\(\dfrac{2}{3}\)-\(\dfrac{5}{12}\)x=
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1 tháng 5 2017

a)

\(\dfrac{2}{3}-\dfrac{5}{12}x=\dfrac{-8}{3}\)\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}-\left(-\dfrac{8}{3}\right)\)

\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\)

\(\Rightarrow x=\dfrac{10}{3}:\dfrac{5}{12}=8\)

b) \(3x-2\left(2x-1\right)=1\dfrac{1}{3}\)\(\Rightarrow3x-4x+2=\dfrac{4}{3}\)

\(\Rightarrow3x-4x=\dfrac{4}{3}-2\)

\(\Rightarrow-x=-\dfrac{2}{3}\)\(\Rightarrow x=\dfrac{2}{3}\)

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\Rightarrow\left(x+4\right)\left(x+4\right)=20.5\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow\left(x+4\right)^2=10^2\) hoặc \(\left(x+4\right)^2=\left(-10\right)^2\)

=> x+4=10 => x+4=-10

=> x=6 => x=-14

1 tháng 5 2017

Thanks

2 tháng 11 2017

1. đề bạn ghi rõ lại giúp mình đc ko r mình giải lại cho

2. Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x^2}{2.3^2}=\dfrac{y^2}{5^2}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)

\(\dfrac{x}{3}=4\Rightarrow x=12\)

\(\dfrac{y}{5}=4\Rightarrow y=20\)

Vậy x=12 và y=20

Kêu người ta giúp mà ói vào mặt người ta vậy à?

10 tháng 8 2017

Bất lịch sự ucche

3 tháng 8 2017

a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)

\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)

\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)

\(\Leftrightarrow50x+\dfrac{99}{100}=1\)

\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)

b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)

\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)

3 tháng 8 2017

a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)

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

Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)

\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)

\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)

\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)

\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)

b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)

\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)

\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)

\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)

Bài 2: 

a: \(\left|x\right|=-x\)

nên x<=0

b: \(\left|x\right|>x\)

=>x<0

29 tháng 10 2017

a)hình như đề sai thì phải

sửa lại

\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)

=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)

=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)

29 tháng 9 2017

a) \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)

\(\Rightarrow\) \(\dfrac{1}{6}=\dfrac{1}{30}:x\)

\(\Rightarrow\) \(x=\dfrac{1}{5}\)

29 tháng 9 2017

a. \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)

\(\Leftrightarrow\dfrac{1}{15}:\left(2x\right)=0,75:4,5\)

\(\Rightarrow\dfrac{1}{15}:\left(2x\right)=\dfrac{1}{6}\)

\(\Rightarrow2x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{2}{5}:2=\dfrac{1}{5}\)

Vậy...

b. \(\dfrac{-5}{x-2}=\dfrac{3}{-9}\)

\(\Leftrightarrow\left(x-2\right).3=\left(-5\right).\left(-9\right)\)

\(\Rightarrow\left(x-2\right).3=45\)

\(\Rightarrow\left(x-2\right)=45:3=15\)

\(\Rightarrow x=15+2=17\)

Vậy...

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

\(\Rightarrow\dfrac{-2}{3}:x=\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}:\dfrac{2}{3}=-1\)

Vậy...

11 tháng 9 2017

a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)

\(x=\dfrac{4}{7}-\dfrac{3}{5}\)

\(x=-\dfrac{1}{35}\)

Vậy ....

b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)

\(x=\dfrac{1}{6}+\dfrac{5}{6}\)

\(x=1\)

Vậy ....

c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)

\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)

\(x=\dfrac{13}{70}\)

Vậy .....

d/ \(\dfrac{5}{7}-x=10\)

\(x=\dfrac{5}{7}-10\)

\(x=\dfrac{-65}{7}\)

Vậy ...

11 tháng 9 2017

e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)

\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)

\(x=\dfrac{13}{90}\)

Vậy ....

f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)

\(0,65.x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:0,65\)

\(x=\dfrac{20}{39}\)

Vậy ....

g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)

\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)

\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)

\(\Leftrightarrow x=\dfrac{-35}{12}\)

Vậy ...

7 tháng 8 2017

Bài 2:

a) Ta có : Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)

Theo tính chất dãy tỉ số bằng nhau, ta có :

\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}\left(1\right)\)

\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a-7b}{5c-7d}\left(2\right)\)

Từ (1) và (2)=> \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)Vậy...

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Thay các đẳng thức vừa tìm được , ta có :

\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(1\right)\)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)

\(=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)

từ (1) và (2)=> đpcm

tik mik nha !!!

7 tháng 8 2017

1. Bạn xem lại đề bài nhé! Mình nghĩ là \(2x=3y=5z\) thì đúng hơn!

2.

a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\)

Từ \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)(đpcm)

Vậy \(\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\left(1\right)\)

\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)\)\(\left(2\right)\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)

Vậy \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)