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\(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right)x^2\\\left(x+2\right)\left(x^2-3x+5\right)-\left(x+2\right)x^2=0\\ \left(x+2\right)\left(5-3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
\(\dfrac{-7x^2+4}{x^3+1}=\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}\\ \dfrac{-7x^2+4}{x^3+1}=\dfrac{5\left(x+1\right)-\left(x^2-x+1\right)}{x^3+1}\\ \Rightarrow-7x^2+4=-x^2+6x-4\\ 6x^2+6x-8=0\\ x^2+x-\dfrac{4}{3}=0\\ x^2+x+\dfrac{1}{4}=\dfrac{4}{3}+\dfrac{1}{4}\\ \left(x+\dfrac{1}{2}\right)^2=\dfrac{19}{12}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{19}{12}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{19}{12}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{12}}-\dfrac{1}{2}\\x=-\sqrt{\dfrac{19}{12}}-\dfrac{1}{2}\end{matrix}\right.\)
4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
a: \(\dfrac{x^2-3x+2}{x^2-1}=\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}\)
\(a,VP=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-2}{x+1}=VP\\ b,VT=\dfrac{u\left(4u^2-1\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1-2u\right)\left(1+2u\right)}{5\left(1-2u\right)}=\dfrac{-u\left(1+2u\right)}{5}=-\dfrac{2u^2+u}{5}=VP\)