\(0,75x\) - \(x\) + \(1^1_4\)\(x\)=20%
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a:=>0,75x-x+1,25x=0,2
=>x=0,2
b: =>1/3-x=-3/6+4/6=1/6
=>x=1/3-1/6=1/6
c: =>(x-1)/45=-6/30=-1/5
=>x-1=-9
=>x=-8
d: =>(2/5x-1)=-5/7
=>2/5x=2/7
=>x=5/7
a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
0,75 x - x + 1 2/5x = 45/100
-0,25x + 1,4x = 45/100
1,15x = 45/100
x = 45/100 : 1,15
x = 9/23
\(-\dfrac{2}{3}\left(x^2y^2\right)^2+\dfrac{1}{3}xy^2\cdot0,75x-3xy\cdot\left(-2x\right)\)
\(=-\dfrac{2}{3}x^4y^4+\dfrac{1}{3}xy^2\cdot\dfrac{3}{4}x-3xy\cdot\left(-2x\right)\)
\(=-\dfrac{2}{3}x^4y^4+\left(\dfrac{1}{3}\cdot\dfrac{3}{4}\right)\left(x\cdot x\right)y^2-\left(3\cdot-2\right)\left(x\cdot x\right)y\)
\(=-\dfrac{2}{3}x^4y^4+\dfrac{1}{4}x^2y^2+6x^2y\)
\(\left(-0,75x+\frac{5}{2}\right).\frac{7}{4}-\left(\frac{-1}{3}\right)=\frac{-5}{7}\)
\(\left(-0,75x+\frac{5}{2}\right).\frac{7}{4}=\frac{-5}{7}+\left(\frac{-1}{3}\right)\)
\(\left(-0,75x+\frac{5}{2}\right).\frac{7}{4}=\frac{-22}{21}\)
\(-0,75x+\frac{5}{2}=\frac{-22}{21}:\frac{7}{4}\)
\(-0,75x+\frac{5}{2}=\frac{-88}{147}\)
\(-0,75x=\frac{-88}{147}-\frac{5}{2}\)
\(-0,75x=\frac{-911}{294}\)
\(x=\frac{-911}{294}:\left(-0,75\right)\)
\(x=\frac{1822}{441}\)