Question

giải các phương trình sau
1. x-3/x-2+x-2/x-4=1
2. 2x-5/x-2-3x-5/x-1+1=0
3. x+1/x-1-x-1/x+1=16/x^2-1
4. 12/x²-4-x+1/x-2+x+7/x+2=0
5. x-2/x+2-3/x-2=2(x-11)/x²-4

6. 8x²/3(1-4x²)=2x/6x-3-1+8x/8x+4

7. x+2/x-2-2/x²-2x=1/x

8.x+1/x-2-x-1/x+2=2(x²+2)/x²-4

Answer 1

1: \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=1\)

\(\Leftrightarrow x^2-7x+12+x^2-4x+4=x^2-6x+8\)

\(\Leftrightarrow2x^2-11x+16-x^2+6x-8=0\)

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

\(\text{Δ}=\left(-5\right)^2-4\cdot1\cdot8=-7< 0\)

Do đó: PTVN

2: \(\dfrac{2x-5}{x-2}-\dfrac{3x-5}{x-1}+1=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)+\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow2x^2-2x-5x+5-\left(3x^2-3x-5x+10\right)+x^2-3x+2=0\)

\(\Leftrightarrow3x^2-10x+7-3x^2+8x-10=0\)

=>-2x-3=0

hay x=-3/2(nhận)

3: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=16\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=16\)

=>4x=16

hay x=4