16 x4-49 y2-49 y4-16 x2-2 x y-4 x2-7 y-7 y2-4 x2

The correct option is A True 16x^4/49y^2 + 49y^4/16x^2 + 2xy ⇒ (4x^2/7y)^2 + (7y^2/4x)^2 + 2xy Let a = (4x^2/7y) and b = (7y^2/4x) ⇒ ab = (4x^2/7y) × (7y^2 ...

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Byju's Answer

Standard VIII

Mathematics

Factorisation by Grouping

16 x4/49 y2+4...

Question

$\frac{16 x^{4}}{49 y^{2}} + \frac{49 y^{4}}{16 x^{2}} + 2 x y = (\frac{4 x^{2}}{7 y} + \frac{7 y^{2}}{4 x})^{2}$

True

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False

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Solution

The correct option is A
True

$\frac{16 x^{4}}{49 y^{2}} + \frac{49 y^{4}}{16 x^{2}} + 2 x y$

$\Rightarrow (\frac{4 x^{2}}{7 y})^{2} + (\frac{7 y^{2}}{4 x})^{2} + 2 x y$

$L e t a = (\frac{4 x^{2}}{7 y}) a n d b = (\frac{7 y^{2}}{4 x})$

$\Rightarrow a b = (\frac{4 x^{2}}{7 y}) \times (\frac{7 y^{2}}{4 x}) = x y$

$\Rightarrow a^{2} + b^{2} + 2 a b$

$\Rightarrow (a + b)^{2}$

Substituting the values of a and b, we get,

$(\frac{4 x^{2}}{7 y} + \frac{7 y^{2}}{4 x})^{2}$

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16x449y2+49y416x2+2xy=(4x27y+7y24x)2

The correct option is A True 16x449y2+49y416x2+2xy ⇒(4x27y)2+(7y24x)2+2xy Let a=(4x27y) and b=(7y24x) ⇒ab=(4x27y)×(7y24x)=xy ⇒a2+b2+2ab ⇒(a+b)2 Substituting the values of a and b, we get, (4x27y+7y24x)2

$\frac{16 x^{4}}{49 y^{2}} + \frac{49 y^{4}}{16 x^{2}} + 2 x y = (\frac{4 x^{2}}{7 y} + \frac{7 y^{2}}{4 x})^{2}$