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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.
a) Ta thấy x=-2 thỏa mãn ĐKXĐ của B.
Thay x=-2 và B ta có :
\(B=\frac{2\cdot\left(-2\right)+1}{\left(-2\right)^2-1}=\frac{-3}{3}=-1\)
b) Rút gọn :
\(A=\frac{3x+1}{x^2-1}-\frac{x}{x-1}\)
\(=\frac{3x+1-x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{-x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
Xấu nhỉ ??
\(A=\left(\frac{x-2}{2x-2}+\frac{3}{2x-2}-\frac{x+3}{2x+2}\right):\left(-1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x-2}{2\left(x-1\right)}+\frac{3}{2\left(x-1\right)}+\frac{-\left(x+3\right)}{2\left(x+1\right)}\right):\left(-\frac{1}{1}+\frac{-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{\left(x-2\right)\left(x+1\right)+3\left(x+1\right)-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{-1\left(x+1\right)-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{x^2-x^2+x+3x-2x-6+3+3}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x-1-x+3}{x+1}\right)\)
=\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}:\frac{2}{x+1}\)
\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}.\frac{x+1}{2}\)
\(=\frac{x}{2\left(x-1\right)}\)
b,Thayx=2005
\(\Rightarrow A=\frac{2005}{4008}\)
\(ĐKXĐ:x\ne1\)
a) \(A=\left(1+\frac{x^2}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x^2+1-2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x-1}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x-1}\)
b) Thay \(x=-\frac{1}{2}\)vào A, ta được :
\(A=\frac{2\left(-\frac{1}{2}\right)^2+1}{-\frac{1}{2}-1}\)
\(\Leftrightarrow A=\frac{\frac{3}{2}}{-\frac{3}{2}}\)
\(\Leftrightarrow A=-1\)
c) Để A < 1
\(\Leftrightarrow2x^2+1< x-1\)
\(\Leftrightarrow2x^2-x+2< 0\)
\(\Leftrightarrow2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{15}{8}< 0\)
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}< 0\)
\(\Leftrightarrow x\in\varnothing\)
Vậy để \(A< 1\Leftrightarrow x\in\varnothing\)
d) Để A có giá trị nguyên
\(\Leftrightarrow2x^2+1⋮x-1\)
\(\Leftrightarrow2x^2-2x+2x-2+3⋮x-1\)
\(\Leftrightarrow2x\left(x-1\right)+2\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow2\left(x+1\right)\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow3⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
1: ĐKXĐ: x \(\ne\) 0; 1
\(\frac{1}{x}+\frac{1}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{2x-1}{x\left(x-1\right)}=\frac{3}{2}\)
\(\Leftrightarrow3x^2-3x=4x-2\)
\(\Leftrightarrow3x^2-7x=-2\)
\(\Leftrightarrow9x^2-21x+12,25=6,25\)
\(\Leftrightarrow\left(3x-3,5\right)^2=6,25\)
\(\Leftrightarrow...\left(tl\right)\)