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nguyễn hoàng linh chi
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Aug.21
25 tháng 6 2019 lúc 8:36

\(\frac{1}{7}.\left(\frac{-23}{10}\right)+\frac{77}{10}\)

\(=\frac{-23}{70}+\frac{77}{10}\)

\(=\frac{258}{35}\)

Xyz OLM
25 tháng 6 2019 lúc 8:36

\(\frac{1}{7}.\left(\frac{-23}{10}\right)+\frac{77}{10}\)

\(\frac{-23}{70}+\frac{77}{10}\)

\(\frac{258}{35}\)

︵✰ßล∂ ß๏у®
25 tháng 6 2019 lúc 8:36

\(\frac{1}{7}\cdot\left(\frac{-23}{10}\right)+\left(\frac{77}{10}\right)\)

\(=-\frac{23}{70}+\frac{77}{10}=\frac{-23}{70}+\frac{539}{70}=\frac{516}{70}=\frac{258}{35}\)

Trần Đình Dủng
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Khuất Thị Thảo Nguyên
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Trần Việt Linh
7 tháng 9 2016 lúc 21:12

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}-\frac{1}{x-7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)

\(\Leftrightarrow\)\(\frac{x+7-x-2}{\left(x+2\right)\left(x+7\right)}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)

\(\Leftrightarrow x=5\)

nguyễn thảo hân
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nguyễn thu ngà
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Trân Minh Tân
11 tháng 12 2016 lúc 15:57

bằng 13,34590301 ( mình bấm máy tính bạn nhé :) )

Công Chúa Hoa Hồng
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Nguyễn Huy Tú
25 tháng 9 2016 lúc 15:48

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=15\)

Vậy x = 15

Nguyễn Ngọc Thiên Trang
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Ngân Hoàng Xuân
28 tháng 6 2016 lúc 10:48

a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

 

Ngân Hoàng Xuân
28 tháng 6 2016 lúc 10:52

c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)

d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)

Ngọc Mai
28 tháng 6 2016 lúc 11:02

a)  \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left[\left(\frac{5}{15}-\frac{3}{15}\right):\frac{1}{5}\right]^2=\left[\frac{2}{15}.5\right]^2=\left[\frac{2}{3}\right]^2=\frac{4}{9}\)

b) \(\frac{2^3.5^2.8}{10^{10}}=\frac{2^3.5^2.2^3}{\left(2.5\right)^{10}}=\frac{\left(2^3.2^3\right).5^2}{2^{10}.5^{10}}=\frac{2^6.5^2}{2^{10}.5^{10}}=\frac{1}{2^4.5^8}=\frac{1}{6250000}\)

c) \(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3\)

\(=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3\)

\(=2+1+8\)

\(=11\)

d) \(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2\)

\(=5.\frac{25}{4}+\frac{1}{5}.9\)

\(=\frac{125}{4}+\frac{9}{5}\)

\(=\frac{625}{20}+\frac{36}{20}=\frac{661}{20}\)

Chuk bạn hok tốt ! vui

Trần Ngô Hạ Uyên
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Bùi Thế Hào
27 tháng 3 2018 lúc 10:12

Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\);  \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)

\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\)

=> Phương trình tương đương:

\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

=> x=15

Đáp số: x=15

Matrix
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Cô Hoàng Huyền
21 tháng 6 2016 lúc 15:30

\(pt\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\Leftrightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow x=15\)

Trà My
21 tháng 6 2016 lúc 15:37

\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)

\(\Rightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=15\)