Question

Giải phương trình sau:

lx-2l(x-1)(x+1)(x+2) = 4

Answer 1

TH1 : \(\left|x-2\right|=x-2\)

Ta có :

\(\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=4\)

\(\Leftrightarrow x^4-5x^2=0\)

\(\Leftrightarrow x^2\left(x^2-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

TH2 : \(\left|x-2\right|=2-x\)

Ta có :

\(\left(2-x\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)

\(\Leftrightarrow\left(4-x^2\right)\left(x^2-1\right)=4\)

\(\Leftrightarrow5x^2-x^4=8\)

\(\Leftrightarrow x^4-5x^2+8=0\) (Vô nghiệm)

Vậy...

Answer 2

|x-2|(x-1)(x+1)(x+2) =4;

+) TH1 : (x-2)(x-1)(x+1)(x+2) =4

<=> (x-2)(x+2)(x-1)(x+1) =4;

<=> (x²-4).(x²-1)=4;

<=> x^4-5x²=0;

=> x=√5;

     x=-√5;

    x=0;

+) TH2: (2-x)(x-1)(x+1)(x+2) =4

<=> (2-x)(2+x)(x-1)(x+1) =4;

<=> (4-x²).(x²-1)=4;

=> 5x²-x^4=8;

đặt x²=t;

=> -t²+5t-8=0;

vô nghiệm

vậy x=√5; x=-√5; x=0;