Question

Find the HCF of x³y/m²n⁴ , x²y³/m²n² and x⁴y²/mn³.

Solve this please please.

Answer 1

Hii friend,

X³Y/m²n⁴ = X²×X×Y/m×m×n² × n²

X²Y³/m²n² = X²×Y×Y×Y/m×m×n²

and,

X⁴Y²/mn³ = X²×X² × Y × Y/ m × n² × n

X²Y/mn² is common in all.

Therefore,

HCF of X³Y/m² n⁴, X²Y³/m²n² and X⁴Y²/mn³ = X²Y/mn².

HOPE IT WILL HELP YOU...... :-)

Answer 2

Hey Friend ☺

x^3y/m^2n^4 = x^2 × x × y/m × m × n^2 × n^2

x^2 y^3 /m^2n^2 = x^2 × y × y^3 / m × m × n^2

x^4 y^2 /mn^3 = x^2 × x × x × y × y / m × n^2 × n

Now we have to put common factors of the three equation

Common factors :-

x^2 , y , 1/m ,1/ n^2

So

the HCF = x^2 × y × 1/m × 1/n^2

= x^2y/mn^2

So the HCF of the given equations is x^2y / mn^2

Hope it helps you ..!!

✌