Find the value of $x^{2} y^{2} - x y$ for $x = 1$ and $y = (- 1)$ .

Open in App

Solution

Find the value of $x^{2} y^{2} - x y$ for $x = 1$ and $y = (- 1)$

putting $x = 1$ and $y = (- 1)$ in $x^{2} y^{2} - x y$

we get $(1^{2}) (- 1)^{2} - (1) (- 1)$

$= 1.1 - (1) (- 1)$

$= 2$

Suggest Corrections

Similar questions

x^{2} + y^{2} = 25

x y = 12

x^{- 1} + y^{- 1}

t a n \frac{1}{2} [s i n^{- 1} \frac{2 x}{1 + x^{2}} + c o s^{- 1} \frac{1 - y^{2}}{1 + y^{2}}]

| x | < 1, y > 0 a n d x y < 1

x^{2} + y^{2} - x y = 3

y - x = 1

\frac{x y}{x^{2} + y^{2}}

t a n \frac{1}{2} [s i n^{- 1} \frac{2 x}{1 + x^{2}} + c o s^{- 1} \frac{1 - y^{2}}{1 + y^{2}}]

| x | < 1, y > 0 a n d x y < 1

x = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}

y = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}

x^{2} + y^{2} + x y

Join BYJU'S Learning Program