\(25.\left(\frac{-1^3}{5}\right):\left(\frac{1}{4}-0,75^2\right)\)
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\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
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: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
Em chỉ làm những bài e biết thôi, thông cảm nhs :D
a/ chịu
b/ \(C=1+7+7^2+.........+7^{50}\)
\(\Leftrightarrow7C=7+7^2+...........+7^{50}+7^{51}\)
\(\Leftrightarrow7C-C=\left(7+7^2+.......+7^{51}\right)-\left(1+7+.....+7^{50}\right)\)
\(\Leftrightarrow6C=7^{51}-1\)
\(\Leftrightarrow C=\dfrac{7^{51}-1}{6}\)
c/ \(A=\dfrac{-1}{4}+\dfrac{7}{3}+\dfrac{3}{4}+\dfrac{9}{2}\)
\(=\left(\dfrac{-1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{7}{3}+\dfrac{9}{2}\right)\)
\(=\dfrac{1}{4}+\dfrac{41}{6}\)
\(=\dfrac{85}{12}\)
d/ Thấy phép tính hơi dài
e/ \(C=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.........+\dfrac{1}{2015.2016.2017}\)
\(\Leftrightarrow2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.........+\dfrac{2}{2015.2016.2017}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.......+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\)
\(=\dfrac{1}{2}-\dfrac{1}{4066272}\)
\(=\dfrac{2033136}{4066272}\)
\(\Leftrightarrow C=\dfrac{2033136}{4066272}:2\)
\(\Leftrightarrow C=?\)
a: \(\left(\dfrac{1}{5}\right)^{-2}=25\)
b: \(4^{\dfrac{3}{2}}=8\)
c: \(\left(\dfrac{1}{8}\right)^{-\dfrac{2}{3}}=\left(\dfrac{1}{2}\right)^{3\cdot\dfrac{-2}{3}}=\left(\dfrac{1}{2}\right)^{-2}=4\)
d: \(\left(\dfrac{1}{16}\right)^{-0.75}=\left(\dfrac{1}{2}\right)^{4\cdot\left(-0.75\right)}=\left(\dfrac{1}{2}\right)^{-3}=8\)
25.-1/125 :(1/4-27/64)
=25.-1/125:(16/64+ -27/64)
=25. -1/125:-11/64
=25.64/1375
=64/55
NẾU ĐÚNG THÌ NHỚ TÍCH CHO MÌNH NHÉ