e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)

\(\Rightarrow x=2\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)

\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)

\(\Leftrightarrow\left(\frac{1}{2}.\frac{44}{90}\right)x=\frac{23}{45}\)

\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{48}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{48}{45}\)

\(\Rightarrow\frac{22}{45}x=\frac{48}{45}\)

\(\Rightarrow x=\frac{24}{11}\)

Vậy...

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{22}{45}x=\frac{44}{45}\)

\(\Rightarrow x=\frac{44}{45}:\frac{22}{45}\)

\(\Rightarrow x=2\)

Vậy x = 2

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\frac{22}{45}:2=\frac{22}{45}:x\)

\(\Rightarrow x=2\)

a) \(\left|2x-1\right|=5\)

\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

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

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

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

b) \(\left(5^x-1\right)3-2=70\)

\(\Rightarrow5^x.3-3=72\)

\(\Rightarrow5^x.3=75\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

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

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

............. Làm tiếp nhé!

d) \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

Có \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)+x=\frac{23}{45}\)

Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)

Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)

\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)

Thay vào phép tính trên ta được:

\(\frac{11}{45}\cdot x=\frac{23}{45}\\ x=\frac{23}{45}:\frac{11}{45}\\ x=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\right)\right]x=\frac{23}{45}\)

=>\(\left[\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)

=> \(\left[\frac{1}{2}\cdot\frac{22}{45}\right]x=\frac{23}{45}\)

=> \(\frac{11}{45}x=\frac{23}{45}\)

=> \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{45}\cdot\frac{45}{11}=\frac{23}{11}\)

Vậy x = 23/11

Ez :))

\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

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

\(\Rightarrow x=2\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{44}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{44}{45}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\)

\(\Rightarrow\frac{22}{45}.x=\frac{44}{45}\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

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