Q = (1 - \(\dfrac{\sqrt{a}-4a}{1-4a}\)_{) : \(\left[1-\dfrac{1+2a-2\sqrt{a}\left(2\sqrt{a}+1\right)}{1-4a}\right]\)}

     _{= \(\left(\dfrac{1-4a-\sqrt{a}+4a}{1-4a}\right):\left[\dfrac{1-4a-1-2a+4a+2\sqrt{a}}{1-4a}\right]\)}

    = \(\dfrac{1-\sqrt{a}}{1-4a}:\left(\dfrac{-2a+2\sqrt{a}}{1-4a}\right)\)

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

    = \(\dfrac{1}{2\sqrt{a}}\) = \(\dfrac{\sqrt{a}}{2a}\)

\(\sqrt{4a^2+12a+9}+\sqrt{4a^2-12a+9}\) với \(-\dfrac{3}{2}\le a\le\dfrac{3}{2}\)

\(\sqrt{\left(2a+3\right)^3}+\sqrt{\left(2a-3\right)^3}\)

\(\left|2a+3\right|+\left|2a-3\right|\)

\(2a+3-2a+3\)

\(6\)

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

\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)

....... Tới đây được chưa bạn? 

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

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

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

Lời giải:

$A=\frac{a^2+4a+4}{a^3+2a^2-4a-8}=\frac{(a+2)^2}{a^2(a+2)-4(a+2)}=\frac{(a+2)^2}{(a+2)(a^2-4)}=\frac{(a+2)^2}{(a+2)(a+2)(a-2)}=\frac{(a+2)^2}{(a+2)^2(a-2)}=\frac{1}{a-2}$