\(a,\frac{\sqrt{7x^2y^4}}{\sqrt{28x^4y^4}}\)

\(\frac{\sqrt{7}xy^2}{2\sqrt{7}x^2y^2}=\frac{1}{2x}\)

\(b,\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)

\(\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1+2\sqrt{2x-1}+1}\)

\(\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}+1\right)^2}\)

\(\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}+1\right|\)

\(\sqrt{2x-1}+1+\sqrt{2x-1}+1\)

\(2\sqrt{2x-1}+2\)

\(c,\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)

\(\sqrt{x-3}+\sqrt{2}\sqrt{x-3}-2\sqrt{x-3}=2-\sqrt{2}\)

\(\sqrt{x-3}\left(1+\sqrt{2}-2\right)=2-\sqrt{2}\)

\(\sqrt{x-3}\left(\sqrt{2}-1\right)=\sqrt{2}\left(\sqrt{2}-1\right)\)

\(\sqrt{x-3}=\sqrt{2}\)

\(x-3=2< =>x=5\)

a) \(\frac{\sqrt{7\left(-x^2\right)y^4}}{\sqrt{28x^4y^4}}=\frac{\sqrt{7}xy^2}{2\sqrt{7}x^2y^2}=\frac{1}{2x}\)(vì  x > 0)

b) \(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x+2\sqrt{2x-1}}\)

\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1+2\sqrt{2x-1}}\)

\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x+1}+1\right)^2}=\sqrt{2x-1}+1+\sqrt{2x+1}+1\)

\(=2\sqrt{2x-1}+2\)

c) ĐK: x \(\ge\)3

Ta có:: \(\frac{1}{3}\sqrt{9x-27}+\sqrt{2x-6}-\sqrt{4x-12}=2-\sqrt{2}\)

<=> \(\sqrt{x-3}+\sqrt{2}.\sqrt{x-3}-2\sqrt{x-3}=2-\sqrt{2}\)

<=> \(\sqrt{x-3}.\left(\sqrt{2}-1\right)=\sqrt{2}\left(\sqrt{2}-1\right)\)

<=> \(\sqrt{x-3}=\sqrt{2}\) <=> x - 3 = 2 <=> x = 5 (tm)

ĐK:  x lớn hơn howcj bằng -9/2 và bé hơn hoặc bằng 5/4

<=> -9/2<x<5/4

=>2x+9=5-4x

=>x=-2/3 (TM)

b)

<=> can(2/75)*can(1/2)*can(121/32)

<=>can(2*1*121)/can(2*75*32)=can(121/2400)

c) cau nay sai de bai roi, em check lai ngoac thu 2 xem

Đề bài của bạn thiếu dữ liện , không làm đc nha

ok thanks

\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}+\sqrt{1,6}+3\sqrt{0,4}\right)\)

\(=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}+\frac{2\sqrt{10}}{5}+\frac{3\sqrt{10}}{5}\right)\)

\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}+\frac{5\sqrt{10}}{5}\right)=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}+\sqrt{10}\right)=10-2=8\)

\(25t^2-20t=9-12\)

\(\Leftrightarrow25t^2-20t=-3\)

\(\Leftrightarrow25t^2-20t+3=0\)

\(\Leftrightarrow25t^2-5t-15t+3=0\)

\(\Leftrightarrow5t\left(5t-1\right)-3\left(5t-1\right)=0\)

\(\Leftrightarrow\left(5t-1\right)\left(5t-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5t-1=0\\5t-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=\frac{1}{5}\\t=\frac{3}{5}\end{cases}}}\)

Vì \(t\ge\frac{3}{5}\) nên \(t=\frac{3}{5}\) thoả mãn đề bài.