Решите уравнение номер 576 577 8класс

Question 1

Question 2

№576

$1) \:\:x^2=7\\x=\pm\sqrt{7}\\x_1=\sqrt{7}\\x_2=-\sqrt{7}$

$2) \: \: x^2=11\\x=\pm\sqrt{11}\\x_1=\sqrt{11}\\x_2=-\sqrt{11}$

$3) \: \: x^2+6x=0\\x(x+6)=0\\x_1=0\\x+6=0\\x=-6$

$4) \: \: x^2+5x=0\\x(x+5)=0\\x_1=0\\x+5=0\\x_2=-5$

$5) \: \: x^2=8x\\x^2-8x=0\\x(x-8)=0\\x_1=0\\x-8=0\\x_2=8$

$6) \: \: x^2=12x\\x^2-12x=0\\x(x-12)=0\\x_1=0\\x-12=0\\x_2=12$

№577

$1) \: \: 1,5x-4x^2=6,3x-x^2\\-4x^2+x^2+1,5x-6,3x=0\\-3x^2-4,8x=0\\-x(3x+4,8)=0\\x(3x+4,8)=0\\x_1=0\\3x+4,8=0\\3x=-4,8\\x=-4,8\div3\\x_2=-1,6$

$2) \: \: 11y-15=(y+5)(y-3)\\11y-15=y^2-3y+5y-15\\11y=y^2+2y\\-y^2+11y-2y=0\\9y-y^2=0\\y(9-y)=0\\y_1=0\\9-y=0\\-y=-9\\y_2=9$

$3) \: \: 3x(x+2)=2x(x-2)\\3x^2+6x=2x^2-4x\\3x^2+6x-2x^2+4x=0\\x^2+10x=0\\x(x+10)=0\\x_1=0\\x+10=0\\x_2=-10$

$4) \: \: \dfrac{1}{4}(3x^2+1)-\dfrac{40x+3}{6}=\dfrac{x-3}{12} \: \: \: \: \: \: \bigg|\cdot12\\\\ 3\cdot(3x^2+1)-2\cdot(40x+3)=x-3\\9x^2+3-80x-6=x-3\\9x^2-80x-x+3-6+3=0\\9x^2-81x=0\\9x(x-9)=0\\9x=0\\x_1=0\\x-9=0\\x_2=9$

$5) \: \: \dfrac{y^2-5}{4}-\dfrac{15-y^2}{5}=\dfrac{y^2-4}{3}\:\:\:\:\:\: \bigg|\cdot60\\\\ 15\cdot(y^2-5)-12\cdot(15-y^2)=20\cdot(y^2-4)\\15y^2-75-180+12y^2=20y^2-80\\27y^2-255=20y^2-80\\27y^2-20y^2=-80+255\\7y^2=175\\y^2=175\div7\\y^2=25\\y=\pm\sqrt{25}\\y_1=5\\y_2=-5$

$6) \: \: \dfrac{2x^2-1}{4}=\dfrac{1+1,5x^2}{5} \: \: \: \: \: \: \bigg|\cdot20\\\\5\cdot(2x^2-1)=4\cdot(1+1,5x^2)\\10x^2-5=4+6x^2\\10x^2-6x^2=4+5\\4x^2=9\\x^2=9\div4\\x^2=2,25\\x=\pm\sqrt{2,25}\\x_1=1,5\\x_2=-1,5$