Помогите пожалуйста

Question

Алгебра

Vaschasno

6 год назад

Помогите пожалуйста

ОТВЕТЫ

Monervam

Jun 30, 2019

$(x^2-9)^2-4(x^2-9)+3=0\\(x^2-9)=y\\y^2-4y+3=0\\ \left \{ {y_2y_2=3} \atop {y_1+y_2=4}} \right. [ { {{y=1} \atop {y=3}} \right.\\x^2-9=1\\x^2=10 [{ {{x= \sqrt{10} } \atop {x=- \sqrt{10} }} \right. \\\\x^2-9=3\\x^2=12[{ {{x= \sqrt{12}=2 \sqrt{3} } \atop {x=- \sqrt{12}=-2 \sqrt{3} }} \right.$
Ответ: $-2 \sqrt{3};- \sqrt{10}; \sqrt{10};2 \sqrt{3}$

$(x^2+x+1)(x^2+x+2)=12\\(x^2+x+1)=y\\y(y+1)=12\\y^2+y-12=0\\ \left \{ {{y_1y_2=-12} \atop {y_1+y_2=-1}} \right. [{ {{y=-4} \atop {y=3}} \right. \\x^2+x+1=-4\\x^2+x+5=0\\\\x^2+x+1=3\\x^2+x-2=0\\ \left \{ {{x_1x_2=-2} \atop {x_1+x_2=-1}} \right. [{ {{x=-2} \atop {x=1}} \right.$
Ответ: $-2;1$

$16x^2-25x^2+9=0\\-9x^2=-9\\x^2=1\\ { {{x=-1} \atop {x=1}} \right.$
Ответ: $-1;1$

$x^4+5x^2-36=0\\x^2=y\\y^2+5y-36=0\\ \left \{ {{y_1y_2=-36} \atop {y_1+y_2=-5}} \right. [{ {{y=-9} \atop {y=4}} \right. \\x^2=-9\\\\x^2=4\\ { {{x=2} \atop {x=-2}} \right.$
Ответ: $-2;2$

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