tính nhanh:
37.39+78+37.39
(. là dấu nhân ạ) ^^
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\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)
\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)
\(2C=\dfrac{4}{13}\)
\(C=\dfrac{2}{13}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\frac{12}{39}\)
\(C=\frac{4}{26}=\frac{2}{13}\)
\(37\cdot39+14\cdot79+13\cdot85\)
\(=13\left(37\cdot3+85\right)+14\cdot79\)
\(=13\cdot196+14\cdot79\)
\(=14\left(13\cdot14+79\right)\)
\(=14\cdot261=3654\)
\(\dfrac{11\cdot9^{11}\cdot3^7-27^{10}}{4\cdot81^1}\)
\(=\dfrac{11\cdot3^{22}\cdot3^7-\left(3^3\right)^{10}}{4\cdot3^4}\)
\(=\dfrac{11\cdot3^{29}-3^{30}}{4\cdot3^4}\)
\(=\dfrac{3^{29}\cdot\left(11-3\right)}{4\cdot3^4}\)
\(=\dfrac{3^{29}\cdot8}{4\cdot3^4}\)
\(=3^{25}\cdot2\)
@ Huỳnh Thanh Phong
Sao lại lấy 329 là thừa số chung ạ
0,2.17.9 + 0,18.540 + 24.1,8
= 0,2.9.17 + 0,18.10.54 + 24.1,8
= 1,8.17 + 1,8.54 + 24.1,8
= 1,8.(17 + 54 + 24)
= 1,8.95
= 171
Ủng hộ mk nha ^-^
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{37}-\frac{2}{39}\)
\(=\frac{2}{3}-\frac{2}{39}\)
\(=\frac{8}{13}\)
Ta có:
\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)
\(\frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)
\(\frac{2}{7.9}=\frac{1}{7}-\frac{1}{9}\)
\(......................................\)
\(\frac{2}{37.39}=\frac{1}{37}-\frac{1}{39}\)
nên \(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(C=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)
37.39+78+37.39
=37.39 .(1+1)+78
=37.39.2+78
=37.78+78
=78.(37+1)=78.38=2964