a) \(\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right)\)/\(\left(\frac{2}{x^2-4}+\frac{1}{2x-x}\right)\)
ai làm đúng mình tích cho
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\(C=\left[\frac{x^2.\left(x^2-4\right)+4x^2}{x^2-4}\right].\left[\frac{x}{2.\left(x-2\right)}+\frac{2-2x}{x.\left(x^2-4\right)}.\frac{x^2-4}{x-2}\right]\)
\(C=\frac{x^4-4x^2+4x^2}{x^2-4}.\left[\frac{x}{2.\left(x-2\right)}+\frac{2-2x}{x\left(x-2\right)}\right]\)
\(C=\frac{x^4}{x^2-4}.\left[\frac{x^2}{2x.\left(x-2\right)}+\frac{\left(2-2x\right).2}{2x.\left(x-2\right)}\right]\)
\(C=\frac{x^4}{x^2-4}.\left[\frac{x^2+4-4x}{2x.\left(x-2\right)}\right]\)
\(C=\frac{x^4}{x^2-4}.\frac{\left(x-2\right)^2}{2x.\left(x-2\right)}\)
\(C=\frac{x^4}{\left(x-2\right).\left(x+2\right)}.\frac{\left(x-2\right).\left(x-2\right)}{2x.\left(x-2\right)}\)
\(C=\frac{x^3}{\left(x+2\right).2}\)
đk : \(x\ne4,-4\)
A= \(\frac{8+x-4}{\left(x+4\right)\left(x-4\right)}:\frac{2\left(x-4\right)-x^2}{2x\left(x+4\right)}\)
A = \(\frac{x+4}{\left(x-4\right)\left(x+4\right)}.\frac{2x\left(x+4\right)}{x^2+2x-8}\)
A=\(\frac{1}{x-4}.\frac{2x\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{2x}{\left(x-4\right)\left(x-2\right)}\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(\left(\frac{x-1}{x+2}\right)^2-4\left(\frac{x^2-1}{x^2-4}\right)^2+3\left(\frac{x+1}{x-2}\right)^2=0\left(1\right)\)
\(ĐKXĐ:x\ne\pm2\)
Đặt \(\frac{x-1}{x+2}=a;\frac{x+1}{x-2}=b\)
=> Phương trình (1) <=> \(a^2-4ab+3b^2=0\)
\(\Leftrightarrow a^2-3ab-ab+3b^2=0\)
\(\Leftrightarrow a\left(a-b\right)-3b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-3b=0\\a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3b\\a=b\end{cases}}}\)
=> \(b=0;a=0\)
Bạn cùng trường :">
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
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