\(\Leftrightarrow\)\(\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)+\(\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)= \(\frac{16}{5}\)
\(\Leftrightarrow\)\(\frac{\left(x-3\right)5\left(x-4\right)+5\left(x-2^{ }\right)^2}{\left(x-2\right)5\left(x-4\right)}\)=\(\frac{16\left(x-2\right)\left(x-4\right)}{5\left(x-2\right)\left(x-4\right)}\)
Suy ra :
\(\Leftrightarrow\)(x-3)5(x-4)+5(x-2)2=16(x-2)(x-4)
\(\Leftrightarrow\)15x2-60x-15x+60+5x2-20x+20=16x2-64x-32x+128
\(\Leftrightarrow\)15x2-15x+5x2-20x-16x2+64x+32x-60x=128-20-60
\(\Leftrightarrow\)4x2+x = 48
\(\Leftrightarrow\)
\(\Leftrightarrow\frac{\left(x-3\right)5\left(x-4\right)+\left(x-2\right)^25}{5\left(x-2\right)\left(x-4\right)}=\frac{16\left(x-2\right)\left(x-4\right)}{5\left(x-2\right)\left(x-4\right)}\)ĐKXĐ : x\(\ne2;x\ne4\)
Suy ra :
\(\Leftrightarrow\)(x-3)5(x-4)+(x-2)25= 16 (x-2)(x-4)
\(\Leftrightarrow\)5x2-20x-15x+60+5x2-20x+20=16x2-64x-32x+128
\(\Leftrightarrow\)5x2-20x-15x+5x2-20x-16x2+64x+32x=128-20-60
\(\Leftrightarrow\) -6x2+41x =48
\(\Leftrightarrow\)-6x2+41x-48=0
\(\Leftrightarrow\)6x2-41x+48=0
\(\Leftrightarrow\)6x2-32x-9x+48=0
\(\Leftrightarrow\)(6x2-32x)-(9x-48)=0
\(\Leftrightarrow\)2x(3x-16) - 3(3x -16) =0
\(\Leftrightarrow\)(3x -16 ) (2x -3) =0
\(\Leftrightarrow\left[{}\begin{matrix}3x-16=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{16}{3}\\x=\frac{3}{2}\end{matrix}\right.\)(tm)
Vậy pt có tập nghiệm S = \(\left\{\frac{3}{2};\frac{16}{3}\right\}\)
