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

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

\(\Rightarrow\)\(\frac{91x-100.91}{91.24}+\frac{84x-84.98}{26.84}+\frac{78x-96.78}{78.28}\)

=\(\frac{91x-9100+84x-8232+78x-7488}{2184}\)

=\(\frac{91x+84x+78x-9100-8232-7488}{2184}\)

=\(\frac{x\left(91+84+78\right)-\left(9100+8232+7488\right)}{2184}\)

=\(\frac{x253-24820}{2184}=3\)

\(\Rightarrow\)x253- 24820 =6552

\(\Rightarrow\)x253= 31372

\(\Rightarrow\)x = 124

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

\(=\frac{\left(x-100\right)}{24}+\frac{\left(x-98\right)}{26}+\frac{\left(x-96\right)}{28}-1=0\)

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

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

Vì: \(\frac{1}{24}+\frac{1}{26}+\frac{1}{28}\ne0\)

\(\Rightarrow x-124=0\)

\(\Rightarrow x=124-0\)

\(\Rightarrow x=124\)