\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{9}-\frac{1}{10}\)

\(\frac{2}{5}x+\frac{3}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)

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

Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)

         = \(1-\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

        = 1 - \(\frac{1}{10}\)

        =  \(\frac{9}{10}\)

X = \(\frac{3}{2}\) là đúng đó bạn

\(\Leftrightarrow10\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}\right)=9\)

\(\Leftrightarrow\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=9\div10\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{10}\)

\(\Rightarrow x+1=10\)

\(\Leftrightarrow x=9\)

Vậy x = 9 

1)

\(2\frac{1}{4}x-9\frac{1}{4}=-7\frac{1}{4}\)

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

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

\(x=2:2\frac{1}{4}\)

\(x=\frac{8}{9}\)

Vậy \(x=\frac{8}{9}\)

( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0

= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0

= ( 1 - 1/6 ) x 10 - x = 0

= 5/6 x 10 - x =0

=   25/3 - x =0

               x = 25/3 - 0

                x  = 25/3

\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)

\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)

\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)

\(\frac{5}{6}\times10-x=0\)

\(\frac{25}{3}-x=0\)

x              =\(\frac{25}{3}-0=\frac{25}{3}\)

Ta co:

 \(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:\left[\left(\frac{95}{9}-\frac{29}{4}\right).\frac{36}{17}\right]=\frac{3}{4}\)

\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:\left(\frac{119}{36}.\frac{36}{17}\right)=\frac{3}{4}\)

\(\left[\left(\frac{29}{25}-x\right).\frac{21}{4}\right]:7=\frac{3}{4}\)

\(\left(\frac{29}{25}-x\right).\frac{21}{4}=\frac{3}{4}.7=\frac{21}{4}\)

\(\frac{29}{25}-x=\frac{21}{4}:\frac{21}{4}=1\)

\(x=\frac{29}{25}-1=\frac{4}{25}\)

\(a)\) \(-\left(x+84\right)+213=-16\)

\(\Leftrightarrow\)\(-x-84+213=-16\)

\(\Leftrightarrow\)\(x=213-84+16\)

\(\Leftrightarrow\)\(x=145\)

Vậy \(x=145\)

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

\(\Leftrightarrow\)\(\left(x-1\right)^2=\left|-1\right|\)

\(\Leftrightarrow\)\(\left(x-1\right)^2=1\)

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

Vậy \(x=0\) hoặc \(x=2\)

Chúc bạn học tốt ~ 

a) \(-\left(x+84\right)+213=-16\)

                   \(-\left(x+84\right)=-16-213\)

                   \(-\left(x+84\right)=-229\)

\(\Rightarrow x+84=229\)

\(\Rightarrow x=229-84=145\)

Vậy \(x=145\)

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

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

    \(\left(x-1\right)^2=\left|-1\right|\)

    \(\left(x-1\right)^2=1\)

\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

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

Vậy \(x\in\left\{0;2\right\}\)

a) \(-\left(x+84\right)+213=-16\)

\(-x-(84-213)=-16\)

\(-x+129=-16\)

\(-x=-16-129\)

\(-x=-145\)

\(x=145\)

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

\(\left(x-1\right)^2=\left|-1\right|\)

\(\left(x-1\right)^2=1\)

\(\Rightarrow\left(x-1\right)^2=1^2\)

\(\Rightarrow x-1=1\Rightarrow x=2\)

CHÚC BN HỌC TỐT!!

a) \(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^2.2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

                               \(=\frac{\left(3.2^{18}\right)^2}{11.2^{13}.2^{22}-2^{36}}\)

                               \(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)

                                \(=\frac{9.2^{36}}{2^{35}.\left(11-2\right)}\)

                                \(=\frac{9.2^{36}}{2^{35}.9}=2\)

b) \(\frac{3}{2}.x-\left(\frac{4}{5}-2.x\right)=1\frac{3}{10}:\frac{3}{2}\)

\(\frac{3}{2}.x-\frac{4}{5}+2.x=\frac{13}{10}:\frac{3}{2}\)

\(\left(\frac{3}{2}.x+2.x\right)-\frac{4}{5}=\frac{13}{10}.\frac{2}{3}\)

\(x.\left(\frac{3}{2}+2\right)-\frac{4}{5}=\frac{13}{15}\)

\(x.\frac{7}{2}=\frac{13}{15}+\frac{4}{5}\)

\(x.\frac{5}{2}=\frac{13}{15}+\frac{12}{15}\)

\(x.\frac{7}{2}=\frac{25}{15}=\frac{5}{3}\)

\(x=\frac{5}{3}:\frac{7}{2}\)

\(x=\frac{5}{3}.\frac{2}{7}=\frac{10}{21}\)