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\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\left(x\ne-3;x\ne1\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{x+1}{x-1}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2}{\left(x+3\right)\left(x-1\right)}-\frac{x^2+4x+3}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2-x^2-4x-3-4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3x-9}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3}{x-1}=0\)
=> PT vô nghiệm
\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)
<=> 4x-16=-3x+6
<=> 4x-16+3x-6=0
<=> 7x-22=0
<=> 7x=22
<=> \(x=\frac{22}{7}\)(TMĐK)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn
\(\frac{2}{x^2-2x}+\frac{1}{x}=\frac{x+2}{x-2}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}+\frac{x\left(x+2\right)}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2+x-2+x^2+2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x^2+3x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x+3}{x-2}=0\)
\(\Rightarrow x+3=0\left(x-2\ne0\right)\)
\(\Leftrightarrow x=-3\)
\(ĐKXĐ:\) \(x\ne0\)
Đặt \(x+\frac{1}{x}=y\) \(\left(\text{*}\right)\), thì khi đó \(x^2+\frac{1}{x^2}=y^2-2\)
Do đó, \(y^2-2-\frac{9}{2}y+7=0\)
\(\Leftrightarrow\) \(y^2-\frac{9}{2}y+5=0\)
\(\Leftrightarrow\) \(2y^2-9y+10=0\)
\(\Leftrightarrow\) \(2y^2-4y-5y+10=0\)
\(\Leftrightarrow\) \(2y\left(y-2\right)-5\left(y-2\right)=0\)
\(\Leftrightarrow\) \(\left(y-2\right)\left(2y-5\right)=0\)
\(\Leftrightarrow\) \(^{y-2=0}_{2y-5=0}\) \(\Leftrightarrow\) \(^{y=2}_{y=\frac{5}{2}}\)
\(\text{*)}\) Với trường hợp \(y=2\) thì khi đó, \(\left(\text{*}\right)\) \(\Rightarrow\) \(x+\frac{1}{x}=2\) \(\left(1\right)\)
Vì \(x\ne0\) nên từ \(\left(1\right)\) suy ra \(x^2+1=2x\) \(\Leftrightarrow\) \(x^2-2x+1=0\) \(\Leftrightarrow\) \(\left(x-1\right)^2=0\) \(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\) ( thỏa mãn điều kiện xác định)
\(\text{*)}\) Với \(y=\frac{5}{2}\) thì \(\left(\text{*}\right)\) \(\Rightarrow\) \(x+\frac{1}{x}=\frac{5}{2}\) \(\left(2\right)\)
Từ \(\left(2\right)\) \(\Rightarrow\) \(2x^2+2=5x\) (do \(x\ne0\) )
\(\Leftrightarrow\) \(2x^2-5x+2=0\)
\(\Leftrightarrow\) \(2x^2-4x-x+2=0\)
\(\Leftrightarrow\) \(2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\) \(\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\) \(^{x-2=0}_{2x-1=0}\) \(\Leftrightarrow\) \(^{x=2}_{x=\frac{1}{2}}\) (t/mãn điều kiện xác định)
Vậy, \(S=\left\{1;2;\frac{1}{2}\right\}\)
a) Ta có: \(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)
\(\Leftrightarrow\frac{3}{x^2+x-2}-\frac{1}{x-1}-\frac{-7}{x+2}=0\)
\(\Leftrightarrow\frac{3}{\left(x-1\right).\left(x+2\right)}-\left[\frac{\left(x+2\right)+\left(-7\right).\left(x+1\right)}{\left(x-1\right).\left(x+2\right)}\right]=0\)
\(\Leftrightarrow\frac{3}{\left(x-1\right).\left(x+2\right)}-\frac{x+2-7x+7}{\left(x-1\right).\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3-\left(-6x+9\right)}{\left(x-1\right).\left(x+2\right)}=0\)
\(\Rightarrow3+6x-9=0\)
\(\Leftrightarrow6x-6=0\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\left(TM\right)\)
Vậy \(S=\left\{1\right\}\)
b)Ta có: \(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{2}{x^2-2x}-\frac{1}{x}=0\)
\(\Leftrightarrow\left[\frac{x.\left(x+2\right)-\left(x-2\right)}{x.\left(x-2\right)}\right]-\frac{2}{x^2-2x}=0\)
\(\Leftrightarrow\frac{x^2+2x-x+2}{x.\left(x-2\right)}-\frac{2}{x.\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+2-2}{x.\left(x-2\right)}=0\)
\(\Rightarrow x^2+x=0\)
\(\Leftrightarrow x.\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1,0\right\}\)
bạn ơi đây là giải phương trình mình quên ko ghi bạn giúp mình lại với
a) \(\frac{x-1}{x+1}-\frac{x+1}{x-1}+\frac{4}{x^2-1}\left(ĐK:x\ne\pm1\right)\)
\(=\frac{\left(x-1\right)^2-\left(x+1\right)^2+4}{\left(x-1\right)\left(x+1\right)}\)
\(\frac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\frac{-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=-\frac{4}{x+1}\)
b) \(\frac{x^3y+xy^3}{x^4y}:\left(x^2+y^2\right)\left(ĐK:x,y\ne0\right)\)
\(=\frac{xy\left(x^2+y^2\right)}{x^4y}\cdot\frac{1}{x^2+y^2}\)
\(=\frac{1}{x^3}\)