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a) \(x^2-12x+36=x^2-2.x.6+6^2=\left(x-6\right)^2\)
b) \(12x-x^2-36=-\left(x^2-12x+36\right)=-\left(x-6\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4-2x^3-12x^2+12x+36=x^4+x^2+36-2x^3+12x-12x^2-x^2\)
\(=\left(x^2-x-6\right)^2-x^2=\left(x^2-6\right)\left(x^2-2x-6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> \(x^2-12x+36=0\Leftrightarrow\left(x-6\right)^2=0\Rightarrow x=6\)
12x - x2 - 36=0
=>- x2 + 12x - 36 = 0
Giải phương trình trên máy tính ta có :
X= 6
Vậy x = 6
Study well
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\(4\left(6-x\right)+x^2-12x+36=0\)
\(24-4x+x^2-12x+36=0\)
\(x^2-16x+60=0\)
\(x^2-2x8+8^2-8^2+60=0\)
\(\left(x-8\right)^2-4=0\)
\(\left(x-8\right)^2=4\)
\(\left(x-8\right)^2=\left(\pm2\right)^2\)
\(\orbr{\begin{cases}x-8=2\Rightarrow x=10\\x-8=-2\Rightarrow x=6\end{cases}}\)
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Bài 4:
\(x^3-2x^2+x=x\left(x-1\right)^2\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(x^2-12x+36=\left(x-6\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3+8x^2+17x+10\)
\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+7x+10\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
b) \(=x^4-2x^3-12x^2+12x+36\)
\(=x^2\left(x^2-2x-6\right)-2\left(x^2-2x-6\right)\)
\(=\left(x^2-2\right)\left(x^2-2x-6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-12x=-36\)
\(\Rightarrow x^2-12x+36=-36+36\)
\(\Rightarrow\left(x-6\right)^2=0\)
\(\Rightarrow x-6=0\)
\(\Rightarrow x=6\)
\(x^2-12x=36\)
\(\Leftrightarrow\left(x-6\right)^2=72\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{72}+6\\x=6-\sqrt{72}\end{cases}}\)