Math, asked by saurabhyadavy363, 9 months ago

√7-1 upon √7+1 -√7+2upon √7-2=a+b√7​

Answers

Answered by tchaudhari504
0

Answer:

fu65r75

Step-by-step explanation:

fyuy. the the first one of the most yytt

14y

91

Answered by Salmonpanna2022
7

Step-by-step explanation:

Correct Question:—

 \frac{ \sqrt{7}  - 1}{ \sqrt{7} + 1 }  -   \frac{ \sqrt{7}  + 1}{ \sqrt{7} - 1 }  = a + b \sqrt{7}  \\  \\

To find out:—

The value of a and b in expression.

Solution:—

Let's solve the problem

we have,

 \frac{ \sqrt{7}  - 1}{ \sqrt{7} + 1 }  -   \frac{ \sqrt{7}  + 1}{ \sqrt{7} - 1 }  \\  \\

⟹ \frac{ \sqrt{7} - 1 }{{7} + 1 }  \times  \frac{ \sqrt{7} - 1 }{ \sqrt{7}   - 1}  -  \frac{ \sqrt{7}  + 1}{ \sqrt{7}  - 1}  \times  \frac{ \sqrt{7}  + 1}{ \sqrt{7}  + 1}  \\  \\

⟹ \frac{( \sqrt{7}  - 1)( \sqrt{7}  - 1)}{( \sqrt{ 7 }  + 1)( \sqrt{7}  - 1)}  -  \frac{( \sqrt{7}  + 1)( \sqrt{7} + 1) }{( \sqrt{7} - 1)( \sqrt{7} + 1)  }  \\  \\

⟹ \frac{( \sqrt{7} - 1 {)}^{2}  }{( \sqrt{7}  {)}^{2}  - (1 {)}^{2} }  -  \frac{( \sqrt{7}  + 1 {)}^{2} }{( \sqrt{7}  {)}^{2} - (1 {)}^{2}  }  \\  \\

⟹ \frac{( \sqrt{7}  {)}^{2} + (1 {)}^{2}   - 2( \sqrt{7} )(1)}{7 - 1}  -  \frac{( \sqrt{7} {)}^{2}  + (1 {)}^{2}  + 2( \sqrt{7})(1)  }{7 - 1}  \\  \\

⟹ \frac{7  + 1 - 2 \sqrt{7} }{6}  -  \frac{7 + 1 + 2 \sqrt{7} }{6}  \\  \\

⟹ \frac{8 - 2 \sqrt{7} }{6}  -  \frac{8 + 2 \sqrt{7} }{6}  \\  \\

⟹ \frac{1}{6} (8 - 2 \sqrt{7}  - 8 - 2 \sqrt{7} ) \\  \\

⟹0 -  \frac{4 \sqrt{7} }{6}  \\  \\

⟹0 +  \bigg( \frac{ - 2}{3}  \bigg) \sqrt{7}  \\  \\

∴\: \:  0 +  \bigg( \frac{ - 2}{3}  \bigg) \sqrt{7}  = a + b \sqrt{7}  \\  \\

On comparing the value of ;

a = 0 \:  \: and \: \:  b =  -  \frac{2}{3}  \\

Answer:—

 \mathrm{Hence, the  \: value  \: of \: a = 0 \: and \: b =  -  \frac{2}{3}  .} \\  \\

Used formulae:—

  • (a-b)(a-b) = (a-b)² = a² + b² - 2ab
  • (a+b)(a+b) = (a+b)² = a²+b²+2ab
  • (a+b)(a-b) = a² - b²
  • (a-b)(a+b) = a² - b²

Learn more;

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