\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)

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18 tháng 4 2018

\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)

\(\Leftrightarrow\left(x+2\right).2=\left(2x+1\right).0,5\)

\(\Leftrightarrow2x+4=x+0,5\)

\(\Leftrightarrow x=-3,5\)

Vậy...

18 tháng 4 2018

\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)

\(\Leftrightarrow\dfrac{4.\left(x+2\right)}{2}=\dfrac{2x+1}{2}\)

\(\Rightarrow4x+8=2x+1\)

\(\Leftrightarrow4x-2x=1-8\)

\(\Leftrightarrow2x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{2}\)

Vậy \(x=\dfrac{-7}{2}\)

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...

15 tháng 10 2018

Bài 1

a, \(D=1-\left|2x-3\right|\)

Ta có : \(\left|2x-3\right|\ge0\)

\(\Rightarrow1-\left|2x-3\right|\le1\)

Dấu "=" xảy ra khi \(\left|2x-3\right|=0\)

\(\Leftrightarrow2x-3=0\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=3:2=\dfrac{3}{2}\)

15 tháng 10 2018

\(b,\) Ta có : \(\left|10-5x\right|\ge0\Rightarrow\left|10-5x\right|+14,2\ge14,3\Rightarrow-\left|10-5x\right|-14,2\le-14,2\)

Dấu "=" xảy ra khi \(-\left|10-5x\right|=0\)

\(\Leftrightarrow10-5x=0\)

\(\Leftrightarrow5x=10\)

\(\Leftrightarrow x=10:5=2\)

Vậy \(Emax=-14,2\Leftrightarrow x=2\)

\(c,\) Ta có : \(\left|5x-2\right|\ge0\)

\(\left|3y-12\right|\ge0\)

\(\left|5x-2\right|+\left|3y+12\right|-4\ge-4\)

\(4-\left|5x-2\right|-\left|3y+12\right|\le4\)

Dấu "=" xảy ra khi \(\left[{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=2\\3y=-12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

\(d,\) \(A=5-3\left(2x-1\right)^2\)

Ta có : \(\left(2x-1\right)^2\ge0\)

\(\Rightarrow3.\left(2x-1\right)^2\ge0\)

\(\Rightarrow3.\left(2x-1\right)^2-5\ge-5\)

\(\Rightarrow5-3\left(2x-1\right)^2\le5\)

Dấu "=" xảy ra khi \(\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

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

Vậy \(Amax=5\Leftrightarrow x=\dfrac{1}{2}\)

17 tháng 11 2022

Bài 7:

x/1=z/2 nên x/6=z/12

=>x/6=y/9=z/12

=>x/2=y/3=z/4

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{27}{9}=3\)

=>x=6; y=9; z=12

21 tháng 7 2018

a) \(|\dfrac{3}{5}x|=|-\dfrac{1}{6}|\)

\(\Rightarrow|\dfrac{3}{5}x|=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{3}{5}x=\dfrac{1}{6}\) hoặc \(\dfrac{3}{5}x=-\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}:\dfrac{3}{5}\) hoặc \(x=-\dfrac{1}{6}:\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{1}{6}.\dfrac{5}{3}\) hoặc \(x=-\dfrac{1}{6}.\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{5}{18}\) hoặc \(x=-\dfrac{5}{18}\)

b) \(|2x-3|+0,5=\dfrac{1}{3}:2\)

\(\Rightarrow|2x-3|+\dfrac{1}{2}=\dfrac{1}{3}.\dfrac{1}{2}\)

\(\Rightarrow|2x-3|+\dfrac{1}{2}=\dfrac{1}{6}\)

\(\Rightarrow|2x-3|=\dfrac{1}{6}-\dfrac{1}{2}=-\dfrac{1}{3}\)

\(|2x-3|\ge0\forall x\)

=> Không tồn tại x thỏa mãn

c)\(|x-\dfrac{5}{6}|=2\left(x+\dfrac{1}{2}\right)\)

\(\Rightarrow|x-\dfrac{5}{6}|=2x+1\)

\(\Rightarrow x-\dfrac{5}{6}=2x+1\) hoặc \(x-\dfrac{5}{6}=-2x-1\)

\(\Rightarrow-\dfrac{5}{6}-1=2x-x\) hoặc \(-\dfrac{5}{6}+1=-2x-x\)

\(\Rightarrow-\dfrac{11}{6}=x\) hoặc \(\dfrac{1}{6}=-3x\)

\(\Rightarrow x=-\dfrac{11}{6}\) hoặc \(x=\dfrac{1}{6}:\left(-3\right)=\dfrac{1}{6}.\left(\dfrac{1}{-3}\right)=\dfrac{1}{-18}\)

22 tháng 7 2018

Thank bạn, mình tk rùi đó!

\(D=\dfrac{9x^8y^6\cdot\dfrac{1}{6}x^2y+\left(-16\right)}{15x^2y^2\cdot0.4\cdot ax^2y^2z^2}=\dfrac{\dfrac{3}{2}x^{10}y^7-16}{6ax^4y^4z^2}\)

a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)

\(B=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{25}{75}\cdot\dfrac{26}{8}=\dfrac{13}{12}\)

b: Để A<x<B thì -11/12<x<13/12

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

28 tháng 6 2017

\(x^2+5x< 0\)

\(x\left(x+5\right)< 0\)

\(\Leftrightarrow x< 0\)

\(\Leftrightarrow x+5>0\Leftrightarrow x>-5\)

\(-5< x< 0\)

\(x\in\left\{-4;-3;-2;-1\right\}\)

\(\Leftrightarrow x>0\)

\(\Leftrightarrow x-5< 0\Leftrightarrow x< 5\)

\(0< x< 5\)

\(x\in\left\{1;2;3;4\right\}\)

Vậy.......

19 tháng 4 2018

Tao có: \(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)

\(B>1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\right)\)

\(B>1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\right)\)

\(B>1-\left(1-\dfrac{1}{2004}\right)=1-1+\dfrac{1}{2004}=\dfrac{1}{2004}\left(đpcm\right)\)

18 tháng 12 2018

It's show time :)

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

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

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

hay \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (đpcm)