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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
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
1: =>2x-5=4 hoặc 2x-5=-4
=>2x=9 hoặc 2x=1
=>x=9/2hoặc x=1/2
2: \(\Leftrightarrow\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}=\dfrac{-1}{8}\)(vô lý)
3: \(\Leftrightarrow\left|5x-3\right|=x+5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(5x-3-x-5\right)\left(5x-3+x+5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(4x-8\right)\left(6x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)
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bạn nhắc tên của các bạn thành viên có ý gì ko??? -_-