# Bài 20 trang 15 SGK Toán 9 tập 1

##### Hướng dẫn giải

Sử dụng các công thức sau:

+) $$\sqrt{a}.\sqrt{b}=\sqrt{a.b}$$,   với $$a ,\ b \ge 0$$.

+) Với mọi số $$a \ge 0$$, luôn có $$\sqrt{a^2}=a$$.

+) $$(a-b)^2=a^2-2ab+b^2.$$

Lời giải chi tiết

a) Ta có:

$$\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}}=\sqrt{\dfrac{2a.3a}{3.8}}$$

$$=\sqrt{\dfrac{(2.3).(a.a)}{3.8}}=\sqrt{\dfrac{6a^2}{24}}$$

$$=\sqrt{\dfrac{6a^2}{6.4}}=\sqrt{\dfrac{a^2}{4}}=\sqrt{\dfrac{a^2}{2^2}}$$

$$=\sqrt{\left(\dfrac{a}{2}\right)^2}=\left| \dfrac{a}{2}\right|$$ $$= \dfrac{a}{2}$$.

Vì $$a \ge 0$$   nên   $$\dfrac{a}{2} \ge 0$$  $$\Rightarrow \left| \dfrac{a}{2} \right| = \dfrac{a}{2}$$.

b) Ta có:

$$\sqrt{13a}.\sqrt{\dfrac{52}{a}}=\sqrt{13a.\dfrac{52}{a}}=\sqrt{\dfrac{13a.52}{a}}$$

$$=\sqrt{\dfrac{13a.(13.4)}{a}}=\sqrt{\dfrac{(13.13).4.a}{a}}$$

$$=\sqrt{13^2.4}=\sqrt{13^2}.\sqrt{4}$$

$$=\sqrt{13^2}.\sqrt{2^2}=13.2$$

$$=26$$    (vì $$a>0$$)

c)

Do $$a\geq 0$$ nên bài toán luôn được xác định có nghĩa.

Ta có: $$\sqrt{5a}.\sqrt{45a}- 3a=\sqrt{5a.45a}-3a$$

$$=\sqrt{(5.a).(5.9.a)}-3a$$

$$=\sqrt{(5.5).9.(a.a)}-3a$$

$$=\sqrt{5^2.3^2.a^2}-3a$$

$$=\sqrt{5^2}.\sqrt{3^2}.\sqrt{a^2}-3a$$

$$=5.3.\left|a\right|-3a=15 \left|a \right| -3a.$$

$$=15a - 3a = (15-3)a =12a.$$

Vì $$a \ge 0$$   nên  $$\left| a \right| = a.$$

d) Ta có:

$$(3 - a)^{2}- \sqrt{0,2}.\sqrt{180a^{2}}=\sqrt{0,2.180a^2}$$

$$= (3-a)^2-\sqrt{0,2.(10.18).a^2}$$

$$=(3-a)^2-\sqrt{(0,2.10).18.a^2}$$

$$=(3-a)^3-\sqrt{2.18.a^2}$$

$$=(3-a)^2-\sqrt{36a^2}$$

$$=(3-a)^2-\sqrt{36}.\sqrt{a^2}$$

$$=(3-a)^2-\sqrt{6^2}.\sqrt{a^2}$$

$$=(3-a)^2-6.\left|a\right|$$.

+) $$TH1$$: Nếu $$a\geq 0\Rightarrow |a|=a$$.

Do đó: $$(3 - a)^{2}- 6\left|a\right|=(3-a)^2-6a$$

$$=(3^2-2.3.a+a^2)-6a$$

$$=(9-6a+a^2)-6a$$

$$=9-6a+a^2-6a$$

$$=a^2+(-6a-6a)+9$$

$$=a^2+(-12a)+9$$

$$=a^2-12a+9$$.

+) $$TH2$$: Nếu $$a<0\Rightarrow |a|=-a$$.

Do đó: $$(3 - a)^{2}- 6\left|a\right| =(3-a)^2-6.(-a)$$

$$=(3^2-2.3.a+a^2)-(-6a)$$

$$=(9-6a+a^2)+6a$$

$$=9-6a+a^2+6a$$

$$=a^2+(-6a+6a)+9$$

$$=a^2+9$$.

Vậy $$(3 - a)^{2}- \sqrt{0,2}.\sqrt{180a^{2}}=a^2-12a+9$$,   nếu $$a \ge 0$$.

$$(3 - a)^{2}- \sqrt{0,2}.\sqrt{180a^{2}}=a^2+9$$,   nếu   $$a <0$$.