Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)
\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)
\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)
\(a;3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)
\(=3^2\cdot\frac{1}{3^5}\cdot3^4\cdot\frac{1}{3^3}\)
\(=\left(3^2\cdot3^4\right)\cdot\left(\frac{1}{3^5}\cdot\frac{1}{3^3}\right)\)
\(=3^6\cdot\frac{1}{3^8}\)
\(=\frac{3^6}{3^8}\)
\(=\frac{1}{3^2}=\frac{1}{9}\)
\(3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
= \(9.\frac{1}{243}.6561.\frac{1}{27}\)
= \(9\)
b ) \(\left(4,2\right)^5:\left(2^3.\frac{1}{16}\right)\)
= \(\left(\frac{21}{5}\right)^5:\left(8.\frac{1}{16}\right)\)
= \(130691232:\frac{1}{2}\)
= \(130691232\times2\)
= 261382464
Chúc bạn học tốt !!!
=
Câu một \(=3^2.\frac{1}{3^5}.\left(3^4\right)^2.\frac{1}{3^3}=3^{10}.\frac{1}{3^8}=3^2=9\)
Câu hai \(=\left(2^2.2^5\right):\left(2^3.\frac{1}{2^4}\right)=\frac{2^7}{\frac{2^3}{2^4}}=2^8=256\)
Chờ chút nhá :D
Câu 3 \(=9-64-25^2=-680\)
Câu 4 \(=8+1-1+1=9\)
Câu 5 \(=4,75-0,37+0,125-1,28-2,5+3\frac{1}{12}=0,725+3\frac{1}{12}=3\frac{97}{120}\)
Sai thì mình xin lỗi :v, vội quá