Question

Помогите пожалуйста (​

Answer 1

Ответ:

$3; \quad 7;$

Объяснение:

$(x-6)^{4}+(x-4)^{4}=82;$

$((x-6)^{2})^{2}+((x-4)^{2})^{2}=82;$

$((x-6)^{2})^{2}-2 \cdot (x-6)^{2} \cdot (x-4)^{2}+((x-4)^{2})^{2}+2 \cdot (x-6)^{2} \cdot (x-4)^{2}=82;$

$((x-6)^{2}-(x-4)^{2})^{2}+2 \cdot (x-6)^{2} \cdot (x-4)^{2}=82;$

$(x^{2}-12x+36-(x^{2}-8x+16))^{2}+2 \cdot (x-6)^{2} \cdot (x-4)^{2}=82;$

$(x^{2}-12x+36-x^{2}+8x-16)^{2}+2 \cdot (x-6)^{2} \cdot (x-4)^{2}=82;$

$(20-4x)^{2}+2 \cdot (x-6)^{2} \cdot (x-4)^{2}=82;$

$(4x-20)^{2}+2 \cdot (x-6)^{2} \cdot (x-4)^{2}=82;$

$(4(x-5))^{2}+2 \cdot (x-5-1)^{2} \cdot (x-5+1)^{2}=82;$

$4^{2}(x-5)^{2}+2 \cdot (x-5-1)^{2}\cdot (x-5+1)^{2}=82;$

$16(x-5)^{2}+2(x-5-1)^{2}(x-5+1)^{2}=82;$

$t=x-5;$

$16t^{2}+2(t-1)^{2}(t+1)^{2}=82;$

$16t^{2}+2((t-1)(t+1))^{2}=82;$

$16t^{2}+2(t^{2}-1)^{2}=82 \quad |:2$

$t^{4}-2t^{2}+1+8t^{2}-41=0;$

$t^{4}+6t^{2}-40=0;$

$(t^{2})^{2}+6t^{2}-40=0;$

$k=t^{2};$

$k^{2}+6k-40=0;$

$\displaystyle \left \{ {{k_{1}+k_{2}=-6} \atop {k_{1} \cdot k_{2}=-40}} \right. \Leftrightarrow \left \{ {{k_{1}=-10} \atop {k_{2}=4}} \right. ;$

Корень k₁ не имеет смысла.

$t^{2}=4 \Rightarrow t=\pm\sqrt{4}=\pm 2;$

$x-5=\pm 2 \Rightarrow x=5 \pm 2 \Rightarrow x=7 \ , \ x=3 \ ;$