Question

1.Giải pt: \(\frac{1}{x}+\frac{1}{x-1}=\frac{3}{2}\)

2. Cho bt: \(P=\frac{x^2}{x-1}-\frac{2x^2-x}{x^2-x}\)

a) Rút gọn P

b) Tìm gt của x để P < 1

Answer 1

Bài 1:

\(\frac{1}{x}\)+\(\frac{1}{x-1}=\frac{3}{2}\)(x khác 0, x khác 1)

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

(=)2(x-1)+2x=3x(x-1)

(=)2x-2+2x=3x²-3x

(=)4x-2=3x²-3x

(=)4x-2-3x²+3x=0

(=)-3x²+7x-2=0

(=)-3x²+6x+x-2=0

(=)-3x(x-2)+(x-2)=0

(=)(-3x+1)(x-2)=0

(=)TH1:-3x+1=0

(=)-3x=-1

(=)x=1/3 (TMĐKXĐ)

TH2:x-2=0

(=)x=2 (TMĐKXĐ)

Vậy S={1/3;2}

Bài 2:

a)P=\(\frac{x^2}{x-1}-\frac{2x^2-x}{x^2-x}\)(x≠_+1;x≠0)

=\(\frac{x^2}{x-1}-\frac{2x^2-x}{x\left(x-1\right)}\)

=\(\frac{x^3}{x\left(x-1\right)}-\frac{2x^2-x}{x\left(x-1\right)}\)

=\(\frac{x^3-2x^2+x}{x\left(x-1\right)}\)

=\(\frac{x^3-x^2-x^2+x}{x\left(x-1\right)}\)

=\(\frac{x^2\left(x-1\right)-x\left(x-1\right)}{x\left(x-1\right)}\)

=\(\frac{\left(x-1\right)\left(x^2-x\right)}{x\left(x-1\right)}\)

=\(\frac{\left(x-1\right)x\left(x-1\right)}{x\left(x-1\right)}\)

=x-1

b)P<1

(=)P-1<0

(=)x-1-1<0

(=)x-2<0

(=)x<2

Vậy P<1 với x<2 khi x khác 0 và -1.