9) \(\dfrac{x}{4}=\dfrac{9}{x}\)

Theo định nghĩa về hai phân số bằng nhau, ta có:

\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

8)

\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)

7)

\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)

6)

\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)

5)

\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)

4)

\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

3)

\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)

2)

\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)

1)

\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)

a, Ta có:

\(2.x-\frac{1}{2}=\frac{5}{4}-\frac{1}{2}.x\)

\(\Leftrightarrow2.x+\frac{1}{2}.x=\frac{5}{4}+\frac{1}{2}\)

\(\Leftrightarrow\frac{5}{2}.x=\frac{7}{4}\)

\(\Leftrightarrow x=\frac{7}{10}\). Vậy: \(x=\frac{7}{10}\)

b, Ta có:

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

\(\Leftrightarrow2-\frac{3}{4}=3.x-\frac{5}{2}.x\)

\(\Leftrightarrow\frac{5}{4}=\frac{1}{2}.x\)

\(\Leftrightarrow x=\frac{5}{2}\). Vậy: \(x=\frac{5}{2}\)

c, Ta có:

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

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

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

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

\(\Leftrightarrow-\frac{1}{2}.x=\frac{5}{2}\)

\(\Leftrightarrow x=-5\). Vậy: \(x=-5\)

d, Ta có:

\(\frac{4}{3}.\left(x-\frac{3}{2}\right)-3.\left(\frac{1}{3}.x+1\right)=6\)

\(\Leftrightarrow\frac{4}{3}.x-\frac{4}{3}.\frac{3}{2}-3.\frac{1}{3}.x-3=6\)

\(\Leftrightarrow\frac{4}{3}.x-2-x-3=6\)

\(\Leftrightarrow\frac{1}{3}.x-5=6\)

\(\Leftrightarrow\frac{1}{3}.x=11\)

\(\Leftrightarrow x=33\). Vậy: \(x=33\)

Chúc bạn học tốt!Tick cho mình nhé!

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

Bước cưối 58/21 minh man viết nhầm nên sai 

a) \(x=\frac{7}{20}\)

b) \(x=\frac{7}{12}\)

c)\(x=\frac{8}{15}\)

a ) \(\frac{7}{8}:x=3-\frac{1}{2}\)

     \(\frac{7}{8}:x=\frac{5}{2}\)

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

              \(x=0,35\)

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

            \(x+\frac{1}{6}=\frac{3}{4}\)

                        \(x=\frac{3}{4}-\frac{1}{6}\)

                        \(x=\frac{7}{12}\)

c ) \(\frac{3}{2}.\frac{4}{5}-x=\frac{2}{3}\)

          \(\frac{7}{10}-x=\frac{2}{3}\)

                        \(x=\frac{7}{10}-\frac{2}{3}\)

                        \(x=\frac{1}{30}\)

d ) \(x.3\frac{1}{3}=3\frac{1}{3}:4\frac{1}{4}\)

      \(x:\frac{10}{3}=\frac{10}{3}:\frac{17}{4}\)

      \(x:\frac{10}{3}=\frac{40}{51}\)

                \(x=\frac{40}{51}:\frac{10}{3}\)

                \(x=\frac{4}{17}\)

e ) \(5\frac{2}{3}:x=3\frac{2}{3}-2\frac{1}{2}\)

      \(\frac{17}{3}:x=\frac{11}{3}-\frac{5}{2}\)

      \(\frac{17}{3}:x=\frac{7}{6}\)

                \(x=\frac{17}{3}:\frac{7}{6}\)

                \(x=\frac{34}{7}\)

Nếu mình đúng thì các bạn k mình nhé

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

bạn tự làm đi tính toán thôi mà

1) Ta có: \(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|-\frac{3}{2}=\frac{1}{4}\)

⇔\(2\cdot\left|\frac{1}{2}x-\frac{3}{8}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

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

⇔\(\left[{}\begin{matrix}\frac{1}{2}x-\frac{3}{8}=\frac{7}{8}\\\frac{1}{2}x-\frac{3}{8}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{10}{8}\\\frac{1}{2}x=\frac{-4}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{10}{8}:\frac{1}{2}=\frac{10}{8}\cdot2=\frac{20}{8}=\frac{5}{2}\\x=\frac{-4}{8}:\frac{1}{2}=-\frac{4}{8}\cdot2=-\frac{8}{8}=-1\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{5}{2};-1\right\}\)

2) Ta có: \(-5\cdot\left(x+\frac{1}{5}\right)-\frac{1}{2}\cdot\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)

⇔\(-5x-1-\frac{1}{2}x+\frac{1}{3}-\frac{3}{2}x+\frac{5}{6}=0\)

\(\Leftrightarrow-7x+\frac{1}{6}=0\)

\(\Leftrightarrow-7x=-\frac{1}{6}\)

hay \(x=\frac{1}{42}\)

Vậy: \(x=\frac{1}{42}\)

3) Ta có: \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)

\(\Leftrightarrow3x-\frac{3}{2}-5x-3+x-\frac{1}{5}=0\)

\(\Leftrightarrow-x-\frac{47}{10}=0\)

⇔\(-x=\frac{47}{10}\)

hay \(x=\frac{-47}{10}\)

Vậy: \(x=\frac{-47}{10}\)

4) Ta có: \(\frac{3}{4}-2\left|2x-0,125\right|=2\)

\(\Leftrightarrow2\left|2x-\frac{1}{8}\right|=\frac{3}{4}-2=-\frac{5}{4}\)

⇔\(\left|2x-\frac{1}{8}\right|=-\frac{5}{8}\)(vô lý)

Vậy: x∈∅

5) Ta có: \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\\\frac{1}{2}x=-\frac{7}{8}+\frac{1}{3}=-\frac{13}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{24}:\frac{1}{2}=\frac{29}{24}\cdot2=\frac{29}{12}\\x=-\frac{13}{24}:\frac{1}{2}=-\frac{13}{24}\cdot2=-\frac{13}{12}\end{matrix}\right.\)

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

Bài mk sai r nhé!!

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

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

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

...

\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)

Cộng vế theo vế 

\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)

\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)

Lại có \(\frac{7}{8}< 1\)

Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)

\(\Rightarrow B< 1\left(đpcm\right)\)