\(x+7\sqrt{x}+10=\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)\)

\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)

\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)

1: \(x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)

2: \(x-16=\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)\)

3: \(9x-1=\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)\)

4: \(x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

\(1,x-9=\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\\ 2,x-16=\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)\\ 3,9x-1=\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)\\ 4,x\sqrt{x}+1=\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)

 1,x−9=(√x−3)(√x+3)

2,x−16=(√x−4)(√x+4)

3,9x−1=(3√x−1)(3√x+1)

4,x√x+1=(√x+1)(x−√x+1

Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$

$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$

$=(\sqrt{x}-2)(\sqrt{x}-3)$

\(x-5\sqrt{x}+6=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)

\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)

\(a\sqrt{a}+2a+\sqrt{a}+2=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(a+1\right)\left(\sqrt{a}+2\right)\)

\(a\sqrt{a}+2a+\sqrt{a}+2=\left(\sqrt{a}+2\right)\left(a+1\right)\)

\(-25x^2\sqrt{2}+10x+4\sqrt{2}=-\sqrt{2}\left(25x^2-\dfrac{10}{\sqrt{2}}-4\right)=-\sqrt{2}.\left(\left(25x\right)^2-2.5.\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}-\dfrac{5}{2}\right)=-\sqrt{2}\left[\left(5x-\dfrac{1}{\sqrt{2}}\right)^2-\dfrac{5}{2}\right]=-\sqrt{2}.\left(5x-\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{5}}{\sqrt{2}}\right).\left(5x-\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{5}}{\sqrt{2}}\right)=-\sqrt{2}.\left(5x-\dfrac{1+\sqrt{5}}{\sqrt{2}}\right)\left(5x-\dfrac{1-\sqrt{5}}{\sqrt{2}}\right)\)

\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=2+\sqrt{3}+\sqrt{6}+2\sqrt{2}\)

\(=2+\sqrt{3}+\sqrt{2}\left(2+\sqrt{3}\right)=\left(2+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)

\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)\)

Bài làm:

Ta có: \(-6x+5\sqrt{x}+1\)

\(=\left(-6x+6\sqrt{x}\right)-\left(\sqrt{x}-1\right)\)

\(=-6\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\)

\(=\left(-6\sqrt{x}-1\right)\left(\sqrt{x}-1\right)\)

\(=\left(6\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\)