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Question 1

Question 2

№7 1)

$\frac{x^2}{5}-\frac{2x}{3}=\frac{x^2-5}{6} \\6x^2-20x=5x^2-25 \\x^2-20x+25=0 \\D=b^2-4ac=400-100=300 \\x_1=\frac{-b+\sqrt{D}}{2a}=\frac{20+\sqrt{300}}{2}= \frac{20+10\sqrt{3}}{2} =10+5\sqrt{3}\\x_2=10-5\sqrt{3}$

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2)

$\frac{x(x-7)}{3}-1=\frac{11x}{10}-\frac{x-4}{3} \\10x^2-70x-30=33x-10x+40 \\10x^2-93x-70=0 \\D=b^2-4ac=8649+2800=11449=107^2 \\x_1=\frac{-b+\sqrt{D}}{2a}=\frac{93+107}{2*10}=10 \\x_2=\frac{-b+\sqrt{D}}{2a}=-\frac{14}{20}=- 0.7$

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$\frac{(x+3)^2}{5}+1-\frac{(3x-1)^2}{5}=\frac{x(2x-3)}{2} \\\\2x^2+12x+18+10-18x^2+12x-2=10x^2-15x \\-26x^2+39x+26=0 \\2x^2-3x-2=0 \\D=b^2-4ac=9+16=25 \\x_1=\frac{-b+\sqrt{D}}{2a}=\frac{3+5}{4}=2 \\x_2=\frac{-b-0\sqrt{D}}{2a}=\frac{3-5}{4}=-0.5$

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$\frac{5x-0.25x^2}{3}-\frac{(x-11)^2}{4}=6.25-\frac{(7-x)^2}{2} \\20x-x^2-3x^2+66x-363=75-294+84x-6x^2 \\2x^2+2x-144=0 \\x^2+x-72=0 \\D=b^2-4ac=1+288=289=17^2 \\x_1=\frac{ -b+\sqrt{D}}{2a}=\frac{-1+17}{2}=8 \\x_2=\frac{ -b-\sqrt{D}}{2a}=\frac{-1-17}{2}=-9$

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$\frac{2x^2}{5}-\frac{4x-2}{3}=-0.2x \\6x^2-20x+10=-3x \\6x^2-17x+10=0 \\D=b^2-4ac=289-240=49=7^2 \\x_1=\frac{-b+\sqrt{D}}{2a}=\frac{17+7}{12}=2 \\x_2=\frac{-b-\sqrt{D}}{2a}=\frac{10}{12}=\frac{5}{6}$

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$\frac{x^2-x}{3}-\frac{4+x}{5}=0.2x \\\\5x^2-5x-12-3x=3x \\5x^2-11x-12=0\\D=b^2-4ac=121+240=361=19^2 \\x_1=\frac{-b+\sqrt{D}}{2a}=\frac{11+19}{10}=3 \\x_2=\frac{-b-\sqrt{D}}{2a}=\frac{11-19}{10}=-0.8$

Question 3

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AlbuRin_zn · Answer 1 · 2018-03-11T22:58:32+0000