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a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=0\\-x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
\(ĐK:x\ne0;x\ne1\\ PT\Leftrightarrow\left(\dfrac{1}{x}+2\right)\left(2+\dfrac{x+1}{x-1}-x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{x}=-2\\\dfrac{x+1}{x-1}=x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x+1=x^2-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x^2-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1+\sqrt{2}\\x=1-\sqrt{2}\end{matrix}\right.\)
Câu a bạn sửa lại đề 11→1
\(a,VT=\dfrac{a^2-2a+1}{\left(a-1\right)\left(a^2+1\right)}\cdot\dfrac{a^2+1}{a^2+a+1}\\ =\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}=\dfrac{a-1}{a^2+a+1}=VP\)
\(b,=\left[\dfrac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}-x\right]\cdot\dfrac{\left(1+x\right)\left(1-x^2\right)}{1+x}\\ =\dfrac{\left(x^2+1\right)\left(1+x\right)\left(1-x^2\right)}{1+x}=\left(x^2+1\right)\left(1-x^2\right)=VP\)
Ta có : \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{3}{10}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{3}{10}\)
\(\Leftrightarrow10\left(x+3\right)-10x=3x\left(x+3\right)\)
\(\Leftrightarrow-3x^2-9x+30=0\)
\(\Delta=\left(-9\right)^2+4.3.30=81+360=441>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{9+\sqrt{441}}{-6}=-5\\x_2=\dfrac{9-\sqrt{441}}{-6}=2\end{matrix}\right.\)
Vậy \(S=\left\{-5;2\right\}\)
Lời giải:
Đặt biểu thức trên là $A$ thì:
\(A=\frac{1}{x+1}:\frac{x^2+3x+2-2}{(x-1)(x+1)(x+2)}=\frac{1}{x+1}:\frac{x(x+3)}{(x-1)(x+1)(x+2)}\)
\(=\frac{1}{x+1}.\frac{(x-1)(x+1)(x+2)}{x(x+3)}=\frac{(x-1)(x+2)}{x(x+3)}\)