Question

Ради бога помогите пожалуйста

Answer 1

$1)\; \; (4a^2b^4)^3+(2ab^2)^6+(8a^3b^5)^2=64a^6b^{12}+64a^6b^{12}+64a^6b^{10}=\\\\=128a^6b^{12}+64a^6b^{10}$

$2)\; \; (ax^2+bx+c)+(dx^2+nx+m)=-2x+6\; \; \Rightarrow \\\\(a+d)x^2+(b+n)x+(c+m)=0\cdot x^2-2x+6\; \; \Rightarrow \\\\a+d=0\; ,\; b+n=-2\; ,\; \; c+m=6\\\\\\(ax^2+bx+c)-(dx^2+nx+m)=4x^2\; \; \Rightarrow \\\\(a-d)x^2+(b-n)x+(c-m)=4x^2+0\cdot x+0\; \; \Rightarrow \\\\a-d=4\; \; ,\; \; b-n=0\; ,\; \; c-m=0\\\\\left \{ {{a+d=0} \atop {a-d=4}} \right.\; \; \left \{ {{d=-a} \atop {2a=4}} \righ\; \left \{ {{d=-2} \atop {a=2}} \right.$

$\left \{ {{b+n=-2} \atop {b-n=0}} \right.\; \left \{ {{2n=-2} \atop {2b=-2}} \right.\; \left \{ {{n=-1} \atop {b=-1}} \right.\\\\\left \{ {{c+m=6} \atop {c-m=0}} \right.\; \left \{ {{2c=6} \atop {2m=6}} \right. \left \{ {{c=3} \atop {m=3}} \right.\\\\(2x^2-x+3)+(-2x^2-x+3)=-2x+6\\\\(2x^2-x+3)-(-2x^2-x+3)=4x^2$

$3)\; \; ax^{k}\cdot bx^{m}\cdot cx^{n}=65x^4\; \; \Rightarrow \; \; \; a\cdot b\cdot c=65\; \; ,\; \; k+n+m=4\\\\65=1\cdot 5\cdot 13\; \; ,\; \; \; 1+1+2=4\\\\x^2\cdot 5x\cdot 13x=65x^4\\\\5x^2\cdot x\cdot 13x=65x^4\\\\5x^2\cdot 13x\cdot x=65x^4\\\\x^2\cdot 13x\cdot 5x=65x^4\\\\65x^2\cdot x\cdot x=65x^4\\\\x^2\cdot \; 65x\cdot x=65x^4\\\\x^2\cdot x\cdot \; 65x=65x^4$

$65x^2+x+x=65x^2+2x\\\\x^2+65x+x=x^2+66x\\\\x^2+x+65x=x^2+66x\\\\x^2+5x+13x=x^2+18x\\\\5x^2+x+13x=5x^2+14x\\\\5x^2+13x+x=5x^2+14x\\\\x^2+13x+5x=x^2+18x$