\(\frac{40}{9}\cdot\frac{4}{3}-\frac{4}{9}:\frac{13}{3}=....\)
Bài 1: Tính(hợp lý nếu có thể) e) \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\) f)\(\frac{12}{19}\cdot\frac{7}{15}\cdot\frac{-13}{17}\cdot\frac{19}{12}\cdot\frac{17}{13}\) g) \(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
Tính nhanh giá trị biểu thức sau:
a) \(-\frac{9}{10}\cdot\frac{5}{14}+\frac{1}{10}\cdot\left(-\frac{9}{2}\right)+\frac{1}{7}\cdot\left(-\frac{9}{10}\right)\)
b)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right)\cdot132\)
c)\(-\frac{2}{3}\cdot\left(\frac{8}{9}\cdot\frac{8}{13}-\frac{8}{27}\cdot\frac{3}{13}+\frac{4}{3}\cdot\frac{22}{39}\right)\)
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
Tính giá trị biểu thức
a,\(A=\frac{24\cdot47-23}{24+47-23}\cdot\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)
b,\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
c,\(A=81\cdot\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right]:\frac{158158158}{711711711}\)
d,\(A=\frac{5\cdot\left(2^2.3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
Chọn dấu " "=", " \( \ne \) " thích hợp cho dấu “?” :
a) \(\frac{{28}}{9} \cdot 0,7 + \frac{{28}}{9} \cdot 0,5\) ? \(\frac{{28}}{9} \cdot (0,7 + 0,5)\);
b) \(\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\) ? \(\frac{{36}}{{13}}:(4 + 9)\).
a)
\(\frac{{28}}{9} \cdot 0,7 + \frac{{28}}{9} \cdot 0,5 = \frac{{28}}{9}.\left( {0,7 + 0,5} \right)\)
b)
\(\begin{array}{l}\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\\ = \frac{{36}}{{13}}.\frac{1}{4} + \frac{{36}}{{13}}.\frac{1}{9}\\ = \frac{{36}}{{13}}.\left( {\frac{1}{4} + \frac{1}{9}} \right)\\ = \frac{{36}}{{13}}.\frac{{13}}{{36}} = 1\end{array}\)
\(\begin{array}{l}\frac{{36}}{{13}}:(4 + 9)\\ = \frac{{36}}{{13}}:13\\ = \frac{{36}}{{13}}.\frac{1}{{13}}\\ = \frac{{36}}{{169}}\end{array}\)
Suy ra \(\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\) \( \ne \) \(\frac{{36}}{{13}}:(4 + 9)\).
tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Mk ko biết lm nhưng cứ k thoải mái nha
SORRY
\(\frac{3}{13}\cdot\frac{7}{9}+\frac{4}{15}\cdot\frac{7}{9}=\)
\(a,\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(b,5\cdot\frac{7}{5}\) \(c,\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(d,4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(e,\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(g,2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
) Ta có: 1615⋅−514⋅5424⋅56211615⋅−514⋅5424⋅5621
=1615⋅−514⋅94⋅83=1615⋅−514⋅94⋅83
=4⋅−13⋅47⋅3=4⋅−13⋅47⋅3
=12⋅−421=−4821=−167=12⋅−421=−4821=−167
b) Ta có: 5⋅75=355=75⋅75=355=7
c) Ta có: 17⋅59+59⋅17+59⋅3717⋅59+59⋅17+59⋅37
=59(17+17+37)=59(17+17+37)
=59⋅57=2563=59⋅57=2563
d) Ta có: 4⋅11⋅34⋅91214⋅11⋅34⋅9121
=4⋅11⋅3⋅94⋅121=2711=4⋅11⋅3⋅94⋅121=2711
e) Ta có: 34⋅169−75:−212034⋅169−75:−2120
=43+43=83=43+43=83
g) Ta có: 213−13⋅[−32+(23+0,4⋅5)]213−13⋅[−32+(23+0,4⋅5)]
=73−13⋅[−32+23+2]=73−13⋅[−32+23+2]
=73−13⋅76=73−13⋅76
=73−718=4218−718=3518
tính:
A=\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}...\frac{8^2}{8\cdot9}\cdot\frac{9^2}{9\cdot10}\)
B=\(\frac{2^2}{3}\cdot\frac{^{3^2}}{8}\cdot\frac{4^2}{15}\cdot\frac{6^2}{35}\cdot\frac{7^2}{48}\cdot\frac{8^2}{63}\cdot\frac{9^2}{80}\)
A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)
A=\(\frac{1}{10}\)
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