Giải câu 1 trang 28 toán VNEN 9 tập 1

a) $\frac{1}{4}$$\sqrt{180}$ + $\sqrt{20}$ - $\sqrt{45}$ + 5 = $\frac{1}{4}$.6$\sqrt{5}$ + 2$\sqrt{5}$ - 3$\sqrt{5}$ + 5 = $\frac{\sqrt{5}}{2}$ + 5

b) 3$\sqrt{\frac{1}{3}}$ + $\frac{1}{4}$$\sqrt{48}$ - 2$\sqrt{3}$ = 3$\frac{\sqrt{3}}{3}$ + $\frac{1}{4}$.4$\sqrt{3}$ - 2$\sqrt{3}$ = 0

c) $\sqrt{2a}$ - $\sqrt{18a^{3}}$ + 4$\sqrt{\frac{a}{2}}$ = $\sqrt{2a}$ - 3a$\sqrt{2a}$ + 4.$\frac{\sqrt{2a}}{2}$ = 3$\sqrt{2a}$ - 3a$\sqrt{2a}$ = 3$\sqrt{2a}$(1 - a)

d) $\sqrt{\frac{a}{1 + 2b + b^{2}}}$.$\sqrt{\frac{4a + 8ab + 4ab^{2}}{225}}$ = $\sqrt{\frac{a}{(1 + 2b + b^{2}}}$.$\sqrt{\frac{4a(1 + 2ab + b^{2})}{225}}$ = $\frac{2a}{15}$.