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

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

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

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

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

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

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

Viết rõ đề bài ra đc không ạ

đấy là phân số

Bài làm:

a) đkxđ: \(2a\ne\pm b\)

Ta có: \(M=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right)\div\left(\frac{1+4a^2+4b^2}{4a^2-b^2}\right)\)

\(M=\left[\frac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right].\left(\frac{\left(2a-b\right)\left(2a+b\right)}{4a^2+4b^2+1}\right)\)

\(M=\frac{4a+2b-6b-8a+4b}{4a^2+4b^2+1}\)

\(M=\frac{-4a}{4a^2+4b^2+1}\)

b) +Nếu: \(a=\frac{1}{3}\)và \(b=2\)

Khi đó GT của M là: \(M=\frac{-4.\frac{1}{3}}{4.\frac{1}{3^2}+4.2^2+1}=-\frac{12}{157}\)

Viết rõ đề ra nhá

\(P=\dfrac{9\sqrt{a}-\sqrt{25a}+\sqrt{4a^3}}{a^2+2a}=\dfrac{9\sqrt{a}-5\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{4\sqrt{a}+2a\sqrt{a}}{a\left(a+2\right)}=\dfrac{2\sqrt{a}\left(2+a\right)}{a\left(2+a\right)}=\dfrac{2\sqrt{a}}{a}=\dfrac{2.\sqrt{a}}{\sqrt{a}.\sqrt{a}}=\dfrac{2}{\sqrt{a}}\)