3 nhân[ 30 -2 nhân (x -1 )+5= 11
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d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)
\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)
\(=\sqrt{2}\)

d: Ta có: \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\)
\(=\dfrac{\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11}+1-\sqrt{11}+1}{\sqrt{2}}\)
\(=\sqrt{2}\)

\(\frac{10}{3}.x+\frac{67}{4}=\frac{53}{4}\)
\(\frac{10}{3}.x=\frac{53}{4}-\frac{67}{4}\)
\(\frac{10}{3}.x=-\frac{7}{2}\)
\(x=-\frac{7}{2}:\frac{10}{3}\)
\(x=-\frac{21}{20}\)

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)

\(25\frac{2}{13}-\left(\frac{15}{17}+15\frac{2}{3}\right)=\frac{327}{13}-\left(\frac{15}{17}+\frac{47}{3}\right)\)
\(=\frac{327}{13}-\frac{844}{51}\)
\(=\frac{5705}{663}\)
\(\frac{5}{30}+\frac{15}{90}+\frac{250}{150}+\frac{350}{210}+\frac{45}{270}=\frac{1}{6}+\frac{1}{6}+\frac{5}{3}+\frac{5}{3}+\frac{1}{6}\)
\(=\frac{3}{6}+\frac{10}{3}\)
\(=\frac{1}{2}+\frac{10}{3}\)
\(=\frac{23}{6}\)
\(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{17}.2\frac{4}{9}=\frac{34}{11}.\frac{27}{46}.\frac{23}{17}.\frac{22}{9}\)
\(=\frac{68}{9}.\frac{27}{34}\)
\(=6\)

1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
3 x [ 30 - 2 x ( x - 1 ) ] + 5 = 11
3 x [ 30 - 2 x ( x - 1 ) ] = 11 - 5
3 x [ 30 - 2 x ( x - 1 ) ] = 6
30 - 2 x ( x - 1 ) = 6 : 3
30 - 2 x ( x - 1 ) = 2
2 x ( x - 1 ) = 30 - 2
2 x ( x - 1 ) = 28
x - 1 = 28 : 2
x - 1 = 14
x = 14 + 1
x = 15
thiếu một cái ngoặc vuông nữa