\(\sqrt{13-4\sqrt{3}}=\sqrt{12+1-2\sqrt{12}}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1\)

\(\frac{\sqrt{4+\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{8+2\sqrt{7}}}{2}=\frac{\sqrt{7+1+2\sqrt{7}}}{2}=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{2}=\frac{\sqrt{7}+1}{2}\)

\(\frac{\sqrt{10+3\sqrt{11}}}{2\sqrt{2}}=\frac{\sqrt{20+2\sqrt{99}}}{2}=\frac{\sqrt{9+11+2\sqrt{99}}}{2}=\frac{\sqrt{\left(\sqrt{9}+\sqrt{11}\right)^2}}{2}=\frac{\sqrt{9}+\sqrt{11}}{2}\)

a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)

\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)

\(=\frac{-5}{9}.\frac{-1}{10}\)

\(=\frac{5}{90}\)

\(=\frac{1}{18}\)

b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)

\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)

\(=\frac{12}{15}\)

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

c, \(\frac{3}{8}.3\frac{1}{3}\)

\(=\frac{3}{8}.\frac{10}{3}\)

\(=\frac{10}{8}\)

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

d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)

\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)

\(=\frac{-3}{5}+\frac{-60}{10}\)

\(=\frac{-3}{5}+\frac{-30}{5}\)

\(=\frac{-33}{5}\)

e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)

\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)

\(=\frac{2}{5}.10\)

\(=4\)

f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)

\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)

\(=\frac{3}{7}.-14\)

\(=-6\)

~Study well~

#KSJ

|1/2x| = 3 - 2x

ĐKXĐ : 3 - 2x \(\ge\)0 => 2x \(\ge\) 3 => x \(\ge\)3/2

Ta có: |1/2x| = 3 - 2x

=> \(\orbr{\begin{cases}\frac{1}{2}x=3-2x\\\frac{1}{2}x=-3+2x\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{1}{2}x+2x=3\\\frac{1}{2}x-2x=-3\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{2}x=3\\-\frac{3}{2}x=-3\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{6}{5}\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

=> x = 2 

|5x| = x - 12

ĐKXĐ : x - 12 \(\ge\)0 => x \(\ge\)12

Ta có: |5x| = x - 12

=> \(\orbr{\begin{cases}5x=x-12\\5x=-x+12\end{cases}}\)

=> \(\orbr{\begin{cases}5x-x=-12\\5x+x=12\end{cases}}\)

=> \(\orbr{\begin{cases}4x=-12\\6x=12\end{cases}}\)

=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)(ktm)

=> pt vô nghiệm

|2x - 5| = x + 1

ĐKXĐ: x + 1 \(\ge\)0 => x \(\ge\)-1

Ta có: |2x - 5| = x + 1

=> \(\orbr{\begin{cases}2x-5=x+1\\2x-5=-x-1\end{cases}}\)

=> \(\orbr{\begin{cases}2x-x=1+5\\2x+x=-1+5\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\3x=4\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\x=\frac{4}{3}\end{cases}}\)(tm)

Vậy ...

|7 - 2x| + 7 = 2x

=> |7 - 2x| = 2x - 7

ĐKXĐ: 2x - 7 \(\ge\)0 => 2x \(\ge\) 7 => x \(\ge\) 7/2

Ta có: |7 - 2x| = 2x - 7

=> \(\orbr{\begin{cases}7-2x=2x-7\\7-2x=7-2x\end{cases}}\)

=> 7 + 7 = 2x + 2x

hoặc x tùy ý (TMĐK)

=> 4x = 14 => x = 7/2

hoặc x tùy ý (Tm ĐK)

Vậy ...

1 you 2 been 3 weather 4 fresh 5 people

1) you

2) go

3) weather

4) good

5) people

MÌNH CHƯA CHẮC CÂU 2 LẮM ĐÂU NHƯNG MẤY CÂU CÒN LẠI THÌ LÀM ĐC :>>>>>