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1 tháng 8 2018

\(91924\)\(:1372=67\)

\(163840:98304=\frac{5}{3}\)

k mk nha

1 tháng 8 2018

cần gấp mn

2 tháng 2 2020

Ta có

  \(2^8.4.13=2^9.26\)

  \(2^7.8.65=2^9.130\)

Nên \(2^8.4.13+2^7.8.65=2^9.156\)

  Mà \(2^9:2^9=1\)

        \(156:3=52\)

Do đó A = 52

27 tháng 4 2017

1)

a)

\(\dfrac{-5}{11}\cdot\dfrac{4}{7}+\dfrac{-5}{11}\cdot\dfrac{3}{7}-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot1-\dfrac{8}{11}\\ =\dfrac{-5}{11}-\dfrac{8}{11}\\ =\dfrac{-5}{11}+\dfrac{-8}{11}\\ =\dfrac{-13}{11}\)

b)

\(\left(\dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3}\right)^2-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =\left(\dfrac{2}{9}\cdot\dfrac{3}{5}+\dfrac{1}{3}\cdot\dfrac{3}{5}\right)^2-\left(\dfrac{-7}{24}\right)\\ =\left[\dfrac{3}{5}\cdot\left(\dfrac{2}{9}+\dfrac{1}{3}\right)\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{3}{5}\cdot\dfrac{5}{9}\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{1}{3}\right]^2+\dfrac{7}{24}\\ =\dfrac{1}{9}+\dfrac{7}{24}\\ =\dfrac{29}{72}\)

c) \(14-\left|\dfrac{-3}{4}\right|-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =14-\dfrac{3}{4}-\left(\dfrac{-7}{24}\right)\\ =14+\dfrac{-3}{4}+\dfrac{7}{24}\\ =13\dfrac{13}{24}\)

3 tháng 4 2017

a)\(\frac{5}{7}:X=1-\frac{4}{7}=\frac{3}{7}\)

\(X=\frac{5}{7}:\frac{3}{7}=\frac{5}{3}\)

b)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{19}-\frac{1}{21}\right)-X=\frac{4}{3}-\frac{221}{231}=\frac{29}{77}\)

\(\left(\frac{1}{11}-\frac{1}{21}\right)-X=\frac{29}{77}\)

\(X=\frac{10}{231}-\frac{29}{77}=-\frac{1}{3}\)

xem xong nhớ tích

23 tháng 7 2023

a) \(\dfrac{-5}{11}+\left(\dfrac{-6}{11}+1\right)\)

\(=\dfrac{-5}{11}+\left(\dfrac{-6}{11}+\dfrac{11}{11}\right)\)

\(=\dfrac{-5}{11}+\dfrac{5}{11}\)

\(=0\)

b) \(\dfrac{2}{3}+\left(\dfrac{5}{7}+\dfrac{-2}{3}\right)\)

\(=\dfrac{2}{3}+\dfrac{-2}{3}+\dfrac{5}{7}\)

\(=0+\dfrac{5}{7}\)

\(=\dfrac{5}{7}\)

c) \(\left(\dfrac{-1}{4}+\dfrac{5}{8}\right)+\dfrac{-3}{8}\)

\(=\dfrac{-1}{4}+\dfrac{-3}{8}+\dfrac{5}{8}\)

\(=\dfrac{-2}{8}+\dfrac{-3}{8}+\dfrac{5}{8}\)

\(=0\)

d) \(\dfrac{3}{4}.\dfrac{7}{25}+\dfrac{3}{4}.\dfrac{18}{25}\)

\(=\dfrac{3}{4}.\left(\dfrac{7}{25}+\dfrac{18}{25}\right)\)

\(=\dfrac{3}{4}.1\)

\(=\dfrac{3}{4}\)

Chúc bạn học tốt

14 tháng 12 2022

a. 600 - 30 =570

b. 72 + (-3) = 69 

c. 2021 + 33:(9 +2x12)

=2021 + 33: 33

=2021+1

=2022

d. 186: (44+2x9)

=186 : 62

=3

9 tháng 4 2017

a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)

\(P=1-\dfrac{1}{56}\)

\(P=\dfrac{55}{56}\)

b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)

\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=3\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}\)

\(A=\dfrac{297}{100}\)

c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}\)

\(B=\dfrac{102}{103}\)

d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)

\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}.\dfrac{102}{103}\)

\(C=\dfrac{170}{103}\)

e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)