50%x+1/8x=0,75
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1, (0,25 - \(x\)) : - \(\dfrac{3}{5}\) = - \(\dfrac{3}{4}\)
0,25 - \(x\) = - \(\dfrac{3}{4}\) x (- \(\dfrac{3}{5}\))
0,25 - \(x\) = \(\dfrac{9}{20}\)
\(x\) = 0,25 - 0,45
\(x\) = - 0,2
2, - \(\dfrac{3}{8}\)\(x\) - 0,75 = - 1\(\dfrac{1}{2}\)
- \(\dfrac{3}{8}\)\(x\) - 0,75 = -1,5
\(\dfrac{3}{8}\)\(x\) = - 0,75 + 1,5
\(\dfrac{3}{8}\)\(x\) = 0,75
\(x\) = 0,75 : \(\dfrac{3}{8}\)
\(x\) = 2
(25% : 0,45 + 50% x 50) x (0,15% - 0,75 x 0,2%)
=(25% : 0,45 + 50% x 50) x (0,15% - 0,15%)
=(25% : 0,45 + 50% x 50) x 0
=0
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(\Leftrightarrow x^2+x+3x+3-x^2+5x=11\)
\(\Leftrightarrow9x+3=11\)
\(\Leftrightarrow9x=11-3\)
\(\Leftrightarrow9x=8\)
\(\Leftrightarrow x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow\left(8x-24x^2+2-6x\right)+\left(24x^2-60x-4x+10\right)=-50\)
\(\Leftrightarrow2x-24x^2+2+24x^2-64x+10=-50\)
\(\Leftrightarrow-62x+12=-50\)
\(\Leftrightarrow-62x=-50-12\)
\(\Leftrightarrow-62x=-62\)
\(\Leftrightarrow x=\dfrac{-62}{-62}\)
\(\Leftrightarrow x=1\)
a) \(\left(x+3\right)\left(x+1\right)-x\left(x-5\right)=11\)
\(x^2+x+3x+3-x^2+5x=11\)
\(x+8x+3=11\)
\(x+8x=8\)
\(x\left(8+1\right)=8\)
\(x=\dfrac{8}{9}\)
b) \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(8x-24x^2+2-6x+24x^2-60x-4x+10=-50\)
\(-62x+12=-50\)
\(-62x=-62\)
\(x=1\)
ta có 50%x+12,5%x=0,75
62,5%x=0,75
x=0,75:62,5.100
x=1,2
x = 6-400%x