Tag: hcf-lcm

Given numbers are $1.08$, $0.36$ and $0.90$.

Let the numbers be $3x$, $4x$ and $5x$.
Then, their L.C.M. $= 60x$.
So, $60x = 2400$ or $x = 40$.
$\therefore $ The numbers are $\left( 3\times 40 \right) $, $\left( 4\times 40 \right) $ and $\left( 5\times 40 \right) $.
Hence, required H.C.F. $= 40$.

Required H.C.F. $=\dfrac { H.C.F.\quad of\quad 9,12,18,21 }{ L.C.M.\quad of\quad 10,25,35,40 } = \dfrac { 3 }{ 1400 } $

The given terms are $\dfrac {3}{1}, \dfrac {6}{5}$ and $\dfrac {3}{50}$.
H.C.F. of these terms $= \dfrac {\text{H.C.F. of}\ 3, 6, 3}{\text{L.C.M. of}\ 1, 5, 50}$
$= \dfrac {3}{50} = 0.06$

$x^{4} - 3x + 2 = x^{4} - x^{3} + x^{3} - x^{2} + x^{2} - x - 2x + 2$
$= x^{3} (x - 1) + x^{2} (x - 1) + x(x - 1) - 2(x - 1)$
$= (x - 1) [x^{3} + x^{2} + x - 2]$
$x^{3} - 3x^{2} + 3x - 1 = (x - 1)^{3}$
$x^{4} - 1 = (x - 1) (x + 1) (x^{2} + 1)$
$HCF = x - 1$

$5xy = 1*5*x*y$
$28ab = 1*7*2*2*a*b$
G.C.D = $1$

We know that