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a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
=
\(\left(x^2+\frac{2}{5}y\right)\left(x^2+\frac{2}{5}y\right)=x^4+\frac{4}{5}x^2y+\frac{4}{25}y\)
b\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2-\frac{1}{3}\right)^3-x^2-\frac{1}{3}\)
\(\frac{x^2}{9}+\frac{1}{x^2}=\frac{5}{3}\left(\frac{x}{3}-\frac{1}{x}\right)\)
đặt \(\left(\frac{x}{3}-\frac{1}{x}\right)=t\Rightarrow t^2=\left(\frac{x^2}{9}-\frac{2}{3}+\frac{1}{x^2}\right)\Rightarrow\frac{x^2}{9}+\frac{1}{x^2}=t^2+\frac{2}{3}\)
\(\Leftrightarrow t^2-\frac{5}{3}t+\frac{2}{3}=0\Leftrightarrow t^2-2.\frac{5}{6}t+\left(\frac{5}{6}\right)^2=\frac{25}{36}-\frac{24}{36}=\frac{1}{36}=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\left[\begin{matrix}t-\frac{5}{6}=\frac{1}{6}\\t-\frac{5}{6}=-\frac{1}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}t=1\\t=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\frac{x}{3}-\frac{1}{x}=1\\\frac{x}{3}-\frac{1}{x}=\frac{2}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\ne0\\x^2-x-3=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ne0\\\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\\\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\end{matrix}\right.\)
Kết luận: \(\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\\x=-1\\x=3\end{matrix}\right.\)