(x-2\right)\left(x+2\right)}:\dfrac{4}{x-2}\)
\(=\dfrac{4\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}.\dfrac{x-2}{4}\)
\(=\dfrac{x+3}{x+2}\)
\(b,x=-1\Rightarrow A=\dfrac{\left(-1\right)+3}{\left(-1\right)+2}=2\)
\(c,A=\dfrac{x+3}{x+2}=\dfrac{x+2+1}{x+2}=1+\dfrac{1}{x+2}\)
\(A\in Z\Leftrightarrow x+2\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow x\in\left\{-1;-3\right\}\) (thỏa mãn điều kiện)
a) ĐKXĐ:
\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\)
b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)
\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)
\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)
c) Thay x = - 1 vào A ta có:
\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)
a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)
\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)
\(=\dfrac{x^2+5x+100}{5x}\)
b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:
\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)
Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)
