Question

a.(4x - 1)\(^2\) = ( 1 -4x)\(^2\)

b. \(\dfrac{x-100}{24}+\dfrac{x-98}{26}+\dfrac{x-96}{28}=3\)

 

Answer 1

a) Ta có: \(\left(4x-1\right)^2=\left(1-4x\right)^2\)

\(\Leftrightarrow\left(4x-1\right)^2-\left(1-4x\right)^2=0\)

\(\Leftrightarrow\left(4x-1-1+4x\right)\left(4x-1+1-4x\right)=0\)

\(\Leftrightarrow0\cdot x=0\)(luôn đúng)

Vậy: \(x\in R\)

b) Ta có: \(\dfrac{x-100}{24}+\dfrac{x-98}{26}+\dfrac{x-96}{28}=3\)

\(\Leftrightarrow\dfrac{x-100}{24}-1+\dfrac{x-98}{26}-1+\dfrac{x-96}{28}-1=0\)

\(\Leftrightarrow\dfrac{x-124}{24}+\dfrac{x-124}{26}+\dfrac{x-124}{28}=0\)

\(\Leftrightarrow\left(x-124\right)\cdot\left(\dfrac{1}{24}+\dfrac{1}{26}+\dfrac{1}{28}\right)=0\)

mà \(\dfrac{1}{24}+\dfrac{1}{16}+\dfrac{1}{28}>0\)

nên x-124=0

hay x=124

Vậy: x=124