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rut gon bieu thuc :
\(\left(x+1\right)^4-6\left(x+1\right)^2-\left(x^2-2\right).\left(x^2+2\right)\)
\(\left(x+1\right)^4-6\left(x+1\right)^2-\left(x^2-2\right)\left(x^2+2\right)\\ =x^4+4x^3+6x^2+4x+1-6x^2-12x-6-x^4+4\\ =4x^3-8x+5\)
1) \(\dfrac{x^2-18x-19}{x^2-1}=\dfrac{x^2-19x+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{x\left(x-19\right)+x-19}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-19\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-19}{x-1}\)
2) \(\dfrac{x\left(4x^2-8x+4\right)}{2x^3-2x^2}=\dfrac{4x\left(x^2-2x+1\right)}{2x^2\left(x-1\right)}=\dfrac{4x\left(x-1\right)^2}{2x^2\left(x-1\right)}=\dfrac{2\left(x-1\right)}{x}\)
rút gọn phân thức:
\(\dfrac{\left(-x\right)^5.a^2}{x^2.\left(-a\right)^3}=\dfrac{x^2.\left(-x\right)^3.a^2}{x^2.\left(-a\right).a^2}=\dfrac{-x^3}{-a}=\dfrac{x^3}{a}\)
\(\dfrac{\left(-x\right)^5.a^2}{x^2.\left(-a\right)^3}\\ =\dfrac{\left(-x\right)^3x^2.a^2}{x^2.\left(-a\right).a^2}\\ =\dfrac{\left(-x\right)^3}{a}\)
\(A-1=\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^2-1\right)\left(x^2+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^4-1\right)\left(x^4+1\right)...\left(x^{256}+1\right)\)
\(\Rightarrow\left(A-1\right)\left(x-1\right)=\left(x^{256}-1\right)\left(x^{256}+1\right)=x^{512}-1\)
\(\Rightarrow A-1=\dfrac{x^{512}-1}{x-1}\)
\(\Rightarrow A=\dfrac{x^{512}-1}{x-1}+1=\dfrac{x^{512}+x-2}{x-1}\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
\(=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{x+2-x^2}{x+2}-\dfrac{x^2+6x+4}{x}\)
\(=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)-x^2-6x-4}{x}\)
\(=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-6x-4}{x}\)
\(=\dfrac{-x^3-2x^2-2x}{x}=-x^2-2x-2\)