Question

розв'яжіть рівняння ​

Answer 1

$\displaystyle 1)x^4-17x^2+16=0\\t^2-17t+16=0\\t=1,t=16\\x^2=1,x^2=16\\x_1=-4,x_2=-1,x_3=1,x_4=4\\\\2)\frac{2x+3}{x+2}=\frac{3x+2}{3x},x\neq -2,x\neq 0\\ \\(2x+3)*3x=(3x+2)(x+2)\\(2x+3)*3x-(3x+2)(x+2)=0\\6x^2+9x-(3x^2+6x+2x+4)=0\\6x^2+9x-(3x^2+8x+4)=0\\6x^2+9x-3x^2-8x-4=0\\3x^2+x-4=0\\3x^2+4x-3x-4=0\\x(3x+4)-(3x+4)=0\\(3x+4)(x-1)=0\\3x+4=0,x-1=0\\\\x_1=-\frac{4}{3},x_2=1\\ \\3)(x-1)^4+21(x-1)^2-100=0\\t^2+21t-100=0\\t=-25,t=4\\(x-1)^2=-25,(x-1)^2=4\\x \notin R, x=-1,x=3\\x_1=-1,x_2=3$

Answer 2

Ответ:

а) (±1; ±4) б) (1; -4/3) в) (-1; 3)

Объяснение:

$a)x^{4}-17x^{2} +16=0\\ x^{2} =y\\y^{2}-17y+16=0\\ y^{2}-16y-y+16=0\\ y(y-16)-1(y-16)=0\\(y-1)(y-16)=0\\y-1=0\\ y-16=0\\y_{1}=1\\y_{2} =16\\x^{2} =y_{1}\\ x^{2} =1\\ x=+-1\\ x^{2} =y_{2}\\ x^{2} =16\\x=+-4$

$\frac{2x+3}{x+2}=\frac{3x+2}{3x}\\ (2x+3)*3x=(3x+2) *(x+2)\\6x^{2} +9x=3x^{2} +6x+2x+4\\6x^{2} +9x-3x^{2} -8x-4=0\\3x^{2} +x-4=0\\3x^{2} -3x+4x-4=0\\3x(x-1)+4(x-1)=0\\(3x+4)(x-1)=0\\x-1=0\\3x+4=0\\x_{1}=1\\ x_{2}=-\frac{4}{3}$

$(x-1)^{4} +21(x-1)^{2} -100=0\\(x-1)^{2}=y\\ y^{2} +21y-100=0\\y^{2} -4y+25y-100=0\\y(y-4)+25(y-4)=0\\(y+25)(y-4)=0\\y+25=0\\y-4=0\\y_{1}=-25\\ y_{2}=4\\ (x-1)^{2} =y_{1} \\(x-1)^{2}=-25\\\neq \\(x-1)^{2}=y_{2}\\ (x-1)^{2}=4\\x-1=4^{2}\\ x-1=+-2\\x-1=2\\x_{1} =3\\x-1=-2\\x_{2}=-1$