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theo công thức đã cho => 1^3+2^3+..+10^3=(1+2+3+..+10)^2
=55^2 = (x+1)^2
=> x= 55-1=54
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1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\
\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\
\left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\
\frac{98}{99}-x=\frac{7}{2}\\
x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)
2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)
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\(A=3+3^2+....+3^{99}\)
\(3A=3^2+3^3+...+3^{100}\)
\(3A-A=3^2+3^3+...+3^{100}-3-3^2-...-3^{99}\)
\(2A=3^{100}-3\)
\(A=\dfrac{3^{100}-3}{2}\)
\(\Rightarrow2A+3=9^{2x+6}\)
\(\Rightarrow2\cdot\dfrac{3^{100}-3}{2}+3=\left(3^2\right)^{2x+6}\)
\(\Rightarrow3^{100}-3+3=3^{2\left(2x+6\right)}\)
\(\Rightarrow3^{100}=3^{4x+12}\)
\(\Rightarrow4x+12=100\)
\(\Rightarrow4x=88\)
\(\Rightarrow x=22\)
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