a: Ta có: \(4\sqrt{3a}-3\sqrt{12a}+\dfrac{6\sqrt{a}}{3}-2\sqrt{20a}\)

\(=4\sqrt{3a}-6\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)

\(=-2\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)

a) \(6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9x-9}+\dfrac{7}{2}\sqrt{4x-4}=24\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9\left(x-1\right)}+\dfrac{7}{2}\sqrt{4\left(x-1\right)}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot3\sqrt{x-1}+\dfrac{7}{2}\cdot2\sqrt{x-1}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\sqrt{x-1}+7\sqrt{x-1}=24\)

\(\Leftrightarrow12\sqrt{x-1}=24\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{24}{12}\)

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

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=4+1\)

\(\Leftrightarrow x=5\left(tm\right)\)

b) \(\dfrac{1}{2}\sqrt{4x+8}-2\sqrt{x+2}-\dfrac{3}{7}\sqrt{49x+98}=-8\) (ĐK: \(x\ge-2\))

\(\Leftrightarrow\dfrac{1}{2}\cdot2\sqrt{x+2}-2\sqrt{x+2}-\dfrac{3}{7}\cdot7\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}-3\sqrt{x+2}=-8\)

\(\Leftrightarrow-4\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{-8}{-4}\)

\(\Leftrightarrow\sqrt{x+2}=2\)

\(\Leftrightarrow x+2=4\)

\(\Leftrightarrow x=4-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

\(\frac{1}{3}-\sqrt{7}-\frac{1}{3}+\sqrt{7}=0\)

cvfbhm,

Xin lỗi em ko biết làm , em vẫn chưa lên lớp 9

1)\(\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

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

\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)

Bài 2 : 

a) \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+2\sqrt{7}+1}-\sqrt{7}\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\left|\sqrt{7}+1\right|-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)

b) \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{4+4\sqrt{3}+3}-2\sqrt{3}\)

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

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

c) \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)

\(=\sqrt{13-2\sqrt{13}+1}+\sqrt{13+2\sqrt{13}+1}\)

\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)

\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)

\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)

d) \(D=\sqrt{22-2\sqrt{21}}+\sqrt{22+2\sqrt{21}}\)

\(=\sqrt{21-2\sqrt{21}+1}+\sqrt{21+2\sqrt{21}+1}\)

\(=\sqrt{\left(\sqrt{21}-1\right)^2}+\sqrt{\left(\sqrt{21}+1\right)^2}\)

\(=\left|\sqrt{21}-1\right|+\left|\sqrt{21}+1\right|\)

\(=\sqrt{21}-1+\sqrt{21}+1=2\sqrt{21}\)

bạn j ơi bạn giải đúng k vậy

giúp mình bài  1 ik

a)\(\frac{3}{4}+\frac{1}{5}.\left(-\frac{1}{2}\right)^2=\frac{4}{5}\)

b)\(\left|-\frac{2}{3}\right|.\sqrt{144}+\frac{7}{6}:3=\frac{151}{18}\)

Mình viết cho gọn thôi nhé bạn!

Nhớ tick cho mình nha!

a, \(\frac{3}{4}+\frac{1}{5}.\left(-\frac{1}{2}\right)^2\)

=\(\frac{3}{4}+\frac{1}{5}.\frac{1}{4}\)

=\(\frac{15}{20}+\frac{1}{20}\)

\(=\frac{16}{20}=\frac{4}{5}\)

b, \(\left|-\frac{2}{3}\right|.\sqrt{144}+\frac{7}{6}:3\)

=\(\frac{2}{3}.12+\frac{7}{6}.\frac{1}{3}\)

=\(8+\frac{7}{18}=8\frac{7}{18}\)