a)  $\frac{{\left(2x+3\right)}^2}{2}=12x\mid \cdot 2$  

${\left(2x+3\right)}^2=24x$  

$4x^2+12x+9=24x\mid -24x$  

$4x^2-12x+9=0$  

$\Delta ={\left(-12\right)}^2-4\cdot 4\cdot 9=144-144=0$  

$x=\frac{-\left(-12\right)}{2\cdot 4}=\frac{12}{8}=\frac{3}{2}=1\frac{1}{2}$  

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b)  ${\left(x-\frac{3}{2}\right)}^2-\frac{1}{4}=4-\frac{2}{3}x$  

$x^2-3x+\frac{9}{4}-\frac{1}{4}=4-\frac{2}{3}x$  

$x^2-3x+\frac{8}{4}=4-\frac{2}{3}x$  

$x^2-3x+2-4+\frac{2}{3}x=0$  

$x^2-2\frac{1}{3}x-2=0$  

$x^2-\frac{7}{3}x-2=0$  

$\Delta ={\left(-\frac{7}{3}\right)}^2-4\cdot 1\cdot \left(-2\right)=\frac{49}{9}+8=\frac{49}{9}+\frac{72}{9}=\frac{121}{9}$  

$\sqrt{\Delta }=\frac{11}{3}$  

$x_1=\frac{-\left(-\frac{7}{3}\right)-\frac{11}{3}}{2\cdot 1}=\frac{\frac{7}{3}-\frac{11}{3}}{2}=\frac{-\frac{4}{3}}{2}=\left(-\frac{4}{3}\right):\frac{2}{1}=\left(-\frac{4}{3}\right)\cdot \frac{1}{2}=-\frac{4}{6}=-\frac{2}{3}$  

$x_2=\frac{-\left(-\frac{7}{3}\right)+\frac{11}{3}}{2\cdot 1}=\frac{\frac{7}{3}+\frac{11}{3}}{2}=\frac{\frac{18}{3}}{2}=\frac{6}{2}=3$ $x_2=\frac{-\left(-\frac{7}{3}\right)+\frac{11}{3}}{2\cdot 1}=\frac{\frac{7}{3}+\frac{11}{3}}{2}=\frac{\frac{18}{3}}{2}=\frac{6}{2}=3$  

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c)  $\frac{1}{2}{\left(1-6x\right)}^2=2\left(x-0,5\right)\left(x+0,5\right)$  

$\frac{1}{2}\left(1-12x+36x^2\right)=2\left(x^2-\frac{1}{4}\right)$  

$\frac{1}{2}-6x+18x^2=2x^2-\frac{1}{2}$  

$\frac{1}{2}-6x+18x^2-2x^2+\frac{1}{2}=0$  

$16x^2-6x+1=0$  

$\Delta ={\left(-6\right)}^2-4\cdot 16\cdot 1=36-64=-28$  

Brak rozwiązań.

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d)  $\frac{1}{3}\left(4x-3\right)\left(4x+3\right)=6x$  

$\frac{1}{3}\left(16x^2-9\right)=6x\mid \cdot 3$  

$16x^2-9=18x$  

$16x^2-18x-9=0$  

$\Delta ={\left(-18\right)}^2-4\cdot 16\cdot \left(-9\right)=324+576=900$  

$\sqrt{\Delta }=30$  

$x_1=\frac{-\left(-18\right)-30}{2\cdot 16}=\frac{18-30}{32}=\frac{-12}{32}=-\frac{3}{8}$  

$x_2=\frac{-\left(-18\right)+30}{2\cdot 16}=\frac{18+30}{32}=\frac{48}{32}=\frac{3}{2}=1\frac{1}{2}$  

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e)  $4x^2-\frac{6x+8}{3}=\frac{1}{3}\left(x+1\right)\left(2x-3\right)\mid \cdot 3$  

$12x^2-\left(6x+8\right)=\left(x+1\right)\left(2x-3\right)$  

$12x^2-6x-8=2x^2-3x+2x-3$  

$12x^2-6x-8-2x^2+3x-2x+3=0$  

$10x^2-5x-5=0\mid :5$  

$2x^2-1x-1=0$  

$\Delta ={\left(-1\right)}^2-4\cdot 2\cdot \left(-1\right)=1+8=9$  

$\sqrt{\Delta }=3$  

$x_1=\frac{-\left(-1\right)-3}{2\cdot 2}=\frac{1-3}{4}=\frac{-2}{4}=-\frac{1}{2}$  

$x_2=\frac{-\left(-1\right)+3}{2\cdot 2}=\frac{1+3}{4}=\frac{4}{4}=1$  

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f)  $\frac{{\left(x-2\right)}^2}{5}+\frac{x\left(x+3\right)}{10}=0,3x\left(2x-1\right)\mid \cdot 10$  

$2{\left(x-2\right)}^2+x\left(x+3\right)=3x\left(2x-1\right)$  

$2\left(x^2-4x+4\right)+x^2+3x=6x^2-3x$  

$2x^2-8x+8+x^2+3x-6x^2+3x=0$  

$-3x^2-2x+8=0$  

$\Delta ={\left(-2\right)}^2-4\cdot \left(-3\right)\cdot 8=4+96=100$  

$\sqrt{\Delta }=10$  

$x_1=\frac{-\left(-2\right)-10}{2\cdot \left(-3\right)}=\frac{2-10}{-6}=\frac{-8}{-6}=\frac{4}{3}=1\frac{1}{3}$  

$x_2=\frac{-\left(-2\right)+10}{2\cdot \left(-3\right)}=\frac{2+10}{-6}=\frac{12}{-6}=-2$  

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g)  $\frac{{\left(1-3x\right)}^2}{2}-\frac{x\left(4-x\right)}{3}=\frac{5}{6}+\frac{x+9}{2}\mid \cdot 6$  

$3{\left(1-3x\right)}^2-2x\left(4-x\right)=5+3\left(x+9\right)$  

$3\left(1-6x+9x^2\right)-8x+2x^2=5+3x+27$  

$3-18x+27x^2-8x+2x^2=32+3x$  

$29x^2-26x+3-32-3x=0$  

$29x^2-29x-29=0\mid :29$  

$x^2-x-1=0$  

$\Delta ={\left(-1\right)}^2-4\cdot 1\cdot \left(-1\right)=1+4=5$  

$\sqrt{\Delta }=\sqrt{5}$  

$x_1=\frac{-\left(-1\right)-\sqrt{5}}{2\cdot 1}=\frac{1-\sqrt{5}}{2}$  

$x_2=\frac{-\left(-1\right)+\sqrt{5}}{2\cdot 1}=\frac{1+\sqrt{5}}{2}$  

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h)  $\frac{1}{3}{x}^2+\frac{{\left(x+2\right)}^2}{5}=\frac{\left(x-1\right)\left(x+1\right)}{5}+x-2\mid \cdot 15$  

$5x^2+3{\left(x+2\right)}^2=3\left(x-1\right)\left(x+1\right)+15x-30$  

$5x^2+3\left(x^2+4x+4\right)=3\left(x^2-1\right)+15x-30$  

$5x^2+3x^2+12x+12=3x^2-3+15x-30$  

$8x^2+12x+12=3x^2+15x-33$  

$8x^2+12x+12-3x^2-15x+33=0$  

$5x^2-3x+45=0$  

$\Delta ={\left(-3\right)}^2-4\cdot 5\cdot 45=9-900=-891$  

Brak rozwiązań.