Question

Giải các pt sau:

A). (3x-4)² = 9(x-1)(x+1)

B). (4x-5)² -4(x-2)² =0

C). |X² - x| = -2x

D). (X+3)/(x-3)+(48x³)/(9-x²)=(x-3)/(x+3)

Answer 1

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=\(\left\{\dfrac{1}{2};\dfrac{3}{2}\right\}\)

Answer 2

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=\(\left\{0;\pm0,5\right\}\)