Question

Bài 1:giải phương trình:

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

b)1 + \(\dfrac{x}{3-x}\) = \(\dfrac{5x}{\left(x+2\right)\left(3-x\right)}\) + \(\dfrac{2}{x+2}\)

c)\(\dfrac{3x+2}{3x-2}\) - \(\dfrac{6}{2+3x}\) = \(\dfrac{9x^2}{9x^2-4}\)

Answer 1

a) ĐKXĐ: \(x\ne-1,x\ne0\)

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

<=> \(\dfrac{x\left(x+3\right)+\left(x-2\right)\left(x+1\right)-2x\left(x+1\right)}{x\left(x+1\right)}=0\)

<=> \(\dfrac{x^2+3x+x^2-x-2-2x^2-2x}{x\left(x+1\right)}=0\)

<=> \(\dfrac{-2}{x\left(x+1\right)}=0\) (vô lí)

=> pt vô nghiệm

b) ĐKXĐ: \(x\ne3,x\ne-2\)

ta có:\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\)

<=> \(\dfrac{\left(x+2\right)\left(3-x\right)+x\left(x+2\right)-5x-2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\)

<=> \(\dfrac{x-x^2+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\)

<=> \(\dfrac{0}{\left(x+2\right)\left(3-x\right)}=0\) (luôn đúng)

Vậy pt trên luôn đúng với mọi x khác 3 và -2

Answer 2

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

(đk: x\(\ne\); x\(\ne\)-1)

<=> \(x^2\)+3x + \(x^2\)-x -2 =\(2x^2\)+2x

<=> 2x -2 =2x

<=>0x=2

=>Pt vô nghiệm.

b) 1+ \(\dfrac{x}{3-x}\)= \(\dfrac{5x}{\left(x+2\right)\left(3-x\right)}\)+\(\dfrac{2}{x+2}\)

(đk:x\(\ne\)3; x\(\ne\)-2)

<=> 3x +6=3x+6

<=>0x=0

=> Pt vô số n_o.

c)\(\dfrac{3x+2}{3x-2}\)-\(\dfrac{6}{2+3x}\)=\(\dfrac{9x^2}{9x^2-4}\)

(đk: x\(\ne\)\(\pm\)\(\dfrac{2}{3}\))

<=>\((3x+2)^2\)-6(3x-2)=\(9x^2\)

<=>\(9x^2 \)+12x +4 -18x+12=\(9x^2\)

<=>16-6x=0

<=>6x=16

<=> x=\(\dfrac{8}{3}\)(t/m)

Vậy pt có n_o duy nhất là x=\(\dfrac{8}{3}\)

Answer 3

MÌnh nghĩ rằng toán học chặt chặt chẽ nhưng không hề gò bó ép buộc

(Mấu chốt đi đến kết quả sao nhanh và nhàn nhất)

a)

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

\(\left(\dfrac{x+3}{x+1}-1\right)+\left(\dfrac{x-2}{x}-1\right)=0\)

\(\left(\dfrac{2}{x+1}\right)-\left(\dfrac{2}{x}\right)=0\Rightarrow\left(x+1\right)=x\Rightarrow VôN_0\)