A=\(\dfrac{1}{2}\) nhân \(\dfrac{3}{4}\)nhân \(\dfrac{5}{6}\)nhân ..... nhân \(\dfrac{99}{100}\)và B=\(\dfrac{1}{10}\)
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27 tháng 4 2023
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
31 tháng 10 2023
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
18 tháng 6 2018
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
HQ
Hà Quang Minh
Giáo viên
27 tháng 11 2023
a) $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ ; $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$
Vậy $\frac{1}{2} \times \frac{1}{3} = \frac{1}{3} \times \frac{1}{2}$
$\frac{3}{5} \times \frac{1}{6} = \frac{3}{{30}} = \frac{1}{{10}}$ ; $\frac{1}{6} \times \frac{3}{5} = \frac{3}{{30}} = \frac{1}{{10}}$
Vậy $\frac{3}{5} \times \frac{1}{6} = \frac{1}{6} \times \frac{3}{5}$
b) Học sinh tự thực hiện
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ < \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{97}{98}.\dfrac{98}{99}< \dfrac{1}{99}\\ < \dfrac{1}{10}.\\\\ =>A< \dfrac{1}{10}\)