Question

Решите упражнение, даю 50 баллов!

Answer 1

$a^{3}+a^{2}b+ax+bx=a^{2}(a+b)+x(a+b)=(a+b)(a^{2}+x)$

a⁶+a⁵-a⁴-a³=a³(a+1)²(a-1)

-0,9x²-0,6xy-0,1y²=$-\frac{9}{10}x^{2}$$-\frac{3}{5}xy$$-\frac{1}{10}y^{2}$=$-\frac{1}{10}(9x^{2}+6xy+y^{2})=-\frac{1}{10}(3x+y)^{2}$

$t^{3}-8t^{2}+16t=t(t^{2}-8t+16)=t(t-4)^{2}$

81$-9x^{2}$-6xy-$y^{2}$=81-(9x²+6xy+y²)=81-(3x+y²)=(9-3x-y)(9+3x+y)

Answer 2

$a^6+a^5-a^4-a^3 = \\= a^3\left(a^3+a^2-a-1\right)=\\=a^3\left(a^2(a+1)-(a+1) \right)=\\=a^3\left((a+1)(a^2-1)\right)=\\=a^3\left((a+1)(a-1)(a+1)\right)=\\=a^3(a+1)^2(a-1)$

$a^3+a^2b+ax+bx =\\= a^2\left(a+b\right)+x\left(a+b\right)=\\= (a+b)(a^2+x)$

$-0,9x^2-0,6xy-0,1y^2=\\=-0,1(9x^2+6xy+y^2)=\\=-0,1((3x)^2+2\cdot3x\cdot y+y^2)=\\=-0,1(3x+y)^2$

$t^3-8t^2+16t=\\=t(t^2-8t+16)=\\=t(t^2-2\cdot t \cdot 4+4^2)=\\=t(t-4)^2$

$81-9x^2-6xy-y^2=\\=9^2-\left((3x)^2+2\cdot 3x\cdot y + y^2\right)\\=9^2-(3x+y)^2=\\=\left(9-(3x+y) \right)(9+3x+y)\\=(9-3x-y)(9+3x+y)$