Câu 20:

Rút gọn các biểu thức sau:

a) \(\sqrt{\frac{2 \mathrm{a}}{3}} \sqrt{\frac{3 \mathrm{a}}{8}}\) với \(\mathrm{a} \geq 0\)

b) \(\sqrt{13 a} \cdot \sqrt{\frac{52}{a}}\) với \(a>0\)

c) \(\sqrt{5 \mathrm{a}} \cdot \sqrt{45 \mathrm{a}}-3 \mathrm{a}\) với \(\mathrm{a} \geq 0\)

d) \((3-a)^{2}-\sqrt{0,2} \cdot \sqrt{180 a^{2}}\)

Lời giải:

a) Ta có:

\(\sqrt{\frac{2 a}{3}} \cdot \sqrt{\frac{3 a}{8}}=\sqrt{\frac{2 a}{3} \cdot \frac{3 a}{8}}=\sqrt{\frac{2 a .3 a}{3.8}}\)

\(=\sqrt{\frac{(2.3) \cdot(a . a)}{3.8}}=\sqrt{\frac{6 a^{2}}{24}}\)

\(=\sqrt{\frac{6 a^{2}}{6.4}}=\sqrt{\frac{a^{2}}{4}}=\sqrt{\frac{a^{2}}{2^{2}}}\)

\(=\sqrt{\left(\frac{a}{2}\right)^{2}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)

b) Ta có:

\(\sqrt{13 a} \cdot \sqrt{\frac{52}{a}}=\sqrt{13 a \cdot \frac{52}{a}}=\sqrt{\frac{13 a .52}{a}}\)

\(=\sqrt{\frac{13 a \cdot(13 \cdot 4)}{a}}=\sqrt{\frac{(13.13) \cdot 4 \cdot a}{a}}\)

\(=\sqrt{13^{2} \cdot 4}=\sqrt{13^{2}} \cdot \sqrt{4}\)

\(=\sqrt{13^{2}} \cdot \sqrt{2^{2}}=13.2\)

\(=26 \quad(\) vì \(a>0)\)

c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:

\(\sqrt{(5 a)} \cdot \sqrt{(45 a)} -3 a\)

\(=\sqrt{(5 \cdot a) \cdot(5 \cdot 9 \cdot a)}-3 a\)

\(=\sqrt{(5 \cdot 5) \cdot 9 \cdot(a \cdot a)}-3 a\)

\(=\sqrt{5^{2} \cdot 3^{2} \cdot a^{2}}-3 a\)

\(=\sqrt{5^{2}} \cdot \sqrt{3^{2}} \cdot \sqrt{a^{2}}-3 a\)

\(=5 \cdot 3 \cdot|a|-3 a=15|a|-3 a\)

\(=15 a-3 a=(15-3) a=12 a\)

(Vì a ≥ 0 nên |a| = a)

d) Ta có:

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

\(=(3-a)^{2}-\sqrt{0,2 \cdot(10 \cdot 18) \cdot a^{2}}\)

\(=(3-a)^{2}-\sqrt{(0,2 \cdot 10) \cdot 18 \cdot a^{2}}\)

\(=(3-a)^{3}-\sqrt{2 \cdot 18 \cdot a^{2}}\)

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

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

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

\(=(3-a)^{2}-6 \cdot|a|\)

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

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

\(=\left(3^{2}-2.3 \cdot a+a^{2}\right)-6 a\)

\(=\left(9-6 a+a^{2}\right)-6 a\)

\(=9-6 a+a^{2}-6 a\)

\(=a^{2}+(-6 a-6 a)+9\)

\(=a^{2}+(-12 a)+9\)

\(=a^{2}-12 a+9\)

+ TH2: Nếu \(a<0 \Rightarrow|a|=-a\)

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

\(=\left(3^{2}-2 \cdot 3 \cdot a+a^{2}\right)-(-6 a)\)

\(=\left(9-6 a+a^{2}\right)+6 a\)

\(=9-6 a+a^{2}+6 a\)

\(=a^{2}+(-6 a+6 a)+9\)

\(=a^{2}+9\)

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

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