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Byju's Answer

Standard VIII

Mathematics

Factorisation by Regrouping Terms

ab-mn+an-bm=?...

Question

$a b - m n + a n - b m = ? (a) (a - b) (m - n) (b) (a - m) (b + n) (c) (a - n) (m + b) (d) (m - a) (n - b)$

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Solution

$a b - m n + a n - b m = a b + a n - b m - m n = a (b + n) - m (b + n) = (b + n) (a - m)$

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Similar questions

Q. Tick (✓) the correct answer:
ab − mn + an − bm = ?
(a) (a − b)(m − n)
(b) (a − m)(b + n)
(c) (a − n)(m + b)
(d) (m − a)(n − b)

Q. If

a

and

b

are non-zero integers,

m

and

n

are any integers, then which of the following are true?

Question 20

For non-zero numbers $a$ and $b$ , ${(\frac{a}{b})}^{m} \div {(\frac{a}{b})}^{n},$ where $m > n,$ is equal to:
(a) ${(\frac{a}{b})}^{m n}$
(b) ${(\frac{a}{b})}^{m + n}$
(c) ${(\frac{a}{b})}^{m - n}$
(d) ${({(\frac{a}{b})}^{m})}^{n}$

Q. In

△ A B C

, seg

M N ∥ s e g B C

, If

B M = 6, A N = 5, A M = 2

then find

N C

Q. In

△ A B C

s e g M N ∥ s e g B C

, if

B M = 6

A N = 5

A M = 2

, then find

N C

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ab−mn+an−bm=?(a) (a−b)(m−n)(b) (a−m)(b+n)(c) (a−n)(m+b)(d) (m−a)(n−b)

ab−mn+an−bm=ab+an−bm−mn=a(b+n)−m(b+n)=(b+n)(a−m)

$a b - m n + a n - b m = ? (a) (a - b) (m - n) (b) (a - m) (b + n) (c) (a - n) (m + b) (d) (m - a) (n - b)$

$a b - m n + a n - b m = a b + a n - b m - m n = a (b + n) - m (b + n) = (b + n) (a - m)$