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प्रश्न
Simplify the following:
\[\frac{11}{2} x^2 y - \frac{9}{4}x y^2 + \frac{1}{4}xy - \frac{1}{14} y^2 x + \frac{1}{15}y x^2 + \frac{1}{2}xy\]
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उत्तर
\[ \frac{11}{2} x^2 y - \frac{9}{4}x y^2 + \frac{1}{4}xy - \frac{1}{14} y^2 x + \frac{1}{15}y x^2 + \frac{1}{2}xy\]
\[= \frac{11}{2} x^2 y + \frac{1}{15}y x^2 - \frac{9}{4}x y^2 - \frac{1}{14} y^2 x + \frac{1}{4}xy + \frac{1}{2}xy\] (Collecting like terms)
= \[\left( \frac{165 + 2}{30} \right) x^2 y + \left( \frac{- 63 - 2}{28} \right)x y^2 + \left( \frac{1 + 2}{4} \right)xy\]
\[= \frac{167}{30} x^2 y - \frac{65}{28} y^2 x + \frac{3}{2}xy\] (Combining like terms)
संबंधित प्रश्न
Identify like term in the following:
10pq, 7p, 8q, −p2q2, −7qp, −100q, −23, 12q2p2, −5p2, 41, 2405p, 78qp, 13p2q, qp2, 701p2
Take away:
\[\frac{6}{5} x^2 - \frac{4}{5} x^3 + \frac{5}{6} + \frac{3}{2}x \text { from }\frac{x^3}{3} - \frac{5}{2} x^2 + \frac{3}{5}x + \frac{1}{4}\]
Take away:
\[\frac{5 a^2}{2} + \frac{3 a^3}{2} + \frac{a}{3} - \frac{6}{5} \text { from } \frac{1}{3} a^3 - \frac{3}{4} a^2 - \frac{5}{2}\]
Take away:
\[\frac{7}{4} x^3 + \frac{3}{5} x^2 + \frac{1}{2}x + \frac{9}{2}\text { from } \frac{7}{2} - \frac{x}{3} - \frac{x^2}{5}\]
Take away:
\[\frac{y^3}{3} + \frac{7}{3} y^2 + \frac{1}{2}y + \frac{1}{2} \text { from } \frac{1}{3} - \frac{5}{3} y^2\]
Which of the following is a binomial?
Which of the following are like terms?
The product of one negative and one positive term is a negative term.
Like terms in the expression n(n + 1) + 6(n – 1) are ______ and ______.
In like terms, variables and their powers are the same.
