Question

my challenge who can solve this question ​there are three question any one solve the question I thik no

Answer 1

Challenge Accepted,.

Step-by-step explanation:

Given :

The sum of $\frac{a}{b}$ and it's reciprocal is 1,

a ≠ 0, b ≠ 0,

Then, find a³ + b³

Solution :

$\frac{a}{b}$ 's reciprocal is $\frac{b}{a}$,.

⇒ $\frac{a}{b} + \frac{b}{a} = 1$

⇒ $\frac{a^2+b^2}{ab} = 1$

⇒ $a^2+b^2 = ab$

⇒ $a^2 + b^2 - ab = 0$  ...(i)

We know that,

$a^3 + b^3 = (a + b) (a^2 - ab+b^2)$

⇒ $a^3 + b^3 = (a + b) (0)$   from (i)

⇒ $a^3 + b^3 = 0$

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If x² = y + z,

y² = x + z,

z² = x + y

Then, find $\frac{1}{x+1} +\frac{1}{y+1} +\frac{1}{z+1}$

Solution :

$\frac{1}{x+1} +\frac{1}{y+1} +\frac{1}{z+1}$

⇒ $\frac{x +1 - x}{x+1} +\frac{y+1-y}{y+1} +\frac{z+1-z}{z+1}$

⇒ $(1 - \frac{x}{x+1} )+(1-\frac{y}{y+1} ) + (1-\frac{z}{z+1})$

⇒$3 - (\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} )$

Multiplying bot numerator & denominator by their respective variables,.

⇒ $3 -[ (\frac{x}{x}\times \frac{x}{x+1}) +(\frac{y}{y}\times \frac{y}{y+1}) + (\frac{z}{z}\times\frac{z}{z+1})]$

⇒ $3 -[ \frac{x^2}{x^2+x} + \frac{y^2}{y^2+y} + \frac{z^2}{z^2+z}]$

⇒ $3 -[ \frac{y+z}{(y+z)+x} + \frac{x+z}{(x+z)+y} + \frac{x+y}{(x+y)+z}]$

⇒ $3 -[ \frac{y+z}{x+y+z} + \frac{x+z}{x+y+z} + \frac{x+y}{x+y+z}]$

⇒ $3 - \frac{(y+z)+(x+z)+(x+y)}{x+y+z}+]$

⇒ $3 - \frac{2x+2y+2z}{x+y+z}+]$

⇒ $3 - \frac{2(x+y+z)}{x+y+z}+]$

⇒ $3-2=1$

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Given :

a² + b² = 2,

c² + d² = 1,.

Then, (ad - bc)² + (ac + bd)²

Solution :

(ad - bc)² + (ac + bd)²

⇒ [(ad)² - 2(ad)(bc) + (bc)²] + [(ac)² + 2(ac)(bd) + (bd)²]

⇒ a²d² - 2abcd + b²c² + a²c² + 2abcd + b²d²

⇒ a²d² + b²c² + a²c² + b²d²

⇒ a² (c² + d²) + b² (c² + d²)

⇒ (a² + b²)(c² + d²) = (2)(1) = 2

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Challenge Completed,.