Math, asked by rajeevsaini3958, 27 days ago

1. Find the smallest number, by which, following numbers can be divided to
find perfect cube root of the quotient
15552​

Answers

Answered by ItzShubham95
0

Answer:

Answer:

\underline{\boxed{\gray{\sf Given \: : }}}

\underline{\boxed{\purple{\sf\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3} }}}

\underline{\boxed{\pink{\sf\implies \frac{(x-1)(x-4)+(x-3)(x-2)}{(x-2)(x-4)}=\frac{10}{3}  }}}

\underline{\boxed{\purple{\sf  \implies \frac{x^{2}-4x-x+4+x^{2}-2x-3x+6}{x^{2}-4x-2x+8}=\frac{10}{3}}}}

\underline{\boxed{\pink{\sf \implies \frac{2x^{2}-10x+10}{x^{2}-6x+8}=\frac{10}{3} }}}

\underline{\boxed{\purple{\sf \implies 3(2x^{2}-10x+10)=10(x^{2}-6x+8)}}}

\underline{\boxed{\pink{\sf  \implies 6x^{2}-30x+30=10x^{2}-60x+80}}}

\underline{\boxed{\purple{\sf\implies 0= 10x^{2}-60x+80-6x^{2}+30x-30  }}}

\underline{\boxed{\pink{\sf \implies 4x^{2}-30x+50=0 }}}

\large\mapsto\boxed{\sf\green{Divide \:  each \:  term  \: by  \: 2 ,we \:  get}}

\underline{\boxed{\red{\sf  \implies 2x^{2}-15x+25=0}}}

\large\mapsto\boxed{\sf\orange{Splitting  \: the  \: middle  \: term, \: we  \: get:}}

\underline{\boxed{\green{\sf  \implies 2x^{2}-10x-5x+25=0}}}

\underline{\boxed{\red{\sf \implies 2x(x-5)-5(x-5)=0 }}}

\underline{\boxed{\green{\sf \implies (x-5)(2x-5)=0 }}}

\underline{\boxed{\red{\sf\implies x-5 = 0 \: Or \: 2x-5 = 0 }}}

\underline{\boxed{\green{\sf \implies x = 5 \: Or \: x=\frac{5}{2} }}}

\huge\pink{\boxed{therefore,}}

\underline{\boxed{\blue{\sf x = 5 Or x = \frac{5}{2}25}}}

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