a)\(\left(\frac{2}{3}+\frac{4}{5}+\frac{12}{7}\right)+\left(2^7.5^6-\frac{25^3}{\left(0.125\right)3}\right).\left(0.0001\right)^2\)
b) \(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
K
Khách
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8 tháng 9 2020
Bạn chú ý rằng 315= 3.105 và 650=651-1 nha
Thế 2 cái đó vào A là ra thoi !!
KT
4 tháng 7 2018
1) \(63^2-47^2=\left(63+47\right)\left(63-47\right)=110.16=1760\)
2) \(127^2+146.127+73^2=\left(127+73\right)^2=200^2=40000\)
3) \(215^2-105^2=\left(215-105\right)\left(215+105\right)=110.320=35200\)
4) mk chỉnh lại đề:
\(\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4-1\right)\left(4+1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^4-1\right)\left(4^4+1\right)...\left(4^{256}+1\right)\)
\(=\frac{1}{3}\left(4^{512}-1\right)\)
17 tháng 7 2016
Cũng khuya rồi , mình làm câu 1 thôi nhé !
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{22}-9.5^{21}}{\left(5^2\right)^{10}}\)
\(\frac{5^{21}.\left(2.5-9\right)}{5^{20}}=5.\left(10-9\right)=5\)
2 tháng 7 2020
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
