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a: =>\(4\cdot3^x\cdot\dfrac{1}{3}+2\cdot3^x\cdot9=4\cdot3^6+2\cdot3^9\)
=>3^x(4*1/3+2*9)=3^6(4+2*3^3)
=>3^x*58/3=3^6*58
=>3^x/3^6=3
=>x-6=1
=>x=7
b: =>\(2^x\cdot\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)
=>2^x=2^7
=>x=7
\(\left(3x+1\right)^2=25\)
\(\Rightarrow\left(3x+1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}\Rightarrow\orbr{\begin{cases}3x=5-1=4\\3x=-5-1=-6\end{cases}}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)
\(\left[x-\frac{1}{2}\right]+\frac{1}{2}=\frac{5}{8}\)
\(\Rightarrow x-0=\frac{5}{8}\)
\(x=\frac{5}{8}\)
\(\left[x+\frac{3}{4}\right]-\frac{1}{3}=0\)
\(x+\frac{3}{4}=0+\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{-5}{12}\)
Ta có : \(P\left(0\right)=a_0=2^{10}\)
\(P\left(1\right)=a_0+a_1+a_2+...+a_{30}=\left(2+1+3\right)^{10}=6^{10}\)
Suy ra : \(S=a_1+a_2+...+a_{30}=P\left(1\right)-P\left(0\right)=6^{10}-2^{10}\)
\(\frac{4}{\frac{2}{5}}:\left(-\frac{33}{10}\right)+x=-\frac{1}{\frac{5}{6}}\)
\(10:\left(-\frac{33}{10}\right)+x=-\frac{6}{5}\)
\(-\frac{100}{33}+x=-\frac{6}{5}\)
\(x=\frac{302}{165}\)