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câu đầu tiên bạn tách \(x\left(x-2\right)-4\)
bạn tách ra để có thừa số chung tìm UCLN rồi tính nhé
câu b mình chịu
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+\sqrt{x}+2\sqrt{x}+2\)
= \(\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
= \(\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)\)
\(2x-2\sqrt{x}+3\sqrt{x}-3\)
= \(2\sqrt{x}\left(\sqrt{x}-1\right)+3\left(\sqrt{x}-1\right)\)
= \(\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^7+x^5+1\)
\(\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
k cho mình nha
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(y-2x\sqrt{y}-3x\sqrt{y}+6x^2=\sqrt{y}\left(\sqrt{y}-2x\right)-3x\left(\sqrt{y}-2x\right)=\left(\sqrt{y}-3x\right)\left(\sqrt{y}-2x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2) ta có \(x^3+4x^2-29x+24=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-x-3x+3\right)=\left(x-8\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]=\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
\(=\left(x^2-x+2\right)^2-x^2+x^2-4x+4+x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2\right)+2\left(x^2-2x+2\right)\)
\(=\left(x^2-2x+2\right)\left(x^2+4\right)\)