NQ
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 6 2016
a) \(\left(\frac{3}{5}\right)^{15}:\left(\frac{9}{25}\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\left(\frac{3}{5}\right)^2\right)^5=\left(\frac{3}{5}\right)^{15}:\left(\frac{3}{5}\right)^{10}=\left(\frac{3}{5}\right)^5\)
b) \(5-\left(-\frac{5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{9}:3=4+\frac{1}{27}=4\frac{1}{27}\)
c) \(2^3+3.\left(\frac{1}{2}\right)^0+\left(-2\right)^2:\frac{1}{2}.8=8+3.1+4:\frac{1}{2}.8=8+3+64=75\)
d) \(\left(-1\right)^{-1}-\left(-\frac{3}{5}\right)^0+\left(\frac{1}{2}\right)^{2:2}=-1-1+\left(\frac{1}{2}\right)^1=-2+\frac{1}{2}=-\frac{3}{2}\)
1 tháng 6 2022
1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
28 tháng 1 2018
\(C=25\cdot\left(\frac{-1}{3}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(C=25\cdot\frac{-1}{27}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(C=\frac{-25}{27}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(C=\frac{-125}{135}+\frac{27}{135}-\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(C=\frac{-98}{135}-1\)
\(C=\frac{-98}{135}-\frac{135}{135}\)
\(C=\frac{-233}{135}\)
JK
23 tháng 7 2019
1, \(a,\left(x+1\right)^2=3\)
\(\Rightarrow x+1=\pm\sqrt{3}\)
\(\Rightarrow x=\pm\sqrt{3}-1\)
\(b,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^4-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x-1=\pm1\Rightarrow x=2or\text{ }x=0\end{cases}}\)
\(c,\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow x+\frac{1}{2}=\pm\sqrt{\frac{4}{25}}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}}\)
2, \(a,\sqrt{x}=4\)
\(\Rightarrow\sqrt{x}=\sqrt{16}\)
\(\Rightarrow x=16\)
\(b,\sqrt{x+1}=5\)
\(\Rightarrow\sqrt{x+1}=\sqrt{25}\)
\(\Rightarrow x+1=25\)
\(\Rightarrow x=24\)
\(\Rightarrow5^{\left(x+2\right)\left(x+3\right)}=1\)
\(\Rightarrow5^{\left(x+2\right)\left(x+3\right)}=5^0\)
\(\Rightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)
\(d,\left(2x-1\right)^{12}=\left(x+1\right)^{12}\)
\(\Rightarrow\left(2x-1\right)^{12}\div\left(x+1\right)^{12}=1\)
\(\Rightarrow\)