Phân tích đa thức thành nhân tử
1. \(x\sqrt{x}+\sqrt{x}-x-1\)
2. \(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
3. \(x-\sqrt{x}-2\)
4. \(x-3\sqrt{x}+2\)
5. \(-6x+5\sqrt{x}+1\)
6. \(x+4\sqrt{x}+3\)
7. \(3\sqrt{a}-2a-1\)
8. \(x+2\sqrt{x-1}\)
9. \(7\sqrt{x}-6x-2\)
10. \(x-5\sqrt{x}+6\)
11. \(x-2+\sqrt{x^2-4}\)
1) ta có : \(x\sqrt{x}+\sqrt{x}-x-1=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x+1\right)\)
2) ta có : \(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
3) ta có : \(x-\sqrt{x}-2=x+\sqrt{x}-2\sqrt{x}-2\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)
4) ta có : \(x-3\sqrt{x}+2=x-\sqrt{x}-2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)=\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
5) ta có : \(-6x+5\sqrt{x}+1=-6x+6\sqrt{x}-\sqrt{x}+1\)
\(=6\sqrt{x}\left(1-\sqrt{x}\right)+\left(1-\sqrt{x}\right)=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)
6) ta có : \(x+4\sqrt{x}+3=x+\sqrt{x}+3\sqrt{x}+3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)\)
7) ta có : \(3\sqrt{a}-2a-1=-2a+2\sqrt{a}+\sqrt{a}-1\)
\(=-2\sqrt{a}\left(\sqrt{a}-1\right)+\left(\sqrt{a}-1\right)=\left(1-2\sqrt{a}\right)\left(\sqrt{a}-1\right)\)
8) ta có : \(x+2\sqrt{x-1}=x-1+2\sqrt{x-1}+1\)
\(=\left(\sqrt{x-1}+1\right)^2\)
9) ta có : \(7\sqrt{x}-6x-2=-6x+3\sqrt{x}+4\sqrt{x}-2\)
\(=-3\sqrt{x}\left(2\sqrt{x}-1\right)+2\left(2\sqrt{x}-1\right)=\left(2-3\sqrt{x}\right)\left(2\sqrt{x}-1\right)\)
10) ta có : \(x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
11) ta có : \(x-2+\sqrt{x^2-4}=\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-2\right)\left(x+2\right)}\)
\(=\sqrt{x-2}\left(\sqrt{x-2}+\sqrt{x+2}\right)\)