\(B=\frac{1}{2a-1}.\sqrt{5a^4\left(2a-1\right)^2}=\sqrt{5}a^2.\frac{\left|2a-1\right|}{2a-1}\)

Nếu \(a>\frac{1}{2}\) thì \(B=\sqrt{5}a^2\)

Nếu \(a< \frac{1}{2}\) thì \(B=-\sqrt{5}a^2\)

\(B=\frac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\)

\(B=\frac{2\left|a\right|}{2a-1}\sqrt{5\left[1-2.2a+\left(2a\right)^2\right]}\)

\(B=\frac{2a}{2a-1}\sqrt{5\left(1-2a\right)^2}\)

\(B=\frac{2a\left|1-2a\right|}{2a-1}\sqrt{5}\)

\(=\frac{2a\left(2a-1\right)}{2a-1}\sqrt{5}=2a\sqrt{5}\)

\(ĐKXĐ:a\ne\frac{1}{2}\)

\(B=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)

\(=\frac{1}{2a-1}.\sqrt{5a^4.\left(1-2a\right)^2}\)

\(=\frac{1}{2a-1}.\sqrt{5}.\sqrt{a^4}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{1}{2a-1}.\sqrt{5}.a^2.\left|1-2a\right|=\frac{\sqrt{5}.a^2.\left|1-2a\right|}{2a-1}\)

+) Nếu \(a< \frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=1-2a=-\left(2a-1\right)\)

\(\Rightarrow B=\frac{-\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=-\sqrt{5}.a^2\)

+) Nếu \(a>\frac{1}{2}\)\(\Rightarrow\left|1-2a\right|=-\left(1-2a\right)=-1+2a=2a-1\)

\(\Rightarrow B=\frac{\sqrt{5}.a^2.\left(2a-1\right)}{2a-1}=\sqrt{5}.a^2\)

\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)

<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)

<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)

<=>\(\frac{\sqrt{3}}{x-y}\)

Hỏi nhiều thế.

Giải:

\(\dfrac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)

\(=\dfrac{1}{2a-1}.\sqrt{5a^4}.\sqrt{1-4a+4a^2}\)

\(=\dfrac{1}{2a-1}.a^2\sqrt{5}.\sqrt{\left(1-2a\right)^2}\)

\(=\dfrac{1}{2a-1}.a^2\sqrt{5}.\left|1-2a\right|\)

\(=\dfrac{\left|2a-1\right|.a^2\sqrt{5}}{2a-1}\left(1\right)\)

Chắc đề thiếu điều kiện, mình cho thêm để ra kết quả đẹp

ĐK: \(a\ge1\Leftrightarrow2a\ge2\Leftrightarrow2a-1\ge1>0\)

\(\left(1\right)=\dfrac{\left(2a-1\right).a^2\sqrt{5}}{2a-1}\)

\(=a^2\sqrt{5}\)

Vậy ...

\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(1-4a+4a^2\right)}\left(a\ne\dfrac{1}{2}\right)\)

\(=\dfrac{1}{2a-1}\sqrt{5\left(a^2\right)^2\left(1-2a\right)^2}=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

Xét \(a>\dfrac{1}{2}\Rightarrow1-2a< 0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(2a-1\right)=\sqrt{5}a^2\)

Xét \(a< \dfrac{1}{2}\Rightarrow1-2a>0\Rightarrow\dfrac{1}{2a-1}\sqrt{5}.a^2.\left|1-2a\right|\)

\(=\dfrac{1}{2a-1}\sqrt{5}.a^2.\left(1-2a\right)=-\sqrt{5}a^2\)

\(E=\dfrac{1}{2a-1}\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{a^2.\left|2a-1\right|.\sqrt{5}}{2a-1}\)

- Với \(2a-1>0\Rightarrow a>\dfrac{1}{2}\) thì \(E=\dfrac{a^2\left(2a-1\right).\sqrt{5}}{2a-1}=a^2\sqrt{5}\)

- Với \(a< \dfrac{1}{2}\) thì \(E=\dfrac{-a^2.\left(2a-1\right).\sqrt{5}}{2a-1}=-a^2\sqrt{5}\)

Ta có: \(E=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\cdot\left(4a^2-4a+1\right)}\)

\(=\dfrac{1}{2a-1}\cdot\dfrac{a^2\cdot\sqrt{5}\cdot\left(2a-1\right)}{1}\)

\(=a^2\sqrt{5}\)

a) \(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x+y\right)^2}{2}}=\dfrac{2\cdot\left(x+y\right)\cdot\sqrt{3}}{\left(x+y\right)\cdot\left(x-y\right)\cdot\sqrt{2}}=\dfrac{2\sqrt{3}}{\left(x-y\right)\cdot\sqrt{2}}=\dfrac{2\sqrt{6}}{2\left(x-y\right)}=\dfrac{\sqrt{6}}{x-y}\)

b) \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\dfrac{2}{2a-1}\cdot\sqrt{5a^2\left[\left(2a\right)^2-2\cdot2\cdot a+1^2\right]}=\dfrac{2}{2a-1}\cdot\sqrt{5a^2\left(2a-1\right)^2}=\dfrac{2}{2a-1}\cdot a\cdot\left(2a-1\right)\cdot\sqrt{5}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)

Đính chính lại bài của bạn trước:

a) \(...=\frac{2}{\left(x+y\right)\left(x-y\right)}.\frac{\left|x+y\right|.\sqrt{3}}{\sqrt{2}}\)

ĐK: \(x\ne\pm y\)

Nếu \(x+y>0\) thì biểu thức được rút gọn thành \(\frac{\sqrt{6}}{x-y}\)

Nếu \(x+y< 0\) thì biểu thức được rút gọn thành \(-\frac{\sqrt{6}}{x-y}\)

b) \(...=\frac{2}{2a-1}.\left|a\right|.\left|2a-1\right|.\sqrt{5}\)

ĐK: \(a\ne\frac{1}{2}\)

Nếu \(a< 0\) hoặc \(a>\frac{1}{2}\) thì biểu thức được rút gọn thành \(2\sqrt{5}.a\)

Nếu \(0\le a< \frac{1}{2}\)thì biểu thức được rút gọn thành \(-2\sqrt{5}.a\)

Ta có: \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=\frac{2}{2a-1}.\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.a\sqrt{5}.\left(2a-1\right)=2a\sqrt{5}\)