Equazioni

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129)

$\small \dfrac{2[(x-4)(2x-1)+7]}{3}-\left(x-\dfrac{1}{4}\right)\left(\dfrac{1}{4}+x\right) = \dfrac{1}{3}x^2+\dfrac{22}{3}-x+\dfrac{17}{16}$

$\small \dfrac{2[2x^2-x-8x+4+7]}{3}-\left(\cancel{\dfrac{1}{4}x}+x^2-\dfrac{1}{16}\cancel{-\dfrac{1}{4}x}\right) = \dfrac{1}{3}x^2+\dfrac{22}{3}-x+\dfrac{17}{16}$

$\small \dfrac{2[2x^2-9x+11]}{3}-\left(x^2-\dfrac{1}{16}\right) = \dfrac{1}{3}x^2+\dfrac{22}{3}-x+\dfrac{17}{16}$ 

$\small \dfrac{4x^2-18x+22}{3}-x^2+\dfrac{1}{16} = \dfrac{1}{3}x^2+\dfrac{22}{3}-x+\dfrac{17}{16}$ $\quad \small (mcm=48)$

$\small 16(4x^2-18x+22)-48x^2+3 = 16x^2+352-48x+51$

$\small 64x^2-288x+352-48x^2+3 = 16x^2-48x+403$

$\small 16x^2-288x+355 = 16x^2-48x+403$

$\small \cancel{16x^2}-288x\cancel{-16x^2}+48x = 403-355$

$\small -240x = 48$

$\small 240x = -48$

$\small \dfrac{240x}{240} = \dfrac{-48}{240}$

$\small \dfrac{\cancel{240}x}{\cancel{240}} = -\dfrac{\cancel{48}^1}{\cancel{240}_5}$

$\small x= -\dfrac{1}{5}$

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130)

$\small \left[\left(2-\dfrac{1}{2}x\right)\left(-2-\dfrac{1}{2}x\right)+\left(x-2\right)^2\right]·\left(-\dfrac{4}{5}\right)-\dfrac{1}{5}x = 1-x(x-5)$

$\small \left[\cancel{-4}\cancel{-x}\cancel{+x}+\dfrac{1}{4}x^2+x^2-4x\cancel{+4}\right]·\left(-\dfrac{4}{5}\right)-\dfrac{1}{5}x = 1-x^2+5x$

$\small \left[\dfrac{1}{4}x^2+x^2-4x\right]·\left(-\dfrac{4}{5}\right)-\dfrac{1}{5}x = 1-x^2+5x$

$\small -\dfrac{\cancel4^1}{\cancel{20}_5}x^2-\dfrac{4}{5}x^2+\dfrac{16}{5}x-\dfrac{1}{5}x = 1-x^2+5x$

$\small -\dfrac{1}{5}x^2-\dfrac{4}{5}x^2+\dfrac{\cancel{15}^3}{\cancel5_1}x = 1-x^2+5x$

$\small -\dfrac{\cancel5^1}{\cancel5_1}x^2+3x = 1-x^2+5x$

$\small -x^2+3x+x^2-5x = 1$

$\small \cancel{-x^2}+3x\cancel{+x^2}-5x = 1$

$\small -2x = 1$

$\small 2x = -1$

$\small \dfrac{\cancel2x}{\cancel2} = -\dfrac{1}{2}$

$\small x = -\dfrac{1}{2}$