b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)

\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)

\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)

\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)

\(VT=0=VP\)

a) \(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{\left(y^2\right)^2}}\) 

\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)

\(=\dfrac{1}{y}\)

b) \(\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{\left(x^2y^4\right)^2}}\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{4}{x^2y^4}\)

\(=\dfrac{20x^3y^3}{2x^2y^4}\)

\(=\dfrac{10x}{y}\)

c) \(ab^2\sqrt{\dfrac{3}{a^2b^4}}\)

\(=ab^2\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)

\(=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)

\(=\sqrt{3}\)

\(a,\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\left(y\ge0;x,y\ne0\right)\) (sửa đề)

\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{y^4}}\)

\(=\dfrac{y}{x}\cdot\dfrac{x}{\sqrt{\left(y^2\right)^2}}\)

\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)

\(=\dfrac{1}{y}\)

\(---\)

\(b,\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\left(x,y\ne0\right)\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}\)

\(=\dfrac{5x^3y^3}{2}\cdot\dfrac{4}{x^2y^4}\)

\(=\dfrac{5x\cdot2}{y}\)

\(=\dfrac{10x}{y}\)

\(---\)

\(c,ab^2\sqrt{\dfrac{3}{a^2b^4}}\left(a>0;b\ne0\right)\) (sửa đề)

\(=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\)

\(=\dfrac{ab^2\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)

\(=\dfrac{ab^2\sqrt{3}}{ab^2}\)

\(=\sqrt{3}\)

#\(Toru\)

Ta có VT =\(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)

=\(\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) =\(\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

=\(\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

=\(\dfrac{a-b}{a-b}=1=VP\)

Xét hiệu \(2a^2+2b^2-\left(a^3+ab^2\right)=\left(2a^2-a^3\right)+\left(2b^2-ab^2\right)\)

\(=a^2\left(2-a\right)+b^2\left(2-a\right)\)

\(=\left(a^2+b^2\right)\left(2-a\right)\)

Do \(a^2+b^2\ge0;\forall a;b\) nên:

\(2a^2+2b^2>a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\2-a>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2\ne0\\a< 2\end{matrix}\right.\)

\(2a^2+2b^2=a^3+ab^2\) khi \(\left[{}\begin{matrix}a^2+b^2=0\\2-a=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=b=0\\a=2\end{matrix}\right.\)

\(2a^2+2b^2< a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\a>2\end{matrix}\right.\) \(\Rightarrow a>2\)

\(2a^2+2b^2\ge a^3+ab^2\) khi \(2-a\ge0\Leftrightarrow a\le2\)

a, \(ĐKXĐ:a;b>0;a\ne2b\\ \)

Xét: \(\dfrac{2\left(a+b\right)}{\sqrt{a^3}-2\sqrt{2b^3}}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{2\left(a+b\right)}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}-\dfrac{\sqrt{a}}{a+\sqrt{2ab}+2b}=\dfrac{a+2b+\sqrt{2ab}}{\left(\sqrt{a}-\sqrt{2b}\right)\left(a+\sqrt{2ab}+2b\right)}=\dfrac{1}{\sqrt{a}-\sqrt{2b}}\)\(\dfrac{\sqrt{a^3}+2\sqrt{2b^3}}{2b+\sqrt{2ab}}-\sqrt{a}=\dfrac{\left(\sqrt{a}+\sqrt{2b}\right)\left(a-\sqrt{2ab}+2b\right)}{\sqrt{2b}\left(\sqrt{a}+\sqrt{2b}\right)}-\sqrt{a}=\dfrac{\left(\sqrt{a}-\sqrt{2b}\right)^2}{\sqrt{2b}}\)\(\Rightarrow P=\dfrac{\sqrt{a}-\sqrt{2b}}{\sqrt{2b}}=\sqrt{\dfrac{a}{2b}}-1\)

b, Tự lm nhé.

a, Khi x = 2, ta được: 

\(A=\dfrac{4}{2\sqrt{2}-2}=2+2\sqrt{2}\)

b, \(B=\dfrac{\sqrt{x}-4}{x-2\sqrt{x}}+\dfrac{3}{\sqrt{x}-2}\\ \Rightarrow B=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ \Rightarrow B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(P=B:A=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{4}=-\left(\sqrt{x}-1\right)=1-\sqrt{x}\) (đpcm)

(a) \(\sqrt{\dfrac{a^2}{25+10b+b^2}}=\sqrt{\dfrac{a^2}{\left(5+b\right)^2}}=\dfrac{\sqrt{a^2}}{\sqrt{\left(5+b\right)^2}}\)

\(=\dfrac{\left|a\right|}{\left|5+b\right|}=\dfrac{-a}{b+5}\) (do \(a< 0,b>0\Rightarrow b+5>0\))

(b) \(\left(a-b\right)\sqrt{\dfrac{a^2b^2}{\left(a-b\right)^2}}=\left(a-b\right)\sqrt{\dfrac{\left(ab\right)^2}{\left(a-b\right)^2}}=\left(a-b\right)\cdot\dfrac{\sqrt{\left(ab\right)^2}}{\sqrt{\left(a-b\right)^2}}\)

\(=\left(a-b\right)\cdot\dfrac{\left|ab\right|}{\left|a-b\right|}\).

(c) \(\dfrac{x+4\sqrt{x}+4}{2+\sqrt{x}}=\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}=\sqrt{x}+2.\)

Chọn C