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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 \(B=-\dfrac{47}{5}\)
\(a,ĐK:x\ne\pm2\\ A=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\\ ĐK:x\ne-1;x\ne-2\\ B=\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x+2}\\ b,x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \forall x=0\Leftrightarrow A=\dfrac{1}{0-2}=-\dfrac{1}{2}\\ \forall x=-1\Leftrightarrow A=\dfrac{1}{-1-2}=-\dfrac{1}{3}\)
\(x^2+2x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ \Leftrightarrow B=\dfrac{1}{0+2}=\dfrac{1}{2}\)
a) P(x) = 7x2 . (x2 – 5x + 2 ) – 5x .(x3 – 7x2 + 3x)
= 7x2 . x2 + 7x2 . (-5x) + 7x2 . 2 – [5x. x3 + 5x . (-7x2) + 5x . 3x]
= 7. (x2 . x2) + [7.(-5)] . (x2 . x) + (7.2).x2 – {5. (x.x3) + [5.(-7)]. (x.x2) + (5.3).(x.x)}
= 7x4 + (-35). x3 + 14x2 – [ 5x4 + (-35)x3 + 15x2 ]
= 7x4 + (-35). x3 + 14x2 - 5x4 + 35x3 - 15x2
= (7x4 – 5x4) + [(-35). x3 + 35x3 ] + (14x2 - 15x2 )
= 2x4 + 0 - x2
= 2x4 – x2
b) Thay x = \( - \dfrac{1}{2}\) vào P(x), ta được:
P(\( - \dfrac{1}{2}\)) = 2. (\( - \dfrac{1}{2}\))4 – (\( - \dfrac{1}{2}\))2 \))
\(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)
a) -ĐKXĐ của A:
x+3≠0 ⇔x≠-3.
x2-9≠0 ⇔(x-3)(x+3)≠0 ⇔x-3≠0 hay x+3≠0⇔x≠3 hay x≠-3.
x-3≠0 ⇔x≠3.
b) B=x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)
c) A=\(\dfrac{x}{x+3}-\dfrac{6x}{x^2-9}+\dfrac{2}{x-3}\)=\(\dfrac{x\left(x-3\right)+2\left(x+3\right)-6x}{\left(x+3\right)\left(x-3\right)}\)=\(\dfrac{x^2-3x+2x+6-6x}{\left(x+3\right)\left(x-3\right)}\)=\(\dfrac{x^2-7x+6}{x^2-9}\)
d)- Vì x=37 thỏa mãn ĐKXĐ của A và A=\(\dfrac{x^2-7x+6}{x^2-9}\)nên:
A=\(\dfrac{37^2-7.37+6}{37^2-9}=\dfrac{279}{340}\)
a)
A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)
\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
MTC: 5(x-1)(x+1)
\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)
\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)
\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)
\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)
\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)
\(\Leftrightarrow10x+10\)
a: Khi x=5 thì A=5/(5+3)=5/8
b: \(C=A+B=\dfrac{x}{x+3}+\dfrac{2}{x-3}+\dfrac{3-5x}{x^2-9}\)
\(=\dfrac{x^2-3x+2x+6+3-5x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
c: Để C nguyên thì x+3-6 chia hết cho x+3
=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
=>\(x\in\left\{-2;-4;-1;-5;0;-6;-9\right\}\)
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