Question

please solve this sum fast​

Answer 1

(i)

$\displaystyle\longrightarrow\sf {\dfrac {x^2+7x+12}{x+3}=\dfrac {x^2+3x+4x+12}{x+3}}$

$\displaystyle\longrightarrow\sf {\dfrac {x^2+7x+12}{x+3}=\dfrac {x(x+3)+4(x+3)}{x+3}}$

$\displaystyle\longrightarrow\sf {\dfrac {x^2+7x+12}{x+3}=\dfrac {(x+3)(x+4)}{x+3}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {x^2+7x+12}{x+3}=x+4}}}$

(ii)

$\displaystyle\longrightarrow\sf {\dfrac {x^2-8x+12}{x-6}=\dfrac {x^2-6x-2x+12}{x-6}}$

$\displaystyle\longrightarrow\sf {\dfrac {x^2-8x+12}{x-6}=\dfrac {x(x-6)-2(x-6)}{x-6}}$

$\displaystyle\longrightarrow\sf {\dfrac {x^2-8x+12}{x-6}=\dfrac {(x-6)(x-2)}{x-6}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {x^2-8x+12}{x-6}=x-2}}}$

(iii)

$\displaystyle\longrightarrow\sf {\dfrac {p^2+5p+4}{p+1}=\dfrac {p^2+p+4p+4}{p+1}}$

$\displaystyle\longrightarrow\sf {\dfrac {p^2+5p+4}{p+1}=\dfrac {p(p+1)+4(p+1)}{p+1}}$

$\displaystyle\longrightarrow\sf {\dfrac {p^2+5p+4}{p+1}=\dfrac {(p+1)(p+4)}{p+1}}$

$\displaystyle\longrightarrow\sf {\dfrac {p^2+5p+4}{p+1}=p+4}$

(iv)

$\displaystyle\longrightarrow\sf {\dfrac {15ab(a^2-7a+10)}{3b(a-2)}=5a\cdot\dfrac {a^2-2a-5a+10}{a-2}}$

$\displaystyle\longrightarrow\sf {\dfrac {15ab(a^2-7a+10)}{3b(a-2)}=5a\cdot\dfrac {a(a-2)-5(a-2)}{a-2}}$

$\displaystyle\longrightarrow\sf {\dfrac {15ab(a^2-7a+10)}{3b(a-2)}=5a\cdot\dfrac {(a-2)(a-5)}{a-2}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {15ab(a^2-7a+10)}{3b(a-2)}=5a(a-5)}}}$

(v)

$\displaystyle\longrightarrow\sf {\dfrac {15lm(2p^2-2q^2)}{3l(p+q)}=\dfrac {30lm(p^2-q^2)}{3l(p+q)}}$

$\displaystyle\longrightarrow\sf {\dfrac {15lm(2p^2-2q^2)}{3l(p+q)}=10m\cdot\dfrac {(p+q)(p-q)}{p+q}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {15lm(2p^2-2q^2)}{3l(p+q)}=10m(p-q)}}}$

(vi)

$\displaystyle\longrightarrow\sf {\dfrac {26z^3(32z^2-18)}{13z^2(4z-3)}=2z\cdot\dfrac {2(16z^2-9)}{4z-3}}$

$\displaystyle\longrightarrow\sf {\dfrac {26z^3(32z^2-18)}{13z^2(4z-3)}=4z\cdot\dfrac {(4z-3)(4z+3)}{4z-3}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {\dfrac {26z^3(32z^2-18)}{13z^2(4z-3)}=4z(4z+3)}}}$