Question

Помогите пожалуйста решить уравнения

Answer 1

Ответ:

Пошаговое объяснение:

$25x^2-20x+4=0\\(5x-2)^2=0\\x=\frac{2}{5}$

$7x^2-0,7x-2,1=0\\0,7(10x^2-x-3)=0\\D=1-4*10*(-3)=121=11^2\\x_{1}=\frac{1+11}{20}=0,6\\ x_{2}=\frac{1-11}{20}=-0,5$

$x^4-37x^2+36=0\\x^2=t\\t\geq 0\\t^2-37t+36=0\\(t-1)(t-36)=0\\t_{1}=1\\t_{2}=36\\x^2=1\\x_{1}=1\\x_{2}=-1\\x^2=36\\x{3}=6\\x_{4}=-6$

$3x^4-10x^2+3=0\\x^2=t\\t\geq 0\\3t^2-10t+3=0\\D=100-4*3*3=64=8^2\\t_{1}=\frac{10+8}{6}=3 \\t_{2}=\frac{10-8}{6}=\frac{1}{3} \\x^2=3\\x_{1}=\sqrt{3} \\x_{2}=-\sqrt{3}\\ x^2=\frac{1}{3} \\x_{3}=\frac{1}{\sqrt{3} } \\x_{4}=-\frac{1}{\sqrt{3} }$

$(x^2-2)^2-4(x^2-2)-21=0\\x^2-2=t\\t\geq -2\\t^2-4t-21=0\\D=16-4*1*(-21)=100=10^2\\t_{1}=\frac{4+10}{2}=7\\ t_{2}=\frac{4-10}{2}=-3$

$t_{2}$ не подходит

$x^2-2=7\\x^2=9\\x_{1}=3\\x_{2}=-3$