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a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)
b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)
c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)
2)
\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)
\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)
\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)
1) \(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{\frac{2x}{3}-3}{10}=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{\frac{2x}{3}}{10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{3\times10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{30}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}+\frac{3}{10}=\frac{2}{5}\)
\(\Leftrightarrow\frac{3}{10}-\frac{x}{15}=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{2}{5}-\frac{3}{10}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{1}{10}\)
\(\Leftrightarrow-x=\frac{15}{10}\)
\(\Leftrightarrow-x=\frac{3}{2}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}\)
2) \(\left|2x-1\right|+1=4\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{2;-1\right\}\)
#)Giải :
\(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-3-x+\frac{1}{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\x=-\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}}\)
a) \(2x-3=x+\frac{1}{2}\)
\(\Leftrightarrow2x-x=\frac{1}{2}+3\)
\(\Leftrightarrow x=\frac{7}{2}\)
Vậy...
b) \(4x-\left(2x+1\right)=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-1=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-x=3-\frac{1}{3}+1\)
\(\Leftrightarrow x=\frac{11}{3}\)
Vậy ...
c) \(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=7-\frac{1}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\left(1-\frac{1}{50}\right)=\frac{349}{50}+x\)
\(\Leftrightarrow2x-\frac{49}{50}=\frac{349}{50}+x\)
\(\Leftrightarrow2x-x=\frac{349}{50}+\frac{49}{50}\)
\(\Leftrightarrow x=\frac{199}{25}\)
Vậy ...
bạn ơi phần a là gì ạ + hay - 1 thế
b) Đặt \(B=7+|2x+3|\)
Vì\(|2x+3|\ge0;\forall x\)
\(\Rightarrow7+|2x+3|\ge7+0;\forall x\)
Hay\(B\ge7;\forall x\)
Dấu "=" xảy ra\(\Leftrightarrow|2x+3|=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
Vậy MIN B=7\(\Leftrightarrow x=-\frac{3}{2}\)