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\(\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}\)
\(=\frac{x^3\left(x-1\right)-\left(x-1\right)}{x^4+x^3+x^2+2x^2+2x+2}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2+2\right)}\)
\(=\frac{\left(x-1\right)^2}{\left(x^2+2\right)}\)
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
=
a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\) ( do \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\) )
\(\Leftrightarrow x=-100\)
b) \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)
Câu a đơn giản
b)
\(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{11}{4}x^2+2x+\frac{4}{11}\right)+1-\frac{4}{11}}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x^2+x+0,25\right)+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
Vì \(\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]>0\)và \(\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}>0\)
nên \(A>0\)hay A ko âm
Nhớ k nha !