第9类RD Sharma解决方案–第3章合理化-练习3.1

问题1：简化以下各项：

$\\(i)\sqrt[3]{4} \times \sqrt[3]{16}\\ (ii)\frac{\sqrt[4]{1250}}{\sqrt[4]{2}}$

解决方案：

(i) 

Using the formula: 

$\sqrt[n]{a}\times\sqrt[n]{b}= \sqrt[n]{a \times b}\\$

Here, 

$\\=\sqrt[3]{4\times 16}\\ =\sqrt[3]{64}\\ =\sqrt[3]{4^{3}}\\ =(4^{3})^{\frac{1}{3}}\\ = 4$

(ii)

Using the formula:

$\\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$

Here, 

$\\= \sqrt[4]{\frac{1250}{2}}\\ = \sqrt[4]{\frac{2\times 625}{2}}\\ = \sqrt[4]{625}\\ = \sqrt[4]{5^{4}}\\ =5^{(4\times \frac{1}{4})}\\ = 5$

问题2：简化以下表达式：

(i)(4 +√7)(3 +√2)

(ii)(3 +√3)(5-√2)

(iii)(√5-2)(√3–√5)

解决方案：

(i) (4 + √7) (3 + √2)

= 12 + 4√2 + 3√7 + √14

(ii) (3 + √3)(5- √2)

= 15 – 3√2 + 5√3 – √6

(iii) (√5 – 2)(√3 – √5)

= √15 – √25 – 2√3 + 2√5

= √15 – 5 – 2√3 + 2√5

问题3：简化以下表达式：

(i)(11 +√11)(11 –√11)

(ii)(5 +√7)(5 –√7)

(iii)(√8–√2)(√8+√2)

(iv)(3 +√3)(3 –√3)

(v)(√5–√2)(√5+√2)

解决方案：

Using Identity: (a – b)(a + b) = a² – b²

(i) (11 + √11) (11 – √11)

= 11² – (√11)²

= 121 – 11

= 110

(ii) (5 + √7) (5 –√7)

= (5² – (√7)²)

= 25 – 7 = 18

(iii) (√8 – √2) (√8 + √2)

= (√8)² – (√2)²

= 8 – 2

= 6

(iv) (3 + √3) (3 – √3)

= (3)² – (√3)²

= 9 – 3

= 6

(v) (√5 – √2) (√5 + √2)

= (√5)² – (√2)²

= 5 – 2

= 3

问题4：简化以下表达式：

(i)(√3+√7) ²

(ii)(√5–√3) ²

(iii)(2√5+3√2) ²

解决方案：

Using identities: (a – b)² = a² + b²– 2ab and (a + b)² = a²+ b² + 2ab

(i) (√3 + √7)²

= (√3)² + (√7)² + 2(√3)(√7)

= 3 + 7 + 2√21

= 10 + 2√21

(ii) (√5 – √3)²

= (√5)² + (√3)² – 2(√5)(√3)

= 5 + 3 – 2√15

= 8 – 2√15

(iii) (2√5 + 3√2)²

= (2√5)² + (3√2)² + 2(2√5)( 3√2)

= 20 + 18 + 12√10

= 38 + 12√10