Verified answer

Пояснення:

№1.

[tex]2x^2-7x+5=2x^2-2x-5x+5=2x*(x-1)-5*(x-1)=(x-1)*(2x-5).[/tex]

№2.

[tex]x^4-17x^2+16=0\\\\(x^2)^2-17x^2+15=0.[/tex]

Нехай х²=t ≥0           ⇒

[tex]t^2-17t+16=0\\\\t^2-16t-t+16=0\\\\t*(t-16)-(t-16)=0\\\\(t-16)*(t-1)=0\\\\t-16=0\ \ \ \ \ \ \Rightarrow\\\\x^2-16=0\\\\x^2-4^2=0\\\\(x-4)*(x+4)=0\\\\x-4=0\\\\x_1=4.\\\\x+4=0\\\\x_2=-4.\\\\t-1=0\ \ \ \ \ \ \Rightarrow\\\\x^2-1^2=0\\\\(x-1)*(x+1)=0\\\\x-1=0\\\\x_3=1.\\\\x+1=0\\\\x_4=-1.[/tex]

Відповідь: х₁=4,  х₂=-4,  х₃=1,  х₄=-1.

№3.

ОДЗ: х+3≠0    х=-3   х-3≠0   х≠3          ⇒     х≠±3.

[tex]\displaystyle\\\frac{x-1}{x+3} +\frac{x+1}{x-3} =\frac{2x+18}{x^2-9} \\\\\frac{x-1}{x+3} +\frac{x+1}{x-3} =\frac{2x+18}{(x-3)*(x+3)} \\\\(x-1)*(x-3)+(x+1)*(x+3)=2x+18\\\\x^2-4x+3+x^2+4x+3=2x+18\\\\2x^2+6=2x+18=0\\\\2x^2-2x-12=0\ |:2\\\\x^2-x-6=0\\\\x^2-3x+2x-6=0\\\\x*(x-3)+2*(x-3)=0\\\\(x-3)*(x+2)=0\\\\x-3=0\\\\x_1=3\notin(x\neq 3)\\\\x+2=0\\\\x_2=-2.[/tex]

Відповідь: х=-2.