Giai phương trình: \(\frac{x+2}{x-3}=9+\frac{6}{2-x}\)
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\(ĐKXĐ:x\ne\pm3\)
Đặt \(\frac{x+2}{x-3}=a;\frac{x-2}{x+3}=b\)
Ta có:
\(pt\Leftrightarrow3a^2+8ab=3b^2\)
\(\Leftrightarrow3a^2+8ab-3b^2=0\)
\(\Leftrightarrow\left(3a-b\right)\left(3b+a\right)=0\)
\(\Leftrightarrow3a=b;3b=-a\)
Đến đây bạn thay vào làm nhá,giải như pt bậc 2 thôi
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(\hept{\begin{cases}x\ge0\\x\ne1;y\ne2\end{cases}}\)
pt <=> \(\hept{\begin{cases}\frac{1}{\sqrt{x}-1}+\frac{6}{\left|y-2\right|}=2\\\frac{2-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\frac{3}{3\left|y-2\right|}=-9\end{cases}}\)
<=> \(\hept{\begin{cases}\frac{1}{\sqrt{x}-1}+\frac{6}{\left|y-2\right|}=2\\\frac{2}{\sqrt{x}-1}-\frac{1}{\left|y-2\right|}=-8\end{cases}}\)
Đặt: \(\frac{1}{\sqrt{x}-1}=u;\frac{1}{\left|y-2\right|}=v>0\)ta có pt:
\(\hept{\begin{cases}u+6v=2\\2u-v=-8\end{cases}}\)=> tìm u; v sau đó tìm x; y
Đặt \(\left|y-2\right|=u;\sqrt{x}-1=v\)
Hệ trở thành \(\hept{\begin{cases}\frac{1}{v}+\frac{6}{u}=2\\\frac{2}{v}-u=-8\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2}{v}+\frac{12}{u}=4\\\frac{2}{v}-u=-8\end{cases}}\)
\(\Rightarrow\frac{12}{u}+u=12\Rightarrow\frac{12+u^2}{u}=12\)
\(\Rightarrow u^2-12u+12=0\)
\(\Delta=12^2-4.12=96,\sqrt{\Delta}=4\sqrt{6}\)
\(\Rightarrow\orbr{\begin{cases}u=\frac{12+4\sqrt{6}}{2}=6+2\sqrt{6}\\u=\frac{12-4\sqrt{6}}{2}=6-2\sqrt{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left|y-2\right|=6+2\sqrt{6}\\\left|y-2\right|=6-2\sqrt{6}\end{cases}}\)
\(\Rightarrow y\in\left\{8\pm2\sqrt{6};-4\pm2\sqrt{6}\right\}\)
Thay vào hệ tính được x nha, th nào ko đúng loại
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\)\(x\ne\pm3\)
\(\frac{x}{x+3}-\frac{x-2}{2x-6}=\frac{x+2}{x^2-9}\)
\(\Leftrightarrow\)\(\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{2\left(x+2\right)}{2\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\)\(2x\left(x-3\right)-\left(x-2\right)\left(x+3\right)=2\left(x+2\right)\)
\(\Leftrightarrow\)\(2x^2-6x-x^2-x+6=2x+4\)
\(\Leftrightarrow\)\(x^2-9x+2=0\)
p/s: mk lm đc có đến đấy thôi, bn tham khảo nhé, lm đc thì gửi cho mk nhé
\(\frac{x+2}{x-3}=9+\frac{6}{2-x}ĐKXĐ:\orbr{\begin{cases}x\ne3\\x\ne2\end{cases}}\)
\(\Leftrightarrow\frac{\left(2+x\right)\left(2-x\right)}{\left(x-3\right)\left(2-x\right)}=\frac{9\left(2-x\right)\left(x-3\right)}{\left(2-x\right)\left(x-3\right)}+\frac{6\left(x-3\right)}{\left(2-x\right)\left(x-3\right)}\)
\(\Leftrightarrow\left(2+x\right)\left(2-x\right)=9\left(2-x\right)\left(x-3\right)+6\left(x-3\right)\)
\(\Leftrightarrow4-x^2=\left(18-9x\right)\left(x-3\right)+6x-18\)
\(\Leftrightarrow4-x^2=18x-54-9x^2+27x+6x-18\)
\(\Leftrightarrow4-x^2=51x-72-9x^2\)
\(\Leftrightarrow51x-72-9x^2+x^2-4=0\)
\(\Leftrightarrow-8x^2+51x-76=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-\frac{19}{8}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-\frac{19}{8}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{19}{8}\end{cases}}\)
\(\frac{x+2}{x-3}=9+\frac{6}{2-x}\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=9\left(x-3\right)\left(2-x\right)+6\left(x-3\right)\)
\(\Leftrightarrow4-x^2=51x-9x^2-72\)
\(\Leftrightarrow4-x^2-51x+9x^2+72=0\)
\(\Leftrightarrow76+8x^2-51x=0\)
\(\Leftrightarrow8x^2-19x-32x+76=0\)
\(\Leftrightarrow x\left(8x-19\right)-4\left(8x-19\right)=0\)
\(\Leftrightarrow\left(8x-19\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}8x-19=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{19}{8}\\x=4\end{cases}}\)
Vậy nghiệm phương trình là: \(\left\{\frac{19}{8};4\right\}\)