Calculator search results

Formula

Expand the expression

Answer

Factorize the expression

Answer

$am-bm+an-bn$

$a m + a n - b m - b n$

Organize polynomials

$\color{#FF6800}{ a } \color{#FF6800}{ m } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ m } \color{#FF6800}{ + } \color{#FF6800}{ a } \color{#FF6800}{ n } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ n }$

$ $ Sort the polynomial expressions in descending order $ $

$\color{#FF6800}{ a } \color{#FF6800}{ m } \color{#FF6800}{ + } \color{#FF6800}{ a } \color{#FF6800}{ n } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ m } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ n }$

$\left ( a - b \right ) \left ( m + n \right )$

Arrange the expression in the form of factorization..

$\color{#FF6800}{ a } \color{#FF6800}{ m } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ m } \color{#FF6800}{ + } \color{#FF6800}{ a } \color{#FF6800}{ n } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ n }$

$ $ Expand the expression $ $

$\color{#FF6800}{ a } \color{#FF6800}{ m } \color{#FF6800}{ + } \color{#FF6800}{ a } \color{#FF6800}{ n } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ m } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ n }$

$\color{#FF6800}{ a } \color{#FF6800}{ m } \color{#FF6800}{ + } \color{#FF6800}{ a } \color{#FF6800}{ n } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ m } \color{#FF6800}{ - } \color{#FF6800}{ b } \color{#FF6800}{ n }$

$ $ Do factorization $ $

$\left ( \color{#FF6800}{ a } \color{#FF6800}{ - } \color{#FF6800}{ b } \right ) \left ( \color{#FF6800}{ m } \color{#FF6800}{ + } \color{#FF6800}{ n } \right )$

Solution search results

search-thumbnail-\bar{+a^{m-n}+a^{n-n}}^{+}\bar{1+a^{n-n}+a^{n-0}}^{+}\bar{1+a-n+a^{n-n}}

$\left(0.02-1\right)$ $1$ $1$ $1$ $\bar{+a^{m-n}+a^{n-n}} ^{+}\bar{1+a^{n-n}+a^{n-0}} ^{+}\bar{1+a-n+a^{n-n}} $

Algebra

search-thumbnail-25. On simplification, f(1+a^{++a} is equal to

$1$ $\left(o\longdiv{x+5x-}$ $P\bar{xP} $ $4-$ $25.$ On simplification, $f\left(1+a^{++a}$ $\right)+1/\left(1+a$ $++a$ $9\right)+1/\left(1+a$ $+a$ $\right)$ is equal to $\left(a\right)$ $0$ $0$ (b) $\right)$ $a$ $\left(c\right)$ $1$ $\left(a\right)$ $1/a$

Other

search-thumbnail-a.(8m-6n)+(+m+3n-3)=

$a.\left(8m-6n\right)+\left(+m+3n-3\right)=$

10th-13th grade

Other

search-thumbnail-Tf \leqan and \leqb 2 are +2y0 convergent series.

$Tf$ $\leq an$ and $\leq b$ $2$ are $+2y0$ convergent series. Then $\left(a_{n}$ $-$ $bn^{\right)}$ $18$ $-$ vergent, O $i^{3}$ $7gen^{4}$ $\left(9\right)$ $can$ not be $\left(1\right)$ $con^{~27g}$ (conditionally convergent, $de+a$ ter mine.

Other

search-thumbnail-(i1) \dfrac{m+2}{a^{2}-b^{2}}\times\dfrac{a-b}{m^{2}-4} \times\dfrac{m-2}{a+b}

$\left(i1\right)$ $\dfrac {m+2} {a^{2}-b^{2}}\times \dfrac {a-b} {m^{2}-4}$ $\times \dfrac {m-2} {a+b}$

7th-9th grade

Algebra

search-thumbnail-7m+9n=15

10th-13th grade

Algebra

Have you found the solution you wanted?

Try more features at Qanda!

Search by problem image

Ask 1:1 question to TOP class teachers

AI recommend problems and video lecture