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\(\left(x-3\right).\left(x+3\right)\)\(+\left(x-3\right)\left(x+4\right)\)=\(\left(x-3\right)\left(x+3+x+4\right)=\left(x-3\right)\left(2x+7\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x^2-6x+9\right)-4y^2\)
\(=\left(x-3\right)^2-\left(2y\right)^2\)
\(=\left(x-3-2y\right)\left(x-3+2y\right)\)
= ( x^2 - 4y^2 ) + ( 9 - 6x)
= [ x^2 - (2y)^2 ] + 3( 3 - 2x )
= (x - 2y)(x + 2y)+ 3(3 - 2x)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)-3\)
\(=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)-3\)
\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)-3\)
\(=\left(a^2+5a+5\right)^2-1-3\)
\(=\left(a^2+5a+5\right)^2-4\)
\(=\left(a^2+5a+7\right)\left(a^2+5a+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(x\sqrt{y}+y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\)
2) \(9-6\sqrt{a}+a=\left(\sqrt{a}-3\right)^2\)
3) \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)
4) \(x-y+\sqrt{x}+\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+1\right)\)
5) \(a+2\sqrt{ab}+b-1=\left(\sqrt{a}+\sqrt{b}\right)^2-1=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)
1) \(x\sqrt{y}+y\sqrt{x}=\sqrt{x}\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)
2) \(9-6\sqrt{a}+a=\left(3-\sqrt{a}\right)^2\)
3) \(a+2\sqrt{ab}+b=\left(\sqrt{a}+\sqrt{b}\right)^2\)
4) \(x-y+\sqrt{x}+\sqrt{y}=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}+1\right)\)
5) \(a+2\sqrt{ab}+b-1=\left(\sqrt{a}+\sqrt{b}\right)^2-1^2=\left(\sqrt{a}+\sqrt{b}-1\right)\left(\sqrt{a}+\sqrt{b}+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)
\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
xét \(x\ne0\)ta có :
\(M=\)\(^{x^2\cdot\left(x^2+6x+7-\frac{6}{x}+\frac{1}{x^2}\right)}\)
Đặt \(x-\frac{1}{x}=t\Rightarrow t^2=x^2-2+\frac{1}{x^2}\Leftrightarrow t^2+2=x^2+\frac{1}{x^2}\)
Do đó \(M=x^2\cdot\left(t^2+2+6t+7\right)\Leftrightarrow x^2\cdot\left(t^2+6t+9\right)\)
\(\Leftrightarrow M=x^2\cdot\left(t+3\right)^2\)
M=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2(x^2+3x-1)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
\(x-5\sqrt{x}+6=x-3\sqrt{x}-2\sqrt{x}+6\)
\(=\sqrt{x}.\left(\sqrt{x}-3\right)-2.\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right).\left(\sqrt{x}-2\right)\)
\(x-9+y-2\sqrt{xy}=\left(x-2\sqrt{xy}+y\right)-9\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2-9\)
\(=\left(\sqrt{x}-\sqrt{y}-3\right).\left(\sqrt{x}-\sqrt{y}+3\right)\)
\(x-2\sqrt{x}-3=x-3\sqrt{x}+\sqrt{x}-3\)
\(=\sqrt{x}.\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right).\left(\sqrt{x}+1\right)\)
Học tốt
A=m3-m2+6m2-6m+9m-9
=m2(m-1)+6m(m-1)+9(m-1)
=(m-1)(m2+6m+9)=(m-1)(m+3)2
(m+1)(m+3)2