Question

Помогите пожалуйста буду благодарна

Answer 1

1a)

$(5x-1)(5x+2)-2=(5x+3)^2\\25x^2+10x-5x-2-2-(5x+3)^2=0\\25x^2+5x-4-(25x^2+30x+9)=0\\25x^2+5x-4-25x^2-30x-9=0\\-25x=13\\x=-\frac{13}{25}$

2б)

$x^4-24x^2+44=0$

замена  $x^2=t$

$t^2-24t+44=0\\D=576-176=400=20^2\\t_1=\frac{24+20}{2}= 22\\t_2=\frac{24-20}{2}=2$

обратная замена:

$\left \{ {{x^2=22} \atop {x^2=2}} \right. =>\left \{ {{x=\pm\sqrt{22} } \atop {x=\pm\sqrt{2} }} \right.$

x1=√22; x2=-√22

x3=√2;  x4=-√2

3в)

$(x^2-8)^2-3(x^2-8)+2=0$

замена  $x^2-8=t$

$t^2-3t+2=0\\D=9-8=1\\t_1=(3-1)/2=1\\t_2=(3+1)/2=2$

обратная замена:

$\left \{ {{x^2-8=1} \atop {x^2-8=2}} \right. =>\left \{ {{x^2=9} \atop {x^2=10}} \right. =>\left \{ {{x=\pm3} \atop {x=\pm\sqrt{10} }} \right.$

x1=-3; x2=3

x3=-√10; x4=√10

2)

$x^3+16x^2=4x+64\\x^3+16x^2-4x-64=0\\x^2(x+16)-4(x+16)=0\\(x+16)(x^2-4)=0\\(x+16)(x-2)(x+2)=\\x_1=-1\\x_2=2\\x_3=-2$