c) \(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{3}{2};3\right\}\)

e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)

Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm

a) Ta có:

\(\frac{3}{x+2}=\frac{5}{2x+1}\)

\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)

\(\Rightarrow6x+3=5x+10\)

\(\Rightarrow6x-5x=10-3\)

\(\Rightarrow x=7\)

b)Ta có:

\(\frac{5}{8x-2}=\frac{-4}{7-x}\)

\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)

\(\Rightarrow35-5x=-32x+8\)

\(\Rightarrow-5x+32x=8-35\)

\(\Rightarrow27x=-27\)

\(\Rightarrow x=-1\)

c) Ta có:

\(\frac{4}{3}=\frac{2x-1}{x}\)

\(\Rightarrow4x=3\left(2x-1\right)\)

\(\Rightarrow4x=6x-3\)

\(\Rightarrow3=6x-4x=2x\)

\(\Rightarrow x=\frac{3}{2}\)

d)Ta có:

\(\frac{2x-1}{3}=\frac{3x+1}{4}\)

\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)

\(\Rightarrow8x-4=9x+3\)

\(\Rightarrow8x-9x=3+4\)

\(\Rightarrow-x=7\Rightarrow x=-7\)

e)Ta có:

\(\frac{4}{x+2}=\frac{7}{3x+1}\)

\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)

\(\Rightarrow12x+4=7x+14\)

\(\Rightarrow12x-7x=14-4\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

f)Ta có:

\(\frac{-3}{x+1}=\frac{4}{2-2x}\)

\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)

\(\Rightarrow-6+6x=4x+4\)

\(\Rightarrow6x-4x=4+6\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

\(a,\frac{3}{7}+\left|2x-\frac{1}{2}\right|=\frac{4}{5}\)

\(\Rightarrow\left|2x-\frac{1}{2}\right|=\frac{13}{35}\)

\(\Rightarrow2x-\frac{1}{2}=\pm\left(\frac{13}{35}\right)\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=\frac{13}{35}\\2x-\frac{1}{2}=\frac{-13}{35}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{61}{70}\\2x=\frac{9}{70}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{61}{140}\\x=\frac{9}{140}\end{cases}}\)

~Study well~

#KSJ

\(b,\frac{3}{4}-4\times\left|2x+1\right|=\frac{1}{2}\)

\(\Rightarrow4\times\left|2x+1\right|=\frac{1}{4}\)

\(\Rightarrow\left|2x+1\right|=\frac{1}{16}\)

\(\Rightarrow2x+1=\pm\left(\frac{1}{16}\right)\)

\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{16}\\2x+1=\frac{-1}{16}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{-15}{16}\\2x=\frac{-17}{16}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{32}\\x=\frac{-17}{32}\end{cases}}\)

~Study well~

#KSJ

g)=>x+1/2=0

x=0-1/2

x=-1/2

hoặc 2/3-2x=0

2x=2/3-0

2x=2/3

x=2/3:2

x=1/3

nhìn @_@ hoa cả mắt đăng từng bài thôi bạn

a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)

b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)

c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)

d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)

e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)

Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((

xin lỗi bạn nên hỏi từng câu cho dễ bạn à

Câu a \(\left|2x-\frac{1}{3}\right|+\frac{5}{6}=1\)

g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)

Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

\(2x+\frac{1}{4}=\frac{3}{2}\)

\(\Rightarrow2x=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)

\(\Rightarrow x=\frac{5}{4}:2=\frac{5}{8}\)

\(\left(x-5\right)-\frac{1}{3}=\frac{2}{5}\)

\(\Rightarrow x-5=\frac{2}{5}+\frac{1}{3}=\frac{11}{15}\)

\(\Rightarrow x=\frac{11}{15}+5=5\frac{11}{15}\)

a. \(\frac{3}{5}+x=\frac{5}{6}\)

=> \(x=\frac{5}{6}-\frac{3}{5}\)

=> \(x=\frac{7}{30}\)

b. \(\left(3\frac{1}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)

=> \(3\frac{1}{2}+2x=\frac{16}{3}:\frac{8}{3}=2\)

=> \(2x=2-3\frac{1}{2}=2-\frac{7}{2}=-\frac{3}{2}\)

=> \(x=-\frac{3}{2}:2\)

=> \(x=-\frac{3}{4}\)

Mấy câu trên dễ , bạn có thể tự làm được 

Chứng minh \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< 1\)

Đặt  \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)

...

\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1}-\frac{1}{10}\)

=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{9}{10}\)

Lại có : \(\frac{9}{10}< 1\)

=> \(A< \frac{9}{10}< 1\)

=> \(A< 1\left(đpcm\right)\)