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Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
Ta có: B = 22010 - 22009 - 22008 -......- 2 -1
=> B = 22010 - (1 + 2 + 22 + ..... + 22009)
Đặt A = 1 + 2 + 22 + .... + 22009
=> 2A = 2 + 22 + .... + 22010
=> 2A - A = 22010 - 1
=> A = 22010 - 1
Vậy B = 22010 - (22010 - 1)
=> B = 22010 - 22010 + 1
=> B = 1
câu 1 :
\(A=\frac{-7}{12}:\frac{49}{11}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:8-3:\frac{3}{4}\cdot-2^3\)
\(A=\frac{-11}{84}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:\frac{8}{1}-\frac{3}{1}:\frac{3}{4}\cdot\left(-2^3\right)\)
\(A=\frac{-5}{924}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{1}{7}-\left(-32\right)\)
\(A=\frac{-361}{308}\) \(B=\frac{-1}{56}-\left(-32\right)\)
\(B=\frac{1791}{56}\)
Câu 2 :
a)\(\frac{22}{7}:x=\frac{11}{7}\) b)\(\left(1-3x\right)\cdot\frac{4}{3}=-2^3\)
\(x=\frac{22}{7}:\frac{11}{7}\) \(\left(1-3x\right)\cdot\frac{4}{3}=-8\)
\(x=2\) \(\left(1-3x\right)=-8:\frac{4}{3}\)
\(\left(1-3x\right)=-6\)
\(3x=-6-1=7\)
\(3x=7:3=\frac{7}{3}\)
c ) bằng \(\frac{27}{5}\)nhé
khó quá bn ơi
2b,