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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, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
Giải:
a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)
b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)
c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)
Đáp số: a) 2700; b) 810000; c) 64.
Chúc bạn học tốt!!!
a) \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
b) \(\left(0,125\right)^3.512=\left(0,512\right)^3.8^3=\left(0,512.8\right)^3=1^3=1\)
c) \(\left(0,25\right)^4.1024=\left[\left(0,25\right)^2\right]^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=2^2=4\)
d) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=3^3=64\)
e) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=3^4=81\)
g) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^3=8^3=512\)
a) $\dfrac{120^3}{40^3}=(\dfrac{120}{40})^3=3^3=27$
b) $\dfrac{390^4}{130^4}=(\dfrac{390}{130})^4=3^4=81$
c) $\dfrac{3^2}{(0,375)^2}=(3:0,375)^2=(3:\dfrac{3}{8})^2=8^2=64$
\(\frac{120^3}{40^3}\)= (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
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: \(x\in\left\{-\frac{7}{4};\frac{1}{4}\right\}\)
\(1,a,\left(-\frac{3}{4}\right)^0=1\)
\(b,\left(-2\frac{1}{3}\right)^4=\left[-\left(\frac{2\cdot3+1}{3}\right)\right]^4=\left(\frac{-7}{3}\right)^4=\frac{2401}{256}\)
\(c,\left(2,5\right)^3=15,625\)
\(d,25^3:5^2=5^6:5^2=5^4=625\)
\(e,2^2\cdot4^3=2^2\cdot2^6=2^8=256\)
\(f,\left(\frac{1}{5}\right)^5\cdot5^3=\left(\frac{1}{5}\right)^5:\frac{1}{5^3}=\left(\frac{1}{5}\right)^5:\left(\frac{1}{5}\right)^3=\left(\frac{1}{5}\right)^2=\frac{1}{25}\)
\(g,\left(\frac{1}{5}\right)^3\cdot10^3=\left(\frac{1}{5}\cdot10\right)^3=2^3=8\)
\(h,\left(-\frac{2}{3}\right)^4:2^4=\left(-\frac{2}{3}:2\right)^4=\left(-\frac{1}{3}\right)^4=\frac{1}{81}\)
\(i,\left(\frac{2}{3}\right)^4:9^2=\left(\frac{2}{3}\right)^4:3^4=\left(\frac{2}{3}:3\right)^4=\left(\frac{2}{9}\right)^4=\frac{16}{6561}\)
\(k,\left(\frac{1}{2}\right)^3.\left(\frac{1}{4}^2\right)=\left(\frac{1}{2}\right)^3.\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^7=\frac{1}{128}\)
\(m,\left(\frac{120}{40}\right)^3=3^3=27\)
\(n,\frac{390^4}{130^4}=\left(\frac{390}{130}\right)^4=3^4=81\)
\(2,a,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(b,\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=-8+4+1+1\)
\(=-2\)
\(c,\left(3^2\right)^2-\left(\left(-5\right)^2\right)^2+\left(\left(-2\right)^3\right)^2\)
\(=3^4-\left(-5\right)^4+\left(-2\right)^6\)
\(=81-625+64\)
\(=-480\)
\(3,a,\left(x-1\right)^3=27\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
\(b,x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\Rightarrow x=-1\end{cases}}\)
\(c,\left(2x+1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\Rightarrow x-1=\pm1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\text{ or }x=0\end{cases}}\)
\(4,M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(M=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(M=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(M=100+99+98+97+...+2+1\)
\(M=\left(100+1\right)\cdot100:2\)
\(M=101\cdot50=5050\)
B và C tượng trưng cho j zậy?
hai y khac nhau a b c y