Câu 20: Trang 15 - sgk toán 9 tập 1
Rút gọn các biểu thức sau :
a. $\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}} (a\geq 0)$
b. $\sqrt{13a}.\sqrt{\frac{52}{a}} (a> 0)$
c. $\sqrt{5a}.\sqrt{45a}-3a (a \geq 0)$
d. $(3-a)^{2}-\sqrt{0,2}.\sqrt{180a^{2}}$
Bài Làm:
Ta có :
a. $\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{2a}{3}.\frac{3a}{8}}=\sqrt{\frac{a^{2}}{4}}=\sqrt{(\frac{a}{2})^{2}}=\left | \frac{a}{2} \right |(a\geq 0)$
Vậy $\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\left | \frac{a}{2} \right |$
b. $\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{13a.\frac{52}{a}}=\sqrt{13.13.4}=\sqrt{26^{2}}=26 (a> 0)$
Vậy $\sqrt{13a}.\sqrt{\frac{52}{a}}=26 $
c. $\sqrt{5a}.\sqrt{45a}-3a =\sqrt{5a.45a}-3a=\sqrt{(15a)^{2}}-3a=\left | 15a \right |-3a=15a-3a=12a(a \geq 0)$
Vậy $\sqrt{5a}.\sqrt{45a}-3a =12a$
d. $(3-a)^{2}-\sqrt{0,2}.\sqrt{180a^{2}}=(3-a)^{2}-\sqrt{0,2.180a^{2}}=(3-a)^{2}-\sqrt{(6a)^{2}}$
<=> $9-6a+a^{2}-\left | 6a \right |=\left\{\begin{matrix}9-6a+a^{2}-6a(a\geq 0) & \\ 9-6a+a^{2}+6a(a<0) & \end{matrix}\right.=\left\{\begin{matrix}9-12a+a^{2}(a\geq 0) & \\ 9+a^{2}(a<0) & \end{matrix}\right.$
Vậy $(3-a)^{2}-\sqrt{0,2}.\sqrt{180a^{2}}=\left\{\begin{matrix}9-12a+a^{2}(a\geq 0) & \\ 9+a^{2}(a<0) & \end{matrix}\right.$