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26 tháng 6 2022
\(D=\dfrac{9x^8y^6\cdot\dfrac{1}{6}x^2y+\left(-16\right)}{15x^2y^2\cdot0.4\cdot ax^2y^2z^2}=\dfrac{\dfrac{3}{2}x^{10}y^7-16}{6ax^4y^4z^2}\)
29 tháng 9 2017
a) \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Rightarrow\) \(\dfrac{1}{6}=\dfrac{1}{30}:x\)
\(\Rightarrow\) \(x=\dfrac{1}{5}\)
TT
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...
19 tháng 4 2018
Tao có: \(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)
\(B>1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\right)\)
\(B>1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\right)\)
\(B>1-\left(1-\dfrac{1}{2004}\right)=1-1+\dfrac{1}{2004}=\dfrac{1}{2004}\left(đpcm\right)\)
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 \(Amax=5\Leftrightarrow x=\dfrac{1}{2}\)
27 tháng 12 2018
\(\dfrac{x+2}{x-4}=\dfrac{2x+3}{x-6}\) (ĐKXĐ: \(x\ne4;x\ne6\))
\(\Rightarrow\left(x+2\right)\left(x-6\right)=\left(2x+3\right)\left(x-4\right)\)
\(\Rightarrow x^2-6x+2x-12=2x^2-8x+3x-12\)
\(\Rightarrow x^2-x=0\)
\(\Rightarrow x\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
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.......
26 tháng 5 2022
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 \(x\in\left\{0;1\right\}\)
25 tháng 7 2022
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)
\(\Leftrightarrow\left(x+2\right).2=\left(2x+1\right).0,5\)
\(\Leftrightarrow2x+4=x+0,5\)
\(\Leftrightarrow x=-3,5\)
Vậy...
\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)
\(\Leftrightarrow\dfrac{4.\left(x+2\right)}{2}=\dfrac{2x+1}{2}\)
\(\Rightarrow4x+8=2x+1\)
\(\Leftrightarrow4x-2x=1-8\)
\(\Leftrightarrow2x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(x=\dfrac{-7}{2}\)