B=\(\left(\frac{x^2-2}{x^2+2x}+\frac{1}{x+2}\right).\frac{x+1}{x-1}\)
a, CMR B= \(\frac{x+1}{x}\)v ới x#0,x#-2 ,x#1
b. tìm x để 2B=2x+5
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\(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}-\frac{1}{x}\) (ĐKXĐ: x \(\ne\) 0 và x \(\ne\) a + b)
<=> \(\frac{1}{a+b-x}+\frac{1}{x}-\frac{1}{a}-\frac{1}{b}=0\)
<=> \(\frac{x}{x\left(a+b-x\right)}+\frac{a+b-x}{x\left(a+b-x\right)}-\frac{b}{ab}-\frac{a}{ab}\)
<=> \(\frac{a+b}{x\left(a+b-x\right)}-\frac{a+b}{ab}=0\)
<=> \(\left(a+b\right)\left(\frac{1}{x\left(a+b-x\right)}-\frac{1}{ab}\right)=0\)
* Nếu a = - b thì tập nghiệm cuả pt là S = R
* Nếu a \(\ne\) b thì \(\frac{1}{x\left(a+b-x\right)}-\frac{1}{ab}=0\)
<=> \(\frac{ab}{abx\left(a+b-x\right)}-\frac{x\left(a+b-x\right)}{abx\left(a+b-x\right)}=0\)
<=> \(\frac{ab-\text{ax}-bx+x^2}{abx\left(a+b-x\right)}=0\)
<=> \(\frac{b\left(a-x\right)-x\left(a-x\right)}{abx\left(a+b-x\right)}=0\)
<=> \(\frac{\left(a-x\right)\left(b-x\right)}{abx\left(a+b-x\right)}=0\)
<=> \(\left[\begin{matrix}a-x=0\\b-x=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=a\\x=b\end{matrix}\right.\)
Vậy tập nghiệm của pt là S = {a ; b}
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (ĐKXĐ: x \(\ne\) 0
<=> \(\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}=\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=> \(\left(x^4+x\right)-\left(x^4-x\right)=3\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\) (nhận)
Vậy S = {1,5}
b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
<=> x2+2x-x+2=2
<=> x2+x=2-2
<=> x2+x=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x = -1
a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
<=> x-3 =10x-15
<=> x-10x= -15+3
<=> -9x = -12
<=> x = \(\frac{-12}{-9}\)
<=> x = \(\frac{4}{3}\)
a)\(B=\left(\frac{x-2}{x^2+2x}+\frac{1}{x+2}\right).\frac{x+1}{x-1}=\left(\frac{x^2-2}{x^2+2x}+\frac{x}{x^2+2x}\right).\frac{x+1}{x-1}=\frac{x^2+x-2}{x^2+2x}.\frac{x+1}{x-1}\)
\(=\frac{x^2-x+2x-2}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x\left(x-1\right)+2\left(x-1\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x+1}{x}\)
b)\(2B=2x+5\Leftrightarrow2.\frac{x+1}{x}=2x+5\Leftrightarrow\frac{2x+2}{x}=2x+5\Leftrightarrow2x+2=2x^2+5x\)
\(\Leftrightarrow0=2x^2+3x-2\Leftrightarrow2x^2+4x-x-2=0\Leftrightarrow2x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
cảm ơn bạn nhé