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![](https://rs.olm.vn/images/avt/0.png?1311)
Cách 1:
\(\frac{2}{x\left(x^2+2\right)}=\frac{a}{x}+\frac{bx+c}{x^2+2}=\frac{a\left(x^2+2\right)+x\left(bx+c\right)}{x\left(x^2+2\right)}=\frac{\left(a+b\right)x^2+cx+2a}{x\left(x^2+2\right)}\)
Đồng nhất hệ số:
\(\hept{\begin{cases}a+b=0\\c=0\\2a=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=-1\\c=0\end{cases}}\).
Cách 2:
\(\frac{2}{x\left(x^2+2\right)}=\frac{x^2+2-x^2}{x\left(x^2+2\right)}=\frac{1}{x}-\frac{x}{x^2+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(x\left(x+1\right)-3=\left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+x-3=2x^2-3x+1\)
\(\Leftrightarrow2x^2-3x+1-x^2-x+3=0\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
2) \(\left(x+2\right)\left(3x+1\right)-x^2+4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{-3}{2}\end{cases}}}\)
3) \(x\left(x-3\right)+2=x\left(1-x\right)+2x-2\)
\(\Leftrightarrow x^2-3x+2=x-x^2+2x-2\)
\(\Leftrightarrow x^2-3x+2=-x^2+3x-2\)
\(\Leftrightarrow x^2-3x+2+x^2-3x+2=0\)
\(\Leftrightarrow2\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có SABO = OE.AB : 2
Vì \(\hept{\begin{cases}AB//CD\\\widehat{AEO}=90^{\text{o}}\end{cases}}\Rightarrow\widehat{CFO}=90^{\text{o}}\)
=> SCDO = OF.CD : 2 = OF.AB : 2
=> SABO + SCDO = EF.AB : 2 = \(\frac{1}{2}S_{ABCD}\)(Vì EF là đường cao hình bình hành ABCD => SABCD = EF.AB)
Tương tự ta được
SBCO + SDAO = HK.BC : 2 = \(\frac{1}{2}S_{ABCD}\)(HK đường cao hình bình hành ABCD => SABCD = HK.BC)
=> SABO + SCDO = SBCO + SDAO (= \(\frac{1}{2}S_{ABCD}\)) => ĐPCM
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t\Rightarrow\hept{\begin{cases}x=at\\y=bt\\z=ct\end{cases}}\).
\(4=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=4+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca=0\)
\(P=xy+yz+zx=abt^2+bct^2+cat^2=t^2\left(ab+bc+ca\right)=0\)