Tính: a/\(\dfrac{1 \dfrac{1}{2} \dfrac{1}{3} \dfrac{1}{4}}{1-\dfrac{1}{2} \dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{3 \dfrac{3}{2} \dfrac{3}{3} \dfrac{3}{4}}{2-\dfrac{2}{2} \dfrac{2}{3}-\dfrac{2}{4}}\) b/\(...

2

NH

Nịna Hatori

28 tháng 6 2017

a

= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}

Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.

Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .

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M

☼™Mặt☼Nạ™☼

20 tháng 7 2017

Kết quả là : C1=\(\dfrac{2}{3}\)

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MG

Mai gia bảo

3 tháng 8 2023

Bài 1: (đề 2) Tính

a) \(2\dfrac{3}{8}:3\dfrac{3}{4}x2\dfrac{2}{3}\)=............

b) \(1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\)=............

c) \(4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\)=............

#Toán lớp 5

2

ND

Nguyễn Đức Trí

VIP

3 tháng 8 2023

a) \(...=\dfrac{19}{8}:\dfrac{15}{4}x\dfrac{8}{3}=\dfrac{19}{8}x\dfrac{4}{15}x\dfrac{8}{3}=\dfrac{76}{45}\)

b) \(...=\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{17}{6}=\dfrac{3}{2}x\dfrac{3}{7}x\dfrac{6}{17}=\dfrac{27}{119}\)

c) \(...=\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}=\dfrac{14}{3}-\dfrac{7}{4}x\dfrac{5}{12}=\dfrac{14}{3}-\dfrac{35}{48}=\dfrac{14x16}{48}-\dfrac{35}{48}=\dfrac{224}{48}-\dfrac{35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

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GN

GV Nguyễn Trần Thành Đạt

Giáo viên

3 tháng 8 2023

\(a,2\dfrac{3}{8}:3\dfrac{3}{4}\times2\dfrac{2}{3}\\ =\dfrac{2\times8+3}{8}:\dfrac{3\times4+3}{4}\times\dfrac{2\times3+2}{3}\\ =\dfrac{19}{8}:\dfrac{15}{4}\times\dfrac{8}{3}\\ =\dfrac{19\times4\times8}{8\times15\times3}=\dfrac{76}{45}\)

\(b,1\dfrac{1}{2}:\dfrac{7}{3}:2\dfrac{5}{6}\\ =\dfrac{3}{2}:\dfrac{7}{3}:\dfrac{2\times6+5}{6}\\ =\dfrac{3}{2}\times\dfrac{3}{7}\times\dfrac{6}{17}\\ =\dfrac{54}{238}=\dfrac{27}{119}\)

\(c,4\dfrac{2}{3}-1\dfrac{3}{4}:2\dfrac{2}{5}\\ =\dfrac{4\times3+2}{3}-\dfrac{1\times4+3}{4}:\dfrac{2\times5+2}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}:\dfrac{12}{5}\\ =\dfrac{14}{3}-\dfrac{7}{4}.\dfrac{5}{12}\\ =\dfrac{14}{3}-\dfrac{35}{48}\\ =\dfrac{14\times16-35}{48}=\dfrac{189}{48}=\dfrac{63}{16}\)

Bài 1: tính a) 3\(\dfrac{1}{2}\) + 4\(\dfrac{5}{7}\) - 5\(\dfrac{5}{14}\) b) 4\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) : \(5\dfrac{1}{2}\) bài 2: tìm X a) X x \(3\dfrac{1}{3}\) = \(3\dfrac{1}{3}\) : \(4\dfrac{1}{4}\) b) \(5\dfrac{2}{3}\) : X = \(3\dfrac{2}{3}\) - \(2\dfrac{1}{2}\) các giáo viên olm giúp e vs, e cần gấp...

PP

Phượng Phạm

5 tháng 6 2023

a) 0,25-\(\dfrac{2}{3}\)+1\(\dfrac{1}{4}\)b) \(\dfrac{3^2}{2}\):\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\).2010c) {[(\(\dfrac{1}{25}\)-0,6)2:\(\dfrac{49}{125}\)].\(\dfrac{5}{6}\)}-[(\(\dfrac{-1}{3}\))+\(\dfrac{1}{2}\)]d) (-\(\dfrac{1}{2}\)-\(\dfrac{1^{ }}{3}\))2:[(\(\dfrac{-5}{36}\))-(\(\dfrac{-5}{36}\))0]Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like...

#Toán lớp 5

1

LM

Lê Minh Vũ

CTVHS VIP

5 tháng 6 2023

\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

= \(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)

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MG

Mai gia bảo

31 tháng 7 2023

Bài 7: Tính

a) \(4\dfrac{2}{5}x8\dfrac{3}{4}-2\dfrac{3}{4}=...............................\)

b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)=..............................................

c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}=...............................\)

#Toán lớp 5

1

LM

Lê Minh Vũ

CTVHS VIP

31 tháng 7 2023

Bài 7: Tính

a) \(4\dfrac{2}{5}\times8\dfrac{3}{4}-2\dfrac{3}{4}\)

\(=\dfrac{22}{5}\times\dfrac{35}{4}-\dfrac{11}{4}\)

\(=\dfrac{77}{2}-\dfrac{11}{4}\)

\(=\dfrac{143}{4}\)

b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)

\(=\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{2}{15}\)

\(=\dfrac{47}{15}-\dfrac{2}{15}\)

\(=\dfrac{45}{15}=3\)

c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}\)

\(=\dfrac{10}{3}-\dfrac{8}{3}+\dfrac{11}{6}\)

\(=\dfrac{2}{3}+\dfrac{11}{6}\)

\(=\dfrac{15}{6}\)

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BM

bé mèo miu

1 tháng 10 2023

#Toán lớp 6

1

T

Toru

1 tháng 10 2023

a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)

\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)

\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)

\(=\dfrac{10}{12}\)

\(=\dfrac{5}{6}\)

\(---\)

b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)

\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)

\(=18+\dfrac{3015}{2}\)

\(=\dfrac{36}{2}+\dfrac{3015}{2}\)

\(=\dfrac{3051}{2}\)

\(---\)

c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)

\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(---\)

d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)

\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)

\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)

\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)

\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)

\(=-\dfrac{25}{41}\)

#\(Toru\)

Tìm số nguyên x, biết: a) \(-4\dfrac{3}{5}\). \(2\dfrac{4}{3}\) < x < \(-2\dfrac{3}{5}\) : \(1\dfrac{6}{15}\) b) \(-4\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{6}\)) < x <...

BM

bé mèo miu

1 tháng 10 2023

cảm ơn nhiều nha vừa kịp giờ lun

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LP

Lê Phương Linh

23 tháng 9 2023

3

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

23 tháng 9 2023

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)

\(x\) \(\in\) {-15; -14;-13;..; -2}

WT

when the imposter is sus

23 tháng 9 2023

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

TD

Trần Duy Quân

10 tháng 8 2017

1. Tính :

a, \(A=\dfrac{\dfrac{1}{3}-\dfrac{5}{2}}{\dfrac{3}{4}-\dfrac{1}{2}}.\dfrac{\dfrac{5}{6}+\dfrac{7}{3}}{1-\dfrac{5}{6}}.\dfrac{\dfrac{-2}{5}+1}{\dfrac{2}{5}-1}\).

b, \(B=\dfrac{\dfrac{1}{3}-\dfrac{4}{5}}{\dfrac{1}{3}+\dfrac{4}{5}}.\dfrac{\dfrac{3}{4}-\dfrac{5}{3}}{\dfrac{3}{4}+\dfrac{5}{3}}:\dfrac{\dfrac{4}{5}-1}{1-\dfrac{2}{3}}\).

Bài 1. Tính A= \(\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\) B= \(\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)Bài 2. Tính a) \(5\dfrac{1}{2}.3\dfrac{1}{4}\) b) \(6\dfrac{1}{3}:4\dfrac{2}{9}\) ...

0

MN

Miru nèe

16 tháng 6 2021

#Toán lớp 6

3

Y

Yeutoanhoc

16 tháng 6 2021

`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`

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T

꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...

16 tháng 6 2021

Bài 1:

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

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

\(A=4-\dfrac{31}{9}\)

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

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

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

\(B=4+\dfrac{13}{5}\)

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

DD

Đỗ Diệu Linh

25 tháng 9 2017

Tính tổng đại số

\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)

\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)

0

LN

Lê Ngọc Duyên

5 tháng 11 2021

a)Tính tổng\(P=\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+2017}\)

b)CMR\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\)

#Toán lớp 8

1

NH

Nguyễn Hoàng Minh

5 tháng 11 2021

\(a,P=\dfrac{1}{\left(2+1\right)\left(2+1-1\right):2}+\dfrac{1}{\left(3+1\right)\left(3+1-1\right):2}+...+\dfrac{1}{\left(2017+1\right)\left(2017+1-1\right):2}\\ P=\dfrac{1}{2\cdot3:2}+\dfrac{1}{3\cdot4:2}+...+\dfrac{1}{2017\cdot2018:2}\\ P=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2017\cdot2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\\ P=2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=2\cdot\dfrac{504}{1009}=\dfrac{1008}{1009}\)

\(b,\) Ta có \(\dfrac{1}{4^2}< \dfrac{1}{2\cdot4};\dfrac{1}{6^2}< \dfrac{1}{4\cdot6};...;\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right)2n}\)

\(\Leftrightarrow VT< \dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2n-2\right)2n}\\ \Leftrightarrow VT< \dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2n-2\right)2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)\\ \Leftrightarrow VT< \dfrac{1}{2}\left(1-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\)