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a) \(\sqrt{a^3}-\sqrt{b^3}+\sqrt{a^2b}-\sqrt{ab^2}\)
\(=a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)-\left(\sqrt{a}-\sqrt{b}\right)\sqrt{ab}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b-\sqrt{ab}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(a+b\right)\)
b) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)
\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)
1) \(x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)
2) \(x-3=\left(\sqrt{x}-\sqrt{3}\right)\left(\sqrt{x}+\sqrt{3}\right)\)
3) \(a+b=a-\left(-b\right)=\left(\sqrt{a}-\sqrt{-b}\right)\left(\sqrt{a}+\sqrt{-b}\right)\)
p/s: chúc bạn học tốt
a, \(A=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)
\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
\(a,\)\(A=x^2-3x\sqrt{y}+2y\)
\(=x^2-2x\sqrt{y}-x\sqrt{y}+2y\)
\(=x\left(x-2\sqrt{y}\right)-\sqrt{y}\left(x-2\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)
\(b,\)Ta có : \(x=\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow A=\left[\sqrt{5}+2-\sqrt{\left(\sqrt{5}-2\right)^2}\right]\left[\sqrt{5}+2-2\sqrt{\left(\sqrt{5}-2\right)^2}\right]\)
\(=\left(\sqrt{5}+2-\sqrt{5}-2\right)\left(\sqrt{5}+2-2\sqrt{5}+4\right)\)
\(=4\left(6-\sqrt{5}\right)\)
\(=24-4\sqrt{5}\)
a, \(x-\sqrt{x}\)= \(\sqrt{x}.\left(\sqrt{x}-1\right)\)
b, 3x+6\(\sqrt{x}\)= \(\sqrt{x}.\left(3\sqrt{x}+6\right)\)
c, x+2\(\sqrt{x}+1\)= \(\left(\sqrt{x}\right)^2+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
d, \(3x-5\sqrt{x}+2=3x-3\sqrt{x}-2\sqrt{x}+2\)
=\(3\sqrt{x}.\left(\sqrt{x}-1\right)-2.\left(\sqrt{x}-1\right)\)
=\(\left(3\sqrt{x}-2\right).\left(\sqrt{x}-1\right)\)
\(ab+b\sqrt{a}+\sqrt{a}+1\)
(đk: \(a\ge0\))
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\sqrt{a}+1=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
ĐK: \(x,y\ge0\)
\(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=x\left(\sqrt{x}+\sqrt{y}\right)-y\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{x}-\sqrt{y}\right)\)
\(\text{a)}x\sqrt{x}+\sqrt{x}-x-1\)
\(=\left(x\sqrt{x}+\sqrt{x}\right)-\left(x+1\right)\)
\(=\sqrt{x}\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(\sqrt{x}-1\right)\)
\(\text{b)}\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\left(\sqrt{ab}+2\sqrt{a}\right)+\left(3\sqrt{b}+6\right)\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
\(\text{c)}\left(1+\sqrt{x}\right)^2-4\sqrt{x}\)
\(=\left(1+\sqrt{x}\right)^2-\left(2\sqrt{\sqrt{x}}\right)^2\)
\(=\left(1+\sqrt{x}+2\sqrt{\sqrt{x}}\right)\left(1+\sqrt{x}-2\sqrt{\sqrt{x}}\right)\)
\(\text{d)}\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)
\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)
\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)
\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)
\(\text{e)}a+\sqrt{a}+2\sqrt{ab}+2\sqrt{b}\)
\(=\left(a+\sqrt{a}\right)+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\left[\left(\sqrt{a}\right)^2+\sqrt{a}\right]+\left(2\sqrt{ab}+2\sqrt{b}\right)\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)+2\sqrt{b}\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\sqrt{b}\right)\)
\(\text{f)}x-2\sqrt{x-1}-a^2\)
\(=\left(\sqrt{x-2}\right)^2\left(\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2}\sqrt{\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}\right)^2-a^2\)
\(=\left(\sqrt{x-2\sqrt{x-1}}+a\right)\left(\sqrt{x-2\sqrt{x-1}}-a\right)\)
a, \(1-a\sqrt{a}\)
\(=\left[1-\left(\sqrt{a}\right)^3\right]\)
\(=\left(1-\sqrt{a}\right)\left[\left(\sqrt{a}\right)^2+1.\sqrt{a}+1^2\right]\)
\(=\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)\)
b, \(x-2\sqrt{x-1}\)
\(=\left(x-1\right)-2\sqrt{x-1}+1\)
\(=\left[\left(\sqrt{x-1}\right)-1\right]^2\)