a) Ta có: \(\left(3x-1\right)^2+\left(4x+5\right)^2=\left(5x-7\right)^2\)
\(\Leftrightarrow9x^2-6x+1+16x^2+40x+25=25x^2-70x+49\)
\(\Leftrightarrow25x^2+34x+26-25x^2+70x-49=0\)
\(\Leftrightarrow104x-23=0\)
\(\Leftrightarrow104x=23\)
hay \(x=\frac{23}{104}\)
Vậy: \(S=\left\{\frac{23}{104}\right\}\)
b) Ta có: \(\left(x-2\right)^3+\left(x+2\right)^3=2\left(x-3\right)\left(x^2+3x+9\right)\)
\(\Leftrightarrow\left(x-2+x+2\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\right]=2\left(x^3-27\right)\)
\(\Leftrightarrow2x\cdot\left(x^2-4x+4-x^2+4+x^2+4x+4\right)=2x^3-54\)
\(\Leftrightarrow2x\cdot\left(x^2+12\right)-2x^3+54=0\)
\(\Leftrightarrow2x^3+24x-2x^3+54=0\)
\(\Leftrightarrow24x=54\)
hay \(x=\frac{9}{4}\)
Vậy: \(S=\left\{\frac{9}{4}\right\}\)
c) Ta có: \(2014x-10.07=20.14x-1007\)
\(\Leftrightarrow2014x-10.07-20.14x+1007=0\)
\(\Leftrightarrow1993.86x+1017.07=0\)
\(\Leftrightarrow1993.86x=-1017.07\)
\(\Leftrightarrow x=-\frac{101}{198}\)
Vậy: \(S=\left\{-\frac{101}{198}\right\}\)
d) Ta có: \(\frac{x-5}{2}+\frac{x-5}{3}-\frac{1}{4}=\frac{1}{2}+\frac{1}{3}-\frac{x-5}{4}\)
\(\Leftrightarrow\frac{x-5}{2}+\frac{x-5}{3}+\frac{x-5}{4}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(\Leftrightarrow x-5=1\)
hay x=6
Vậy: S={6}