Question

important question please solve
$16(2x - 1) {}^{2} - 25y {}^{2}$
​

Answer 1

$\huge\mathbb{\underline{\blue{SOLUTION:-}}}$

we have given the question :-

$\sf 16(2x-1){}^{2}-25y{}^{2}$

$\sf{\blue{Let (2x-1)= x}}$

The. put this value:-

$\sf 16x{}^{2}-25y{}^{2}$

$\mathtt{\underline{\red{According\:to\: question:-}}}$

identity used :-

$\sf a{}^{2}-b{}^{2}=(a+b)(a-b)$

Now in the place of a=4 and b=5

Then the solution:-

$\sf 16x{}^{2}- 25y{}^{2}\\ \\ \sf\implies(4x){}^{2}-(5y){}^{2}\\ \\ \sf\implies (4x+5y)(4x-5y)$

Now put the value of x which is

(2x-1)

$\sf [4(2x-1)+5y][4(2x-1)-5y]\\ \\ \sf\implies (8x+5y-4)(8x-5y-4)$

$\boxed{\red{\rm{(8x+5y-4)(8x-5y4)}}}$

#BAL

#Answerwithquality

Answer 2

┏─━─━─━─━∞◆∞━─━─━─━─┓

✭✮ӇЄƦЄ ƖƧ ƳƠƲƦ ƛƝƧƜЄƦ✭✮

┗─━─━─━─━∞◆∞━─━─━─━─┛

✨$\mathbb{SOLUTION}$

$16(2x - 1) {}^{2} - 25 {y}^{2}$

$let \: (2x - 1) = x$

$putting \: \: this \: \: value$

$16 {x }^{2} - 25 {y}^{2}$

ACCORDING TO QUESTION,

IDENTITY USED:

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

Now,

$a= 4 \: and \: b = 5$

$= > 16 {x}^{2} - 25 {y}^{2}$

$= > 4 {x}^{2} - 5 {y}^{2}$

$= > (4x + 5y)(4x - 5y)$

PUTTING THE VALUE,

$(2x - 1){4(2x - 1)} + 5y(4(2x - 1)) - 5y$

$= > (8x + 5y - 4)(8x - 5y - 4)$

▀▄ [_ANSWER_]▄▀

$(8x + 5y - 4)(8x - 5y - 4)$

▬▬▬▬▬ஜ۩۞۩ஜ▬▬▬▬▬▬

#BAL

#Answerwithquality