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

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

\(x=\left(-\frac{2}{5}\right):\frac{2}{3}=-\frac{3}{5}\)

a/ \(2^x+2^{x+1}=96\)

\(2^x+2^x.2=96\)

\(2^x\cdot\left(2+1\right)=96\)

\(2^x=\frac{96}{3}=32\)

\(2^x=2^5\)

\(=>x=5\)

b/ \(3^{4x+4}=81^{x+3}\)

\(\Rightarrow3^{4x+4}-81^{x+3}=0\)

\(3^{4x}.3^4-3^{4x}\cdot81^3=0\)

\(3^{4x}\cdot\left(81-81^3\right)=0\)

\(3^{4x}=\frac{0}{81-81^3}\)

\(3^{4x}=0\Rightarrow x=0\)

Hình đâu?

\(x^4-7x^2+1\)

\(=x^2\left(x^2-7\right)+1\)

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\)

\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{10}}\)

\(2A=1-\frac{1}{3^{10}}\)

\(A=\frac{1-\frac{1}{3^{10}}}{2}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{10}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^9}\)

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

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{10}}\)

\(2A=1-\frac{1}{3^{10}}\)

\(\Rightarrow A=\frac{1-\frac{1}{3^{10}}}{2}\)

Dịch: Tổng của hai số liên tiếp là 149. Số lớn là?

Đáp án: 75

Cha chả!!

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