651 Вправа 651Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 193

651. Спростіть вираз:

1) ${\displaystyle {(\sqrt{x}+3)}^2+{(\sqrt{x}-3)}^2=}$${\displaystyle ({\left(\sqrt{x}\right)}^2+2\cdot \sqrt{x}\cdot 3+3^2)+({\left(\sqrt{x}\right)}^2-2\cdot \sqrt{x}\cdot 3+3^2)=}$${\displaystyle (x+6\sqrt{x}+9)+(x-6\sqrt{x}+9)=}$${\displaystyle x+6\sqrt{x}+9+x-6\sqrt{x}+9=}$${\displaystyle 2x+18}$;

2) ${\displaystyle {(\sqrt{a}+5\sqrt{2})}^2-{(\sqrt{a}-5\sqrt{2})}^2=}$${\displaystyle ({\left(\sqrt{a}\right)}^2+2\sqrt{a}\cdot 5\sqrt{2}+{\left(5\sqrt{2}\right)}^2)-({\left(\sqrt{a}\right)}^2-2\sqrt{a}\cdot 5\sqrt{2}+{\left(5\sqrt{2}\right)}^2)=}$${\displaystyle (a+10\sqrt{2a}+50)-(a-10\sqrt{2a}+50)=}$${\displaystyle a+10\sqrt{2a}+50-a+10\sqrt{2a}-50=}$${\displaystyle 20\sqrt{2a}}$;

3) ${\displaystyle {(2+\sqrt{y})}^2-{(1-2\sqrt{y})}^2=}$${\displaystyle (2^2+2\cdot 2\sqrt{y}+{\left(\sqrt{y}\right)}^2)-(1^2-2\cdot 1\cdot 2\sqrt{y}+{\left(2\sqrt{y}\right)}^2)=}$${\displaystyle (4+4\sqrt{y}+y)-(1-4\sqrt{y}+4y)=}$${\displaystyle 4+4\sqrt{y}+y-1+4\sqrt{y}-4y=}$${\displaystyle 3+8\sqrt{y}-3y}$;

4) ${\displaystyle (\sqrt{x}+4)(\sqrt{x}-4)-{(-\sqrt{x}+4)}^2=}$${\displaystyle ({\left(\sqrt{x}\right)}^2-4^2)-{(4-\sqrt{x})}^2=}$${\displaystyle (x-16)-(16-8\sqrt{x}+x)=}$${\displaystyle x-16-16+8\sqrt{x}-x=}$${\displaystyle 8\sqrt{x}-32}$;

5) ${\displaystyle {(\sqrt{m}+\sqrt{n})}^2-(\sqrt{m}+\sqrt{n})(\sqrt{m}-\sqrt{n})=}$${\displaystyle (m+2\sqrt{mn}+n)-({\left(\sqrt{m}\right)}^2-{\left(\sqrt{n}\right)}^2)=}$${\displaystyle m+2\sqrt{mn}+n-(m-n)=}$${\displaystyle m+2\sqrt{mn}+n-m+n=}$${\displaystyle 2\sqrt{mn}+2n}$;

6) ${\displaystyle (\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})-{(-\sqrt{x}-\sqrt{y})}^2=}$${\displaystyle ({\left(\sqrt{x}\right)}^2-{\left(\sqrt{y}\right)}^2)-{(\sqrt{x}+\sqrt{y})}^2=}$${\displaystyle (x-y)-(x+2\sqrt{xy}+y)=}$${\displaystyle x-y-x-2\sqrt{xy}-y=}$${\displaystyle -2y-2\sqrt{xy}}$.