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a) \(0,75:4,5=\dfrac{1}{15}:\left(2x\right)\)
\(\Rightarrow\) \(\dfrac{1}{6}=\dfrac{1}{30}:x\)
\(\Rightarrow\) \(x=\dfrac{1}{5}\)
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...
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 \(\left|2x-3\right|=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=3:2=\dfrac{3}{2}\)
\(b,\) 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 \(Amax=5\Leftrightarrow x=\dfrac{1}{2}\)
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
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) \(|2x-3|+0,5=\dfrac{1}{3}:2\)
\(\Rightarrow|2x-3|+\dfrac{1}{2}=\dfrac{1}{3}.\dfrac{1}{2}\)
\(\Rightarrow|2x-3|+\dfrac{1}{2}=\dfrac{1}{6}\)
\(\Rightarrow|2x-3|=\dfrac{1}{6}-\dfrac{1}{2}=-\dfrac{1}{3}\)
Vì \(|2x-3|\ge0\forall x\)
=> 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 \(x=\dfrac{1}{6}:\left(-3\right)=\dfrac{1}{6}.\left(\dfrac{1}{-3}\right)=\dfrac{1}{-18}\)
\(D=\dfrac{9x^8y^6\cdot\dfrac{1}{6}x^2y+\left(-16\right)}{15x^2y^2\cdot0.4\cdot ax^2y^2z^2}=\dfrac{\dfrac{3}{2}x^{10}y^7-16}{6ax^4y^4z^2}\)
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 \(x\in\left\{0;1\right\}\)
\(x^2+5x< 0\)
\(x\left(x+5\right)< 0\)
\(\Leftrightarrow x< 0\)
\(\Leftrightarrow x+5>0\Leftrightarrow x>-5\)
\(-5< x< 0\)
\(x\in\left\{-4;-3;-2;-1\right\}\)
\(\Leftrightarrow x>0\)
\(\Leftrightarrow x-5< 0\Leftrightarrow x< 5\)
\(0< x< 5\)
\(x\in\left\{1;2;3;4\right\}\)
Vậy.......
Tao có: \(B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2004^2}\)
\(B>1-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2003\cdot2004}\right)\)
\(B>1-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2003}-\dfrac{1}{2004}\right)\)
\(B>1-\left(1-\dfrac{1}{2004}\right)=1-1+\dfrac{1}{2004}=\dfrac{1}{2004}\left(đpcm\right)\)
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)
\(\dfrac{x+2}{0,5}=\dfrac{2x+1}{2}\)
\(\Leftrightarrow\left(x+2\right).2=\left(2x+1\right).0,5\)
\(\Leftrightarrow2x+4=x+0,5\)
\(\Leftrightarrow x=-3,5\)
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 \(x=\dfrac{-7}{2}\)