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Question Number 175801 Answers: 2 Comments: 5

solve the follwing equation x(√(x )) + y(√(y )) = 3 and x(√(y )) + y(√(x )) = 2 someone solve the above equations in the following way x^3 + y^3 + 2xy(√(xy )) = 9.....(1) and x^2 y + y^2 x + 2xy(√(xy )) = 4......(2) (1) − (2) ⇒ (x − y)(x^2 − y^2 ) = 5 hence x = 3 and y = 2 which is obiviusly does not satisfy the original equations. where is the fallacy in the above solution? Please explain.

$s o l v e t h e f o l l w i n g e q u a t i o n$ $x \sqrt{x} + y \sqrt{y} = 3 a n d x \sqrt{y} + y \sqrt{x} = 2$ $s o m e o n e s o l v e t h e a b o v e e q u a t i o n s i n t h e f o l l o w i n g w a y$ $x^{3} + y^{3} + 2 x y \sqrt{x y} = 9 . . . . . (1) a n d x^{2} y + y^{2} x + 2 x y \sqrt{x y} = 4 . . . . . . (2)$ $(1) - (2) \Rightarrow (x - y) (x^{2} - y^{2}) = 5$ $h e n c e x = 3 a n d y = 2 w h i c h i s o b i v i u s l y d o e s n o t s a t i s f y t h e$ $o r i g i n a l e q u a t i o n s .$ $w h e r e i s t h e f a l l a c y i n t h e a b o v e s o l u t i o n ? Please explain .$

Question Number 175800 Answers: 1 Comments: 0

Find prime numbers of 3 digits such that equal to sum of 3 diffrent numbers of prime

$Find prime numbers of$ $3 digits such that equal to$ $sum of 3 diffrent numbers of$ $prime$

Question Number 175793 Answers: 0 Comments: 2

if xy+y^2 +zx = 48; where x,y,z are three positive real numbers then find the maximum possible value of the product (xyz)

$if xy + y^{2} + zx = 48; where x, y, z$ $are three positive real numbers$ $then find the maximum possible$ $value of the product (xyz)$

Question Number 175799 Answers: 2 Comments: 0

For x ,y ε Z^+ such that 7x+9y=405. Find max value of x−y.

$For x, y ε Z^{+} such that$ $7 x + 9 y = 405 . Find \max value$ $of x - y .$

Question Number 175785 Answers: 2 Comments: 3

Question Number 175780 Answers: 0 Comments: 0

find the range of the function f(x) = cosx{sinx + (√(sin^2 x + sin^2 α )) }

$find the range of the function$ $f (x) = cosx {\sin x + \sqrt{\sin^{2} x + \sin^{2} α}}$

Question Number 175770 Answers: 0 Comments: 0

P_n = e^( ((1/1) −(1/2)) +((1/3) −(1/4)) +...+((1/(n−1)) −(1/n))) = e^( (1−(1/2) +(1/3) −(1/4) +...+(1/(n−1)) −(1/n))) = e^( Σ_(k=1) ^n (( (−1 )^( k+1) )/k)) ∴ P = lim_( n→∞) (e^( Σ_(k=1) ^n (((−1)^( k+1) )/k)) ) = e^( lim_( n→∞) ( Σ_(k=1) ^n (((−1)^(k+1) )/k))) = e^( ln(2)) = 2

$P_{n} = e^{(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{4}) + . . . + (\frac{1}{n - 1} - \frac{1}{n})}$ $= e^{(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{n - 1} - \frac{1}{n})}$ $= e^{\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{k}}$ $∴ P = {l i m}_{n \to \infty} (e^{\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{k}})$ $= e^{{l i m}_{n \to \infty} (\underset{k = 1}{\sum^{n}} \frac{{(- 1)}^{k + 1}}{k})}$ $= e^{l n (2)} = 2$