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1: ĐKXĐ: x \(\ne\) 0; 1
\(\frac{1}{x}+\frac{1}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{2x-1}{x\left(x-1\right)}=\frac{3}{2}\)
\(\Leftrightarrow3x^2-3x=4x-2\)
\(\Leftrightarrow3x^2-7x=-2\)
\(\Leftrightarrow9x^2-21x+12,25=6,25\)
\(\Leftrightarrow\left(3x-3,5\right)^2=6,25\)
\(\Leftrightarrow...\left(tl\right)\)
a, tự lm......
P=x2 / x-1
b, P<1
=> x2/x-1 <1
<=>x2/x-1 -1 <0
<=>x2-x+1 / x-1<0
Vi x2-x+1= (x -1/2 )2+3/4 >0
=> Để P<1
x-1 <0
x <1
c, x2/x-1 = x2-1+1/x-1
= x+1 +1/x-1
= 2 +(x-1) + 1/x-1
Áp dụng BDT Cô si ta có :
x-1 + 1/x-1 >hoặc = 2
=> P>= 3
Đầu = xảy ra <=> x=2( x >1)
Vay......
làm đúng nhuwng phần c, phải >=4 cơ vì công cả 2 vế với 2 ta có P>=4
làm a thôi nha :D
a) \(C=\left(\frac{x^2+x}{x^2-2x+1}\right):\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{2-x^2}{x\left(x+1\right)}\right]\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right]\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{x+1}{x}+\frac{x+2-x^2}{x\left(x-1\right)}\right]\)
\(C=\frac{x\left(x+1\right)}{x^2-2x+1}.\left[\frac{\left(x-1\right)\left(x+1\right)+x+2-x^2}{x\left(x-1\right)}\right]\)
\(C=\frac{x+1}{x^2-2x+1}.\frac{x^2-1+x+2-x^2}{x-1}\)
\(C=\frac{x+1}{\left(x^2-2x+1\right)}.\frac{1.x}{x-1}\)
\(C=\frac{\left(x+1\right)^2}{x^3-x^2-2x^2+2x+x-1}\)
\(C=\frac{x^2+2x+1}{x^3-3x^2+3x-1}\)
a)\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x}-\frac{1}{-\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)
\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x^2-1}{x.\left(x-1\right)}+\frac{x}{x.\left(x-1\right)}+\frac{-x^2+2}{x.\left(x-1\right)}\right]\)
\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x^2-1+x-x^2+2}{x.\left(x-1\right)}\right]\)
\(C=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right]:\left[\frac{x+1}{x.\left(x-1\right)}\right]=\left[\frac{x.\left(x+1\right)}{\left(x-1\right)^2}\right].\left[\frac{x.\left(x-1\right)}{x+1}\right]=\frac{x.\left(x+1\right).x}{\left(x-1\right).\left(x+1\right)}=\frac{x^2}{x-1}\)
b)\(\text{Để B nguyên }\Rightarrow x^2⋮x-1\)
\(x^2=x^2-1+1=\left(x-1\right).\left(x+1\right)+1\)
\(\Rightarrow\text{Để }x^2⋮x-1\Rightarrow1⋮x-1\Rightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\Rightarrow x\in\left\{2;0\right\}\)
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.