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\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-\left(x-1\right)\left(x-2\right)=2x^2+4\)
\(\Leftrightarrow2x^2+4=x^2+3x+2-x^2+3x-2\)
\(\Leftrightarrow2x^2-6x+4=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
=>x=1(nhận) hoặc x=2(loại)
\(ĐKXĐ:x\ne\pm2\)
\(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)=2\left(x^2+2\right)\)
\(\Leftrightarrow x^2+x+2x+2+x^2-x-2x+2=2x^2+4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
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a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=x^3-2x^5\)
b: \(=5x^2-x^3+5-x\)
e: \(=\left(x-y\right)^3=x^3-3x^2y+3xy^2-y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(x+2-\dfrac{2}{x^2}-2x=\dfrac{1}{x}\) (ĐKXĐ: \(x\ne0\))
\(\Leftrightarrow\dfrac{x^3}{x^2}+\dfrac{2x^2}{x^2}-\dfrac{2}{x^2}-\dfrac{2x^3}{x^2}=\dfrac{x}{x^2}\)
\(\Rightarrow x^3+2x^2-2-2x^3=x\)
\(\Leftrightarrow-x^3+2x^2-x-2=0\)
-Bạn bấm mode-5-4 là ra nghiệm của nó á, hihi.
-Mà công nhận đề cho số xấu thiệt he.