2:

\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)

=căn ab(6+7/b-5/a)

a: ĐKXĐ: a>=0; b>=0; ab<>1

Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}\)

\(=\frac{\sqrt{a}+a\cdot\sqrt{b}+\sqrt{b}+b\cdot\sqrt{a}+\sqrt{a}-a\cdot\sqrt{b}-\sqrt{b}+b\cdot\sqrt{a}}{1-ab}=\frac{2\cdot\sqrt{a}+2b\cdot\sqrt{a}}{1-ab}\)

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

Ta có: \(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)

\(=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}:\frac{1-ab+a+b+2ab}{1-ab}=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}\cdot\frac{1-ab}{ab+a+b+1}\)

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

b: \(a=\frac{2}{2+\sqrt3}=\frac{2\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)

\(=\frac{4-2\sqrt3}{4-3}=4-2\sqrt3=\left(\sqrt3-1\right)^2\)

Thay \(a=\left(\sqrt3-1\right)^2\) vào D, ta được:

\(D=\frac{2\cdot\sqrt{\left(\sqrt3-1\right)^2}}{\left(\sqrt3-1\right)^2+1}\)

\(=\frac{2\left(\sqrt3-1\right)}{4-2\sqrt3+1}=\frac{2\sqrt3-2}{5-2\sqrt3}=\frac{\left(2\sqrt3-2\right)\left(5+2\sqrt3\right)}{\left(5-2\sqrt3\right)\left(5+2\sqrt3\right)}\)

\(=\frac{10\sqrt3+12-10-4\sqrt3}{25-12}=\frac{6\sqrt3+2}{13}\)

c: \(\frac{1}{D}=\frac{a+1}{2\sqrt{a}}\)

=>\(\frac{1}{D}-1=\frac{a+1-2\sqrt{a}}{2\sqrt{a}}=\frac{\left(\sqrt{a}-1\right)^2}{2\sqrt{a}}\ge0\forall a\) thỏa mãn ĐKXĐ

=>\(\frac{1}{D}\ge1\forall a\) thỏa mãn ĐKXĐ

=>D<=1∀a thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{a}-1=0\)

=>a=1(nhận)

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

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

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

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)

=0

\(B=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}\)

\(B=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\sqrt{ab}\)

\(B=a\sqrt{b}-b\sqrt{a}\)

Với `a,b > 0` có:

`B=[a\sqrt{b}-b\sqrt{a}]/\sqrt{ab} :1/[\sqrt{a}.\sqrt{b}]`

`B=[a\sqrt{b}-b\sqrt{a}]/[\sqrt{ab}] .\sqrt{ab}`

`B=a\sqrt{b}-b\sqrt{a}`

\(\text{​​}\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}.\sqrt{b}}=\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}.\sqrt{ab}=a\sqrt{b}-b\sqrt{a}\)

=\(\dfrac{\left(a-b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

=\(\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)

=\(0\)

a: ĐKXĐ: x<>1/2

Sửa đề: \(A=\frac{3}{2\left(2x-1\right)}\cdot\sqrt{8x^4\left(4x^2-4x+1\right)}\)

\(=\frac{3}{2\left(2x-1\right)}\cdot\sqrt8\cdot\sqrt{x^4}\cdot\sqrt{\left(2x-1\right)^2}\)

\(=\frac{3}{2\left(2x-1\right)}\cdot2\sqrt2\cdot x^2\cdot\left|2x-1\right|=\frac{6\sqrt2\cdot x^2}{2\left(2x-1\right)}\cdot\left|2x-1\right|\)

=\(\pm3\sqrt2\cdot x^2\)

b: ĐKXĐ: b<>0

\(B=\frac{a-b}{b^2}\cdot\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}\)

\(=\frac{a-b}{b^2}\cdot\sqrt{a^2}\cdot\frac{\sqrt{b^4}}{\sqrt{\left(a-b\right)^2}}\)

\(=\frac{a-b}{b^2}\cdot\left|a\right|\cdot\frac{b^2}{\left|a-b\right|}=\left|a\right|\cdot\frac{a-b}{\left|a-b\right|}=\pm\left|a\right|\)

\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)

\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)

\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)

Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)

\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)

\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)

\(=\pm3\sqrt{2}x^2\)

\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)

\(=\pm\left|a\right|\)