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a) \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)
b) \(\frac{3^2}{0,375^2}=\left(\frac{3}{0,375}\right)^2=8^2=64\)
HỌC TỐT
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
Tính :
a) \(\frac{8^{14}}{4^{12}}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}=\frac{2^{42}}{2^{24}}=2^{18}=262144.\)
b) \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27.\)
Tìm x:
b) \(x^2-0,25=0\)
\(\Rightarrow x^2=0+0,25\)
\(\Rightarrow x^2=0,25\)
\(\Rightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy \(x\in\left\{0,5;-0,5\right\}.\)
c) \(\frac{8}{2^x}=2\)
\(\Rightarrow2^x=8:2\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
a, 2 mũ 17 phần 2 mũ 14
b,=30
mình chỉ làm được 2 câu thôi,chúc cậu học tốt!
\(\frac{120^3}{40^3}\)= (120:40)3 = 33 = 27
\(\frac{390^4}{130^4}\)=(390 : 130)4=34 = 81
\(\frac{3^2}{\left(0,375\right)^2}\)= (3:0,375)2 = 82 =64
\(\left[\left(-\frac{4}{5}\right).\left(\frac{-5}{4}\right)\right]^3=1^3=1\)
\(\frac{3}{5}+\frac{3.\left(-4\right)}{4\cdot5}=\frac{3}{5}+\frac{-3}{5}=0\)
\(\frac{5}{9}-\frac{1}{6}-\frac{4}{9}=\frac{5}{9}-\frac{4}{9}-\frac{1}{6}=\frac{1}{9}-\frac{1}{6}=-\frac{1}{18}\)
a) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{47}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)
b) \(\frac{390}{130}^4=3^4=81\)
c) \(\left(0,125\right)^3.512=\frac{1}{512}.512=1\)
a) Ta có: \(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)=\frac{2}{3}\)
⇒\(\frac{2}{3}:\left(x-1\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
⇒\(x-1=\frac{2}{3}:\frac{1}{6}=\frac{2}{3}\cdot6=4\)
hay x=5
Vậy: x=5
b) \(5,4-3\left[x-120\%\right]=\frac{3}{10}\)
⇔\(\frac{27}{5}-3\cdot\left(x-\frac{6}{5}\right)=\frac{3}{10}\)
⇔\(3\left(x-\frac{6}{5}\right)=\frac{27}{5}-\frac{3}{10}=\frac{51}{10}\)
hay \(x-\frac{6}{5}=\frac{51}{10}\cdot\frac{1}{3}=\frac{17}{10}\)
⇔\(x=\frac{17}{10}+\frac{6}{5}=\frac{29}{10}\)
Vậy: \(x=\frac{29}{10}\)
c) \(10\cdot3^{x+2}-3^x=89\)
\(\Leftrightarrow10\cdot3^2\cdot3^x-3^x=89\)
\(\Leftrightarrow3^x\left(90-1\right)=89\)
\(\Leftrightarrow3^x=1\)
hay x=0
Vậy: x=0
d) \(5\cdot\left(x-0,2\right)=3x+\left(\frac{-2}{3}\right)^3\)
⇒\(5\cdot\left(x-\frac{1}{5}\right)=3x+\frac{-8}{27}\)
\(\Leftrightarrow5x-1-3x-\frac{-8}{27}=0\)
\(\Leftrightarrow2x-\frac{19}{27}=0\)
\(\Leftrightarrow2x=\frac{19}{27}\)
hay \(x=\frac{\frac{19}{27}}{2}=\frac{19}{27}\cdot\frac{1}{2}=\frac{19}{54}\)
Vậy: \(x=\frac{19}{54}\)
e) \(\left(2x+\frac{3}{4}\right)^2-1,5=2\frac{1}{2}\)
\(\Leftrightarrow\left(2x+\frac{3}{4}\right)^2=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{3}{2}=-2\\2x+\frac{3}{2}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2-\frac{3}{2}\\2x=2-\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{7}{2}\\2x=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{2}\cdot\frac{1}{2}\\x=\frac{1}{2}\cdot\frac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
\(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)