34 Вправа 34 (повторення)Алгебра 8 клас Тарасенкова 2025 · Повторення · стор. 320

34. Спростіть вираз:

1) ${\displaystyle (\frac{\sqrt{a}+2}{\sqrt{a}-2}-\frac{4\sqrt{a}}{a-4}):\frac{a+4}{\sqrt{a}+2}=}$${\displaystyle (\frac{{\sqrt{a}+2}^{\mathrm{\setminus }\sqrt{a}+2}}{\sqrt{a}-2}-\frac{4\sqrt{a}}{(\sqrt{a}-2)(\sqrt{a}+2)})\cdot \frac{\sqrt{a}+2}{a+4}=}$${\displaystyle \frac{a+4\sqrt{a}+4-4\sqrt{a}}{(\sqrt{a}-2)\cancel{\left(\sqrt{a}+2\right)}}\cdot \frac{\cancel{\sqrt{a}+2}}{a+4}=}$${\displaystyle \frac{\cancel{a+4}}{(\sqrt{a}-2)\cancel{\left(a+4\right)}}=}$${\displaystyle \frac{1}{\sqrt{a}-2}}$;

2) ${\displaystyle (\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1})\cdot (x-1)=}$${\displaystyle (\frac{{\sqrt{x}+1}^{\mathrm{\setminus }\sqrt{x}+1}}{\sqrt{x}-1}-\frac{{\sqrt{x}-1}^{\mathrm{\setminus }\sqrt{x}-1}}{\sqrt{x}+1})\cdot (x-1)=}$${\displaystyle \frac{{(\sqrt{x}+1)}^2-{(\sqrt{x}-1)}^2}{(\sqrt{x}-1)(\sqrt{x}+1)}\cdot (x-1)=}$${\displaystyle \frac{\cancel{x}+2\sqrt{x}\cancel{+1}-\left(\cancel{x}-2\sqrt{x}\cancel{+1}\right)}{\cancel{x-1}}\cdot \cancel{\left(x-1\right)}=}$${\displaystyle 4\sqrt{x}}$;

3) ${\displaystyle \frac{\sqrt{x}+2}{x-2\sqrt{x}+1}:\frac{x-4}{3\sqrt{x}-3}-\frac{3}{\sqrt{x}-2}=}$${\displaystyle \frac{\cancel{\sqrt{x}+2}}{{(\sqrt{x}-1)}^{\stackrel{1}{\cancel{2}}}}\cdot \frac{3\cancel{\left(\sqrt{x}-1\right)}}{(\sqrt{x}-2)\cancel{\left(\sqrt{x}+2\right)}}-\frac{3}{\sqrt{x}-2}=}$${\displaystyle \frac{3}{(\sqrt{x}-1)(\sqrt{x}-2)}-\frac{3^{\mathrm{\setminus }\sqrt{x}-1}}{\sqrt{x}-2}=}$${\displaystyle \frac{3-3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}-2)}=}$${\displaystyle \frac{3-3\sqrt{x}+3}{(\sqrt{x}-1)(\sqrt{x}-2)}=}$${\displaystyle \frac{6-3\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}-2)}=}$${\displaystyle \frac{-3\cancel{\left(\sqrt{x}-2\right)}}{(\sqrt{x}-1)\cancel{\left(\sqrt{x}-2\right)}}=}$${\displaystyle \frac{-3}{\sqrt{x}-1}=}$${\displaystyle \frac{3}{1-\sqrt{x}}}$.