TÍNH GIÁ TRỊ BIỂU THỨC:
- A= 310 nhân 11 cộng 310 nhân 5 30 nhân 24
- B= 210 nhân 13 cộng 210 nhân 65
28 nhân 104
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\(\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}:\dfrac{22}{28}\)
\(=\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{11}\)
\(=\dfrac{11\times12\times13\times14}{12\times13\times14\times11}\)
\(=1\)
=================
\(3\times\dfrac{7}{10}+\dfrac{7}{10}\times5+2\times\dfrac{7}{10}\)
\(=\dfrac{7}{10}\times\left(3+5+2\right)\)
\(=\dfrac{7}{10}\times10=7\)
\(\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}:\dfrac{22}{28}\)
\(=\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}:\dfrac{11}{14}\)
\(=\dfrac{11}{12}\times\dfrac{12}{13}\times\dfrac{13}{14}\times\dfrac{14}{11}\)
\(=\dfrac{11\times12\times13\times14}{12\times13\times11\times14}\)
\(=1\)
\(3\times\dfrac{7}{10}+\dfrac{7}{10}\times5+2\times\dfrac{7}{10}\)
\(=\dfrac{7}{10}\times\left(3+5+2\right)\)
\(=\dfrac{7}{10}\times10\)
\(=7\)
182 x 13 + 91 x 28 + 182 x 73
= 182 x 13 + 91 x 2 x 14 + 182 x 73
= 182 x 13 + 182 x 14 + 182 x 73
= 182 x ( 13 + 14 + 73 )
= 182 x 100
= 18200
= 182 x 13 + 91 x 2 x 14 + 182 x 73
= 182 x 13 + 182 x 14 + 182 x 73
= 182 x ( 13 + 14 + 73 )
= 182 x 100
= 18200
HT
\(25\frac{2}{13}-\left(\frac{15}{17}+15\frac{2}{3}\right)=\frac{327}{13}-\left(\frac{15}{17}+\frac{47}{3}\right)\)
\(=\frac{327}{13}-\frac{844}{51}\)
\(=\frac{5705}{663}\)
\(\frac{5}{30}+\frac{15}{90}+\frac{250}{150}+\frac{350}{210}+\frac{45}{270}=\frac{1}{6}+\frac{1}{6}+\frac{5}{3}+\frac{5}{3}+\frac{1}{6}\)
\(=\frac{3}{6}+\frac{10}{3}\)
\(=\frac{1}{2}+\frac{10}{3}\)
\(=\frac{23}{6}\)
\(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{17}.2\frac{4}{9}=\frac{34}{11}.\frac{27}{46}.\frac{23}{17}.\frac{22}{9}\)
\(=\frac{68}{9}.\frac{27}{34}\)
\(=6\)
a) Ta có:
\(A=\frac{3^{10}\cdot11+3^{10}\cdot5}{3^0\cdot2^4}\)
\(A=\frac{3^{10}\left(11+5\right)}{1\cdot16}\)
\(A=\frac{3^{10}\cdot16}{16}=3^{10}\)
b) Ta có:
\(B=\frac{2^{10}\cdot13+2^{10}\cdot65}{28\cdot104}\)
\(B=\frac{2^{10}\cdot13\cdot\left(1+5\right)}{2^2\cdot7\cdot2^3\cdot13}\)
\(B=\frac{2^{10}\cdot6\cdot13}{2^5\cdot7\cdot13}=\frac{2^{11}\cdot3\cdot13}{2^5\cdot7\cdot13}\)
\(B=\frac{2^6\cdot3}{7}=\frac{192}{7}\)