Question

plz solve this question fastttt... ​

Answer 1

Step-by-step explanation:

1) Given

$x + \frac{1}{x} = 5$

squaring both sides we get

${(x + \frac{1}{x} })^{2} = 25$

${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 25$

${x}^{2} + \frac{1}{ {x}^{2} } = 23$

squaring again, we get

${x}^{4} + \frac{1}{ {x}^{4} } + 2 = 529$

It implies that

${x}^{4} + \frac{1}{ {x}^{4} } = 527$

Ans 2) Given

$\frac{(1.98 \times 1.98 - 1.02 \times 1.02)}{0.96}$

Now

$1.98 = 2 - 0.02$

$1.02 = 1 + 0.02$

$0.96 = 1 - 2 \times 0.02$

It implies that given expression is

$\frac{ {(2 - 0.02)}^{2} - {(1 + 0.02)}^{2} }{(1 - 2 \times 0.02)}$

for simplification let us assume 0.02=x, therefore given expression becomes

$\frac{ {(2 - x)}^{2} - {(1 + x)}^{2} }{(1 - 2x)}$

using

${x}^{2} - {y}^{2} = (x - y) \times (x + y) \: in \: the \: numerator$

$= \frac{(2 - x - 1 - x {) \times (2 - x + 1 + x}) }{( 1- 2x)}$

$\frac{(1 - 2x) \times (3)}{1 - 2x)}$

$= 3 \: \:$

Ans 3

${x}^{2} + {y}^{2} = {(x + y)}^{2} - xy$

Given

$x + y = 10$

$xy = 21$

Therefore

${x}^{2} + {y}^{2} = {(10)}^{2} - 21$

${x}^{2} + {y}^{2} = 100 - 21$

${x}^{2} + {y}^{2} = 79 \: (answer)$

Ans 4)

No of spades= 13

Therefore Probability of a spade is

$\frac{13}{52} = \frac{1}{4}$

No of Ace spades =1

Therefore Probability of an ace spade is

$\frac{1}{52}$