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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) \(x-y+\sqrt{xy^2}-\sqrt{y^3}\)
\(=\left(x-y\right)+\left(y\sqrt{x}-y\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+y\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}+y\right)\)
b4 :
\(a,x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(b,x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)
\(c,x+2\sqrt{xy}+y=\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(d,x-4\sqrt{x}\sqrt{y}+4y=\left(\sqrt{x}-2\sqrt{y}\right)^2\)
b5:
\(a,ĐK:x\ge1\)
\(\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}-\frac{4}{5}\sqrt{25\left(x-1\right)}=1\)
\(\Leftrightarrow3\sqrt{x-1}+2\sqrt{x-1}-4\sqrt{x-1}=1\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
\(b,ĐK:x\ge5\)
\(\frac{1}{3}\sqrt{9\left(x-5\right)}+\frac{1}{2}\sqrt{4\left(x-5\right)}-\frac{7}{5}\sqrt{25\left(x-5\right)}=2\)
\(\Leftrightarrow\sqrt{x-5}+\sqrt{x-5}-7\sqrt{x-5}=2\)
\(\Leftrightarrow-5\sqrt{x-5}=2\)
\(\Leftrightarrow\sqrt{x-5}=-\frac{2}{5}\left(voli\right)\)
\(c,ĐK:x>0\)
\(\sqrt{x}+\frac{9}{\sqrt{x}}=6\)
\(\Leftrightarrow x+9=6\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+9=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)^2=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
a, \(7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}=7\sqrt{B}\left(\sqrt{A}+\sqrt{B}\right)-\left(\sqrt{A}+\sqrt{B}\right)\)\(=\left(\sqrt{A}+\sqrt{B}\right)\left(7\sqrt{B}-1\right)\)
b, \(a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)\)
c,\(\sqrt{x^2-25y^2}-\sqrt{x-5y}=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}\left(\sqrt{x+5y}-1\right)\)
\(a,7\sqrt{AB}+7B-\sqrt{A}-\sqrt{B}\)( Với A>= 0, B>=0)
\(=\left(7\sqrt{AB}-\sqrt{A}\right)+\left(7B-\sqrt{B}\right)\)
\(=7\sqrt{A}\left(\sqrt{B}-1\right)+7\sqrt{B}\left(\sqrt{B}-1\right)\)
\(=\left(\sqrt{B}-1\right)\left(7\sqrt{A}+7\sqrt{B}\right)\)
\(=7\left(\sqrt{B}-1\right)\left(\sqrt{A}+\sqrt{B}\right)\)
\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)Với a>= 0, b>=0)
\(=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)\)
\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)
\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}.\sqrt{x+5y}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}\left(\sqrt{x+5y}-1\right)\)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(a,\)\(7\sqrt{ab}+7b-\sqrt{a}-\sqrt{b}\)
\(=7\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}+\sqrt{b}\right)\left(7\sqrt{b}-1\right)\)
\(b,a\sqrt{b}-b\sqrt{a}+\sqrt{a}-\sqrt{b}\)
\(=\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)+\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}-1\right)\)
\(c,\sqrt{x^2-25y^2}-\sqrt{x-5y}\)
\(=\sqrt{\left(x-5y\right)\left(x+5y\right)}-\sqrt{x-5y}\)
\(=\sqrt{x-5y}\left(\sqrt{x-5y}-1\right)\)
Bài 2b : Biết đc bài nào thì làm bài đó hjhj
\(5+2\sqrt{6}=3+2.\sqrt{2}.\sqrt{3}+2=\left(\sqrt{3}+\sqrt{2}\right)^2\)