Question

Найдите корни уравнения .1. 6x(x+3)=2(x+23)
2. 5y(y-4)=10(y-8)
3.7(6-2t)=21(4-t)
4.4(2z-8)=6(6+2z)

Answer 1

$1. \; \; 6x(x+3)=2(x+23) \; \; \; \mid \; \div 2 \\\\</p><p>3x(x+3)=x+23 \\\\</p><p>3x^2+9x=x+23 \\\\</p><p>3x^2+9x-x-23=0 \\\\</p><p>3x^2+8x-23=0 \\\\</p><p>D=8^2-4\cdot 3\cdot (-23)=64+276=340=2\sqrt{85} \\\\</p><p>x_{1,2} = \dfrac{-8\pm 2\sqrt{85}}{6} \\\\</p><p>x_1 = \dfrac{-8+2\sqrt{85}}{6} = \dfrac{2(-4+\sqrt{85})}{6} = \dfrac{-4+\sqrt{85}}{3} \\\\</p><p>x_2=\dfrac{-8-2\sqrt{85}}{6} = \dfrac{2(-4-\sqrt{85})}{6} = \dfrac{-4-\sqrt{85}}{3}$

$3. \; \; 7(6-2t)=21(4-t) \; \; \; \mid \; \div \; 7 \\\\</p><p>6-2t=3(4-t) \\\\</p><p>6-2t=12-3t \\\\</p><p>-2t+3t=12-6 \\\\</p><p>t=6$

$2. \; \; 5y(y-4)=10(y-8) \; \; \; \mid \; \div \; 5 \\\\</p><p>y(y-4)=2(y-8) \\\\</p><p>y^2-4y=2y-16 \\\\</p><p>y^2-4y-2y+16=0 \\\\</p><p>y^2-6y+16=0 \\\\</p><p>D=(-6)^2-4\cdot 1\cdot 16 = 36-64 = -28<0 \; \; , \; \; net \: korney.$

$4. \; \; 7(6-2t)=21(4-t) \; \; \; \mid \; \div \; 7 \\\\</p><p>6-2t=3(4-t) \\\\</p><p>6-2t=12-3t \\\\</p><p>-2t+3t=12-6 \\\\</p><p>t=6$