Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
\(1,a,\left(-\frac{3}{4}\right)^0=1\)
\(b,\left(-2\frac{1}{3}\right)^4=\left[-\left(\frac{2\cdot3+1}{3}\right)\right]^4=\left(\frac{-7}{3}\right)^4=\frac{2401}{256}\)
\(c,\left(2,5\right)^3=15,625\)
\(d,25^3:5^2=5^6:5^2=5^4=625\)
\(e,2^2\cdot4^3=2^2\cdot2^6=2^8=256\)
\(f,\left(\frac{1}{5}\right)^5\cdot5^3=\left(\frac{1}{5}\right)^5:\frac{1}{5^3}=\left(\frac{1}{5}\right)^5:\left(\frac{1}{5}\right)^3=\left(\frac{1}{5}\right)^2=\frac{1}{25}\)
\(g,\left(\frac{1}{5}\right)^3\cdot10^3=\left(\frac{1}{5}\cdot10\right)^3=2^3=8\)
\(h,\left(-\frac{2}{3}\right)^4:2^4=\left(-\frac{2}{3}:2\right)^4=\left(-\frac{1}{3}\right)^4=\frac{1}{81}\)
\(i,\left(\frac{2}{3}\right)^4:9^2=\left(\frac{2}{3}\right)^4:3^4=\left(\frac{2}{3}:3\right)^4=\left(\frac{2}{9}\right)^4=\frac{16}{6561}\)
\(k,\left(\frac{1}{2}\right)^3.\left(\frac{1}{4}^2\right)=\left(\frac{1}{2}\right)^3.\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^7=\frac{1}{128}\)
\(m,\left(\frac{120}{40}\right)^3=3^3=27\)
\(n,\frac{390^4}{130^4}=\left(\frac{390}{130}\right)^4=3^4=81\)
\(2,a,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(b,\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=-8+4+1+1\)
\(=-2\)
\(c,\left(3^2\right)^2-\left(\left(-5\right)^2\right)^2+\left(\left(-2\right)^3\right)^2\)
\(=3^4-\left(-5\right)^4+\left(-2\right)^6\)
\(=81-625+64\)
\(=-480\)
\(3,a,\left(x-1\right)^3=27\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
\(b,x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\Rightarrow x=-1\end{cases}}\)
\(c,\left(2x+1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\Rightarrow x-1=\pm1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\text{ or }x=0\end{cases}}\)
\(4,M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(M=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(M=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(M=100+99+98+97+...+2+1\)
\(M=\left(100+1\right)\cdot100:2\)
\(M=101\cdot50=5050\)