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Lưu Gia Lợi
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Nguyễn Hà Tuấn Hưng 7A14
20 tháng 5 2022 lúc 10:20

A= 4/7.

Biết có cái

Lê Ngọc Anh
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dâu cute
17 tháng 3 2022 lúc 10:55

\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)

Nguyễn Thị Mỹ Loan♍13/9
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YangSu
21 tháng 6 2023 lúc 19:16

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)

 

títtt
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a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729

b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)

c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)

\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)

\(=\dfrac{2^3}{4^4}\cdot7\)

\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)

d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)

\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)

\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)

e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)

\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)

\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)

f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)

\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)

\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)

g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)

\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)

\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)

Lê Phương Linh
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Miru nèe
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Yeutoanhoc
16 tháng 6 2021 lúc 10:16

`A=(8 2/7-4 2/7)-3 4/9`

`=8+2/7-4-2/7-3-4/9`

`=4-3-4/9`

`=1-4/9=5/9`

`B=(10 2/9-6 2/9)+2 3/5`

`=10+2/9-6-2/9+2+3/5`

`=4+2+3/5`

`=6+3/5=33/5`

Bài 2:

`a)5 1/2*3 1/4`

`=11/2*13/4`

`=143/8`

`b)6 1/3:4 2/9`

`=19/3:38/9`

`=19/3*9/38=3/2`

`c)4 3/7*2`

`=31/7*2`

`=62/7`

Bài 1:

\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\) 

\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\) 

\(A=4-\dfrac{31}{9}\) 

\(A=\dfrac{5}{9}\) 

 

\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\) 

\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\) 

\(B=4+\dfrac{13}{5}\) 

\(B=\dfrac{33}{5}\)

Bài 2:

a) \(5\dfrac{1}{2}.3\dfrac{1}{4}=\dfrac{11}{2}.\dfrac{13}{4}=\dfrac{11.13}{2.4}=\dfrac{143}{8}\) 

b) \(6\dfrac{1}{3}:4\dfrac{2}{9}=\dfrac{19}{3}:\dfrac{38}{9}=\dfrac{19}{3}.\dfrac{9}{38}=\dfrac{3}{2}\) 

c) \(4\dfrac{3}{7}.2=\dfrac{31}{7}.2=\dfrac{31.2}{7}=\dfrac{62}{7}\)

Minh Châu Trần
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⭐Hannie⭐
16 tháng 10 2023 lúc 19:44

`(6/11 +5/11) xx 3/7`

`= 11/11xx 3/7`

`=1xx3/7`

`=3/7`

__

`3/5 xx 7/9 - 3/5 xx 2/9`

`= 3/5 xx (7/9-2/9)`

`= 3/5 xx 5/9`

`= 15/45`

`= 1/3`

__

`(6/7 -4/7):2/5`

`= 2/7 : 2/5`

`= 2/7 xx 5/2`

`= 10/14`

`= 5/7`

Võ Ngọc Phương
16 tháng 10 2023 lúc 19:44

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=1\times\dfrac{3}{7}=\dfrac{3}{7}\)

___________

\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\)

\(=\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{3}{5}\times\dfrac{5}{9}\)

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

__________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\)

\(=\dfrac{2}{7}\times\dfrac{5}{2}\)

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

\(#WendyDang\)

HT.Phong (9A5)
16 tháng 10 2023 lúc 19:45

\(\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}\)

\(=\dfrac{11}{11}\times\dfrac{3}{7}\\ =1\times\dfrac{3}{7}=\dfrac{3}{7}\)

_____

\(\dfrac{3}{5}\times\dfrac{7}{9}-\dfrac{3}{5}\times\dfrac{2}{9}\\ =\dfrac{3}{5}\times\left(\dfrac{7}{9}-\dfrac{2}{9}\right)\\ =\dfrac{3}{5}\times\dfrac{5}{9}\\ =\dfrac{3}{9}\\ =\dfrac{1}{3}\)

_________

\(\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}\\ =\dfrac{2}{7}\times\dfrac{5}{2}\\ =\dfrac{10}{14}\\ =\dfrac{5}{7}\)

Xử Nữ Chính Là Tôi
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Vũ Ngọc Diệp
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Thuỳ Linh Nguyễn
12 tháng 3 2023 lúc 17:53

Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)