Question

Solve Q no. 4, 5, 6, 7, 8, 9 ,10

URGENT!!!!!!

Answer 1

HELLO DEAR,

------(4)
$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = a +7 b \sqrt{5} \\ \\ \\ = > \frac{49+ 5 + 14 \sqrt{5} - 49 - 5 + 14 \sqrt{5} }{49 - 5} \\ \\ = > \frac{ 28 \sqrt{5} }{44} \\ \\ = > \frac{7 \sqrt{5} }{11}$
on comparing with L.H.S

we get,

a = 0 and b = 1/11

-----(5)

given that:-

x = (√3 + √2)/(√3 - √2)

x = (√3 + √2)/(√3 - √2) × (√3 + √2)/(√3 + √2)

x = (3 + 2 + 2√6)/( 3 - 2)

x = (5 + 2√6)/1 = (5 + 2√6)

x² = (25 + 24 + 20√6)

x² = (49 + 20√6)---------(1)

y = (√3 - √2)/(√3 + √2)

y = (√3 - √2)/(√3 + √2) × (√3 - √2)/(√3 - √2)

y = (3 + 2 - 2√6)/(3 - 2)

y = (5 - 2√6)/1 = (5 - 2√6)

y² = (25 + 24 - 20√6)

y² = (49 - 20√6)---------(2)

xy = (√3 + √2)/(√3 - √2) × (√3 - √2)/(√3 + √2)

xy = 1-------(3)

now adding-----(1) , ---(2) & ---(3)

we get,

x² + xy + y² = (49 + 20√6) + 1 + (49 - 20√6)

(x² + y² + xy) = 98 + 1 = 99

---(6)

3/(√3 - √2 + √5)

= 3/[(√3 - √2) + √5] × [(√3 - √2) - √5] /[(√3 - √2) - √5]

= 3[(√3 - √2) - √5] / [ 3 + 2 - 2√6 - 5]

= 3[(√3 - √2) - √5] / (5 - 2√6 - 5)

= 3[(√3 - √2) - √5] / 2√6

= 3[(√3 - √2) - √5] / 2√6 × √6/√6

= 3[(√3 - √2) - √5] ×√6 / 12

= (√18 - √12 - √30)/4

------(7)

x = (1 - √2)

x² = (1 + 2 - 2√2)

x² = (3 - 2√2)

AND,

1/x² = 1/(3 - 2√2)

1/x² = 1/(3 - 2√2) × (3 + 2√2)/(3 + 2√2)

1/x² = (3 + 2√2)/(9 - 8)

1/x² = (3 + 2√2)

now,

(x + 1/x)² = x² + 1/x² + 2×X × 1/X

(x + 1/x)² = (3 - 2√2) + (3 + 2√2) + 2

(x + 1/x)²= 6 + 2 = 8

-----(8)

x = 2 + √5

x² = (4 + 5 + 4√5)

x² = (9 + 4√5)

1/x² = 1/(9 + 4√5) × (9 - 4√5)/(9 - 4√5)

1/x² = (9 - 4√5)/(81 - 80)

1/x² = (9 - 4√5)

now,

(x² - 1/x²) = (9 + 4√5) - (9 - 4√5)

(x² - 1/x²) = 9 + 4√5 - 9 + 4√5

(x² - 1/x²) = 8√5

---------(9)

25^(x - 1) = 5^(2x - 1) - 100

=> 25^x * 1/25 = 25^x * 1/5 - 100

=> 25^x * 1/5 (1/5 - 1) = -100

=> 25^x * 1/5 * (-4/5) = -100

=> 25^x * 1/5 * 4/5 = 100

=> 25^x = (100 × 25) /4

=> 25^x = 25 * 25

=> 25^x = 25^2

=> x = 2

---------(10)

(2/3)^x * (3/2)^2x = 81/16

=> (3/2)^(-x) * (3/2)^2x = 81/16

=> (3/2)^(-x + 2x) = 81/16

=> (3/2)^x = (3/2)⁴

=> x = 4

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hey friend, Harish here.

Here is your answer:(Pls refer to the images)

10)  

→ $( \frac{2}{3} )^{x} \times ( \frac{3}{2} )^{2x}= ( \frac{81}{16} )$

→ $( \frac{3}{2} )^{-x} \times ( \frac{3}{2} )^{2x}= ( \frac{3}{2} )^4$

→ $( \frac{3}{2})^{x} = (\frac{3}{2})^{4}$

Bases are equal, therefore by comparing we get, x  =4

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Hope my answer is helpful to you.