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\(-2=\frac{2}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{2}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{2}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{2}{\left(x^2+2\right)\left(x^2+1\right)}\)
<=>\(\frac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\frac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\frac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\frac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+2}+\frac{1}{x^2+2}-\frac{1}{x^2+3}+...+\frac{1}{x^2+4}-\frac{1}{x^2+5}=-1\)
<=>\(\frac{1}{x^2+1}-\frac{1}{x^2+5}=-1\)
<=>(x2+5)-(x2+1)=-(x2+1)(x2+5)
<=>4=-x4-6x2-5
<=>x4+6x2+9=0
<=>(x2+3)2=0
<=>x2+3=0
Do x2>0
=>x2+3>0 nên PT vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)
\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)
\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)
\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)
\(\Rightarrow-16x+12x+16-3=24x+48-18\)
\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-------------------ko chép đề nha---------
\(\Leftrightarrow\frac{4\left(x^2-4x+4\right)-3\left(2x+1\right)}{12}=\frac{12-2\left(4x^2-12x+9\right)}{12}\)
\(\Rightarrow4x^2+16x+16-6x-3=12-8x^2+24x-18\)
\(\Leftrightarrow4x^2+10x+13=-8x^2+24x-6\)
\(\Leftrightarrow4x^2+8x^2+10x-24x+13+6=0\)
\(\Leftrightarrow12x-14x+19=0\)
Ta có :\(\Delta'=7^2-12.19=-179< 0\)
\(\Rightarrow\)phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ne\pm2\)
Đặt \(\frac{x+3}{x-2}=a,\frac{x-3}{x+2}=b\)
\(PT\Leftrightarrow a^2+6b^2-7ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(a-6b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\a=6b\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=6.\frac{x-3}{x+2}\end{cases}}\)
Đến đây nhân chéo rồi tìm nghiệm nhé :))))
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
![](https://rs.olm.vn/images/avt/0.png?1311)
<=>\(\left(x+2\right)^2-\left(x-2\right)^3=-\left(x-6\right)\left(x^2-x+2\right)\)
=>\(12x\left(x-1\right)-8=4\left(3x^2-3x-2\right)\)
=>\(-\left(x-6\right)\left(x^2-x+2\right)=4\left(3x^2-3x-2\right)\)
=>x=-5;-2 hoặc 2
<=>\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=\frac{1894289\left(x+100\right)}{6370665}\)
=>\(\frac{1894289\left(x+100\right)}{6370665}=0\)(rút gọn)
=>\(\frac{1894289x}{6370665}+\frac{37885780}{1274133}=0\)
=>\(\frac{1894289\left(x+100\right)}{6370665}=0\)(giải PT này )
=>x=-100
\(\frac{3}{4}\left(x^2+1\right)^2+3\left(x^2+x\right)-9=0\)
<=> \(3\left(x^2+1\right)^2.4+3\left(x^2+x\right).4-9.4=0.4\)
<=> \(3\left(x^2+1\right)^2+12\left(x^2+x\right)-36=0\)
<=> \(3x^4+18x^2+12x-33=0\)
<=> \(3\left(x-1\right)\left(x^3+x^2+7x+11\right)=0\)
<=> \(x-1=0\)
<=> \(x=1\)
Mà vì: \(x^3+x^2+7x+11\ne0\)
=> x = 1
\(=>\frac{3}{4}\left[\left(x^2+1\right)^2+4\left(x^2+1\right)+4\right]-12=0\)
\(=>\frac{3}{4}\left(x^2+1+2\right)^2-12=0\)
\(=>\left(x^2+3\right)^2=16\)
Đến đây tự tìm nha
Hok tốt