a. (3x-4)2=9(x-1)(x+1)
<=> 9x2-24x+16=9x2-9
<=> -24x=-25
<=> x=\(\dfrac{25}{24}\)
Vậy S=\(\left\{\dfrac{25}{24}\right\}\)
b. (4x-5)2-4(x-2)2=0
<=> (4x-5)2-(2x-4)2=0
<=> (4x-5-2x+4)(4x-5+2x-4)=0
<=> (2x-1)(6x-9)=0
<=> \(\left[{}\begin{matrix}2x-1=0\\6x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy S=\(\left\{\dfrac{1}{2};\dfrac{3}{2}\right\}\)
c. |x2-x|= -2x
Ta có: |x2-x|=x2-x khi x2-x\(\ge0\) hay x\(\ge1\)
=> x2-x= -2x
<=> x2-x+2x=0
<=> x2+x=0
<=> x(x+1)=0
<=> \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) (không thỏa mãn điều kiện x\(\ge1\))
Lại có: |x2-x|= x-x2 khi x2-x<0 hay x<1
=> x-x2= -2x
<=> x-x2+2x=0
<=> 3x-x2=0
<=> x(3-x)=0
x=0 (thỏa mãn điều kiện x<1)
hoặc: 3-x=0<=> x=3 (không thỏa mãn điều kiện x<1)
Vậy S=\(\left\{0\right\}\)
d. \(\dfrac{x+3}{x-3}+\dfrac{48x^3}{9-x^2}=\dfrac{x-3}{x+3}\)
ĐKXĐ: \(x\ne\pm3\)
Ta có:\(\dfrac{x+3}{x-3}+\dfrac{48x^3}{9-x^2}=\dfrac{x-3}{x+3}\)
<=> \(\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{48x^3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
=> x2+6x+9-48x3=x2-6x+9
<=> 12x-48x3=0
<=> 12x(1-4x2)=0
<=> 12x(1-2x)(1+2x)=0
<=> \(\left[{}\begin{matrix}x=0\\1-2x=0\\1+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\) (thỏa mãn ĐKXĐ)
Vậy S=\(\left\{0;\pm0,5\right\}\)
