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lê văn khải
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Vũ Phong
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A = \(\dfrac{2}{35}\) + \(\dfrac{4}{77}\) + \(\dfrac{2}{143}\) + \(\dfrac{4}{221}\) + \(\dfrac{2}{323}\) + \(\dfrac{4}{437}\) + \(\dfrac{2}{575}\)

A =  \(\dfrac{2}{5\times7}\)+\(\dfrac{4}{7\times11}\)+\(\dfrac{2}{11\times13}\)+\(\dfrac{4}{13\times17}\)+\(\dfrac{2}{17\times19}\)+\(\dfrac{4}{19\times23}\)+\(\dfrac{2}{23\times25}\)

A = \(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)\(\dfrac{1}{7}\) - \(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{17}\)+\(\dfrac{1}{17}\)-\(\dfrac{1}{19}\)+\(\dfrac{1}{19}\)-\(\dfrac{1}{23}\)+\(\dfrac{1}{23}\)-\(\dfrac{1}{25}\)

A = \(\dfrac{1}{5}\) - \(\dfrac{1}{25}\)

A = \(\dfrac{4}{25}\)

Hoàng Phương Anh
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Liêu Thị Hường
6 tháng 8 2017 lúc 21:50

Ôn tập toán 7

Ôn tập toán 7

=> A=218.339 B=\(\dfrac{4}{15}\)

vì A>1>B

=> B<A

~ Kammin Meau ~
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Giải:

\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\) 

\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{19.21}=\dfrac{x}{49}\) 

\(2.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\) 

\(2.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=\dfrac{x}{49}\) 

\(2.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{x}{49}\) 

\(2.\dfrac{2}{7}=\dfrac{x}{49}\) 

   \(\dfrac{4}{7}=\dfrac{x}{49}\) 

\(\Rightarrow x=\dfrac{4.49}{7}=28\) 

Chúc bạn học tốt!

MRXBennjamin
15 tháng 5 2021 lúc 16:03

 \(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)

 2 . \(\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{399}=\dfrac{x}{49}\) 

 2 . \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{19.21}=\dfrac{x}{49}\)

 2 . ( \(\dfrac{1}{3}-\dfrac{1}{21}\) ) = \(\dfrac{x}{49}\)

 2 . \(\dfrac{2}{7}\) = \(\dfrac{x}{49}\)

=> \(\dfrac{4}{7}=\dfrac{x}{49}\)

=> \(\dfrac{21}{49}=\dfrac{x}{49}\)

=> \(x=21\)

Vậy \(x=21\)

 

 

 

~ Kammin Meau ~
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✿✿❑ĐạT̐®ŋɢย❐✿✿
15 tháng 5 2021 lúc 16:52

Ta có : \(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)

\(\Leftrightarrow2\cdot\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{x}{98}\)

\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{98}\)

\(\Leftrightarrow\dfrac{2}{7}=\dfrac{x}{98}\Rightarrow x=28\)

Vậy $x=28$

Nguyen Ngoc Linh
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Hoàng Quỳnh Trang
6 tháng 8 2017 lúc 19:55

may mà m đăng để t đỡ phải viết (mỏi tay) Nguyen Ngoc Linh

Nguyễn Thị Hồng Nhung
6 tháng 8 2017 lúc 20:08

\(A=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+2^{12}}\)

\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^{19}+2^{12}}=\dfrac{2^{10}+5.2^8}{3^{10}+2^3}=\dfrac{2^7+5.2^5}{3^{10}}\)

Nguyễn Thị Hồng Nhung
6 tháng 8 2017 lúc 20:11

\(B=\dfrac{4}{35}+\dfrac{4}{63}+\dfrac{4}{99}+\dfrac{4}{143}+\dfrac{4}{195}\)

\(=2\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\right)\)

\(=2.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{15}\right)=2\left(\dfrac{2}{15}\right)\)=\(\dfrac{4}{15}\)

ChaosKiz
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OkeyMan
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Hạnh Huỳnh
4 tháng 2 2018 lúc 16:18

a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)

<=> \(25x+10-80x+10=24x+12-30\)

<=> \(25x-80x-24x=12-30-10-10\)

<=> \(-79x=-38\)

<=> \(x=\dfrac{-38}{-79}\)

\(x=\dfrac{38}{79}\)

b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)

<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)

<=> \(30x-12x+30+5x+40=210+10x-10\)

<=> \(30x-12x+5x-10x=210-10-30-40\)

<=> \(13x=130\)

<=> \(x=\dfrac{130}{13}\)

\(x=10\)

c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)

<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)

<=> \(28x+28+60x+120+105x+420+2520=0\)

<=> \(28x+60x+105x=-28-120-420-2520\)

<=> \(193x=-3088\)

<=> \(x=\dfrac{-3088}{193}\)

\(x=-16\)

d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)

<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)

<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)

<=> \(22968x=8199576\)

<=> \(x=\dfrac{8199576}{22968}\)

\(x=357\)

OkeyMan
4 tháng 2 2018 lúc 14:07

Đề là giải PT nha các bn

Nguyễn Quốc Toàn
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Nguyễn Lê Phước Thịnh
26 tháng 6 2022 lúc 18:14

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)