Question

$questions$solve and factorise in detail ​

Answer 1

Step-by-step explanation:

1)

prove that:

$\frac{ {a}^{ - 1} }{ {a}^{ - 1} + {b}^{ - 1} } + \frac{ {a}^{ - 1} }{ {a}^{ - 1} + {b}^{ - 1} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }$

$\frac{ {a}^{ - 2} - (ab)^{ - 1} + {a}^{ - 2} + {(ab)}^{ - 1} }{ {a}^{ - 2} - {b}^{ - 2} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }$

$\frac{ {a}^{ - 2} + {a}^{ - 2} }{ {a}^{ - 2} - {b}^{ - 2} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }$

$\frac{2 {a}^{ - 2} }{ {a}^{ - 2} - {b}^{ - 2} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }$

cross-multiplying,

$2 {a}^{ - 2} ( {b}^{2} - {a}^{2} ) = 2 {b}^{2} ( {a}^{ - 2} - {b}^{ - 2} )$

$2 {b}^{2} {a}^{ - 2} - 2 = 2 {b}^{2} {a}^{ - 2} - 2$

hence, proved!

2)

prove that,

$\frac{ {2}^{30} + {2}^{29} + {2}^{28} }{ {2}^{31} + {2}^{30} - {2}^{29} } = \frac{7}{10}$

$\frac{ {2}^{28 + 2} + {2}^{28 + 1} + {2}^{28} }{ {2}^{29 + 2} + {2}^{29 + 1} - {2}^{29} } = \frac{7}{10}$

$\frac{ {2}^{28} \times {2}^{2} + {2}^{28} \times {2}^{1} + {2}^{28} }{ {2}^{29} \times {2}^{2} + {2}^{29} \times {2}^{1} - {2}^{29} } = \frac{7}{10}$

$\frac{ {2}^{28}( {2}^{2} + 2 + 1 ) }{ {2}^{29} ( {2}^{2} + 2 + 1 )} = \frac{7}{10}$

$\frac{4 + 2 + 1}{2( 4+ 2 - 1)} = \frac{7}{10}$

$\frac{7}{2(5)} = \frac{7}{10}$

hence, proved!

3)

Factorise:

$a) \: \: {x}^{2} - 4x - 2 1 \\ = {x}^{2} - 7x + 3x - 21 \\ = x(x - 7) + 3(x - 7) \\ = (x - 7)(x + 3)$

$b) \: \: 9x ^ { 2 } -22x+8 \\ = 9 {x}^{2} - 18x - 4x + 8 \\ = 9x(x - 2) - 4(x - 2) \\ = (x - 2)(9x - 4)$

$c) \: \: 2x ^ { 2 } -7x-39 \\ = 2 {x}^{2} + 6x - 13x - 39 \\ = 2x(x + 3) - 13(x + 3) \\ = (x + 3)(2x - 13)$

$d) \: \: 35y ^ { 2 } +13y-12 \\ = 35 {y}^{2} - 15y + 28y - 12 \\ = 5y(7y - 3) + 4(7y - 3) \\ = (7y - 3)(5y + 4)$

4)

Find the value of a and b if:

$\frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2} \\ \frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } = a + b \sqrt{2} \\ \frac{ {3}^{2} + 2 \times 3 \times \sqrt{2} + ( { \sqrt{2} }^{2} ) }{ {3}^{2} - { (\sqrt{2}) }^{2} } = a + b \sqrt{2} \\ \frac{9 + 6 \sqrt{2} + 2}{9 - 2} = a + b \sqrt{2} \\ \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2} \\ \frac{11}{7} + \frac{6}{7} \sqrt{2} = a + b \sqrt{2} \\ \\ hence, \: a = \frac{11}{7} , \: b = \frac{6}{7}$