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\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)
a: \(B=\frac{2\sqrt{x}+13}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}+13+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\sqrt{x}+13+x+\sqrt{x}-6}{\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
b: C=A-B
\(=\frac{2\sqrt{x}-1}{\sqrt{x}+3}-\frac{x+3\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(x+3\sqrt{x}+7\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{2x+3\sqrt{x}-2-x-3\sqrt{x}-7}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-3}{\sqrt{x}+2}\)
C nguyên khi \(\sqrt{x}-3\) ⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2-5\) ⋮\(\sqrt{x}+2\)
=>-5⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2=5\) (Do \(\sqrt{x}+2\ge2\forall x\) thỏa mãn ĐKXĐ)
=>\(\sqrt{x}=3\)
=>x=9(nhận)
Toán lớp 9 như này á hả
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