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`a)ĐK:(x+1)(2x-6) ne 0`
`<=>(x+1)(x-3) ne 0`
`<=> x ne -1,x ne 3`
`b)C=(3x^2+3x)/((x+1)(2x-6))`
`=(3x(x+1))/((x+1)(2x-6))`
`=(3x)/(2x-6)`
`C=1`
`=>3x=2x-6`
`<=>x=-6(tm)`
Vậy `x=-6`
Mk có tâm rút gọn hộ bạn luôn rồi nè =))
a, ĐK : \(x\ne-2;3\)
b, \(A=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2}{x+2}\)
\(=\frac{8-x}{\left(x+2\right)\left(x-3\right)}+\frac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\frac{8-x+2x-6}{\left(x+2\right)\left(x-3\right)}\)
\(=\frac{x-2}{\left(x-2\right)\left(x-3\right)}=\frac{1}{x-3}\)
a, ĐKXĐ: \(x\ne1;x\ne-1\)
b, Với \(x\ne1;x\ne-1\)
\(B=\left[\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\left[\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\dfrac{5}{x^2-1}\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =4\)
=> ĐPCM
a)Đk:\(\left\{{}\begin{matrix}x^2-4\ne0\\2x^2-x^3\ne0\\x^2-3x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)\ne0\\x^2\left(2-x\right)\ne0\\x\left(x-3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow x\ne\left\{2;-2;0;3\right\}\)
b)\(P=\left[\dfrac{\left(2+x\right)^2}{\left(2+x\right)\left(2-x\right)}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}\right]:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2+x\right)\left(2-x\right)}.\dfrac{x\left(2-x\right)}{x-3}\)
\(=\dfrac{x\left(8x-4x^2\right)}{\left(2+x\right)\left(x-3\right)}\) (sai đề chỗ nào ko em)
c)\(\left|x-5\right|=2\Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Thay x=7 vào bt P ta được: \(P=\dfrac{7\left(8.7-4.7^2\right)}{\left(2+7\right)\left(7-3\right)}=-\dfrac{245}{9}\)
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
=
\(a,ĐK:x\ne1\\ b,A=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\\ c,A=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(tm\right)\)
a: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)
b: \(A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)
Để A=0 thì 3x=0
hay x=0