The correct option is B m^2x^2 2mxny+n^2y^2 ( mx ny ) ( mx ny )= ( mx ny )^2 = #160; ( mx )^2 2mx× ny+ ( ny )^2 = #160; m^2x^2 2mxny ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Mathematics

Algebraic Identities

mx-ny mx-ny ...

Question

$(m x - n y) (m x - n y)$ =_____

m^{2} x^{2} + 2 m x n y - n^{2} y^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m^{2} x^{2} - 2 m x n y - n^{2} y^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m^{2} x^{2} - 2 m x n y + n^{2} y^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

m^{2} x^{2} + 2 m x n y + n^{2} y^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $m^{2} x^{2} - 2 m x n y + n^{2} y^{2}$
$(m x - n y) (m x - n y) = {(m x - n y)}^{2}$
= ${(m x)}^{2} - 2 m x \times n y + {(n y)}^{2}$
= $m^{2} x^{2} - 2 m x n y + n^{2} y^{2}$

Suggest Corrections

Similar questions

Q. If

x = sec θ - cos θ

and

y = {sec}^{n} θ - {cos}^{n} θ

, then

{(\frac{d y}{d x})}^{2}

is equal to

Q. If

x = cos e c θ - sin θ, y = cos e c^{n} θ - {sin}^{n} θ

then

(x^{2} + 4) {(\frac{d y}{d x})}^{2} - n^{2} y^{2} =

Q. If

x = sec θ - cos θ; y = {sec}^{n} θ - {cos}^{n} θ

,then

(x^{2} + 4) {(\frac{d y}{d x})}^{2} - n^{2} y^{2} =

y = mx is a chord of a circle of radius a and the diameter of the circle lies along x-axis and one end of this chord in origin .The equation of the circle described on this chord as diameter is

Q. Find the centre and radius of the circle

\sqrt{1 + m^{2}} (x^{2} + y^{2}) - 2 c x - 2 m c y = 0

Join BYJU'S Learning Program

(mx−ny)(mx−ny)=_____

The correct option is B m2x2−2mxny+n2y2(mx−ny)(mx−ny)=(mx−ny)2= (mx)2−2mx×ny+(ny)2= m2x2−2mxny+n2y2

$(m x - n y) (m x - n y)$ =_____

The correct option is B $m^{2} x^{2} - 2 m x n y + n^{2} y^{2}$
$(m x - n y) (m x - n y) = {(m x - n y)}^{2}$
= ${(m x)}^{2} - 2 m x \times n y + {(n y)}^{2}$
= $m^{2} x^{2} - 2 m x n y + n^{2} y^{2}$