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\(M=\dfrac{x-x^2}{1-x^2}+\dfrac{1}{1+x}\cdot\dfrac{3x-1}{x+1}-1\)
\(=\dfrac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{3x-1}{\left(x+1\right)^2}-1\)
\(=\dfrac{x^2+x+3x-1-x^2-2x-1}{\left(x+1\right)^2}\)
\(=\dfrac{2x-2}{\left(x+1\right)^2}\)
c/C=\(\frac{2x^2+2x}{1-x}-\frac{x}{x-1}=\frac{2x^2+2x+x}{1-x}=\frac{2x^2+3x}{1-x}\)
d/C thuộc Z thì C=\(\frac{\left(2x^2-2x\right)+\left(5x-5\right)+5}{1-x}=\frac{-2x\left(1-x\right)-5\left(1-x\right)+5}{1-x}=-2x-5+\frac{5}{1-x}\Rightarrow1-x\in\left(+-1,+-5\right)\Rightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-4\\x=6\end{matrix}\right.\)
a/A đã rút gọn B=\(\frac{1-2x}{x^2-3x+2}+\frac{x+1}{x-2}=\frac{1-2x}{\left(x-1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}=\frac{1-2x+x^2-1}{\left(x-1\right)\left(x-2\right)}=\frac{x\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{x}{x-1}\)b/\(\left|x-2\right|=3\Rightarrow\left\{{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}B=\frac{2.5^2+2.5}{1-5}=-15\\B=\frac{2.\left(-1\right)^2+2\left(-1\right)}{1-\left(-1\right)}=0\end{matrix}\right.\)
,(3x-1) mũ 2=9/16
<=> (3x-1)^2 = ( ±3/4)^2
<=> l3x-1l = 3/4
Hoặc 3x-1 = 3/4
<=> 3x= 3/4 + 1
<=> x = 7/4 : 3
<=> x= 7/1
a) \(x^2-\frac{1}{49}=0\)
<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)
<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)
Vậy x = \(\pm\frac{1}{7}\)
b) \(64-\frac{1}{4}x^2=0\)
<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)
<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)
Vậy \(x=\pm16\)
c) 9x2 + 12x + 4 = 0
<=> (3x + 2)2 = 0
<=> 3x + 2 = 0
<=> x = -2/3
Vậy x = -2/3
e) \(x^2+\frac{1}{4}=x\)
<=> \(x^2-x+\frac{1}{4}=0\)
<=> \(\left(x-\frac{1}{2}\right)^2=0\)
<=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)
Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng
\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{-16}{1-x^2}\left(x\ne\pm1\right)\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{16}{1-x^2}=0\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1-16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{4x-16}{\left(x-1\right)\left(x+1\right)}=0\)
=> 4x-16=0
<=> 4x=16
<=> x=4 (tmđk)
Vậy x=4
\(M=\left(\dfrac{x-x^2}{\left(x-1\right)^2}+\dfrac{1}{1+x}-\dfrac{x}{x-1}\right)\cdot\left(\dfrac{3x-1}{x}+\dfrac{1}{x+1}-1\right)\)
\(=\left(\dfrac{-x}{x-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(3x-1\right)\left(x+1\right)+x-x\left(x+1\right)}{x\left(x+1\right)}\)
\(=\dfrac{-2x\left(x+1\right)+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{3x^2+2x-1+x-x^2-x}{x\left(x+1\right)}\)
\(=\dfrac{-2x^2-2x+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x^2+2x-1}{x\left(x+1\right)}\)
\(=\dfrac{-2x^2-x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{2x^2+2x-1}{x\left(x+1\right)}\)
\(=\dfrac{\left(-2x^2-x-1\right)\left(2x^2+2x-1\right)}{x\left(x+1\right)^2\cdot\left(x-1\right)}\)
\(\left(\dfrac{3x-1}{x+1}-1\right)\)bạn sửa lại đề bào thế này