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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
1/a/
\(A=\frac{2}{xy}+\frac{3}{x^2+y^2}=\left(\frac{1}{xy}+\frac{1}{xy}+\frac{4}{x^2+y^2}\right)-\frac{1}{x^2+y^2}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}-\frac{1}{\frac{\left(x+y\right)^2}{2}}=16-2=14\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
b/
\(4B=\frac{4}{x^2+y^2}+\frac{8}{xy}+16xy=\left(\frac{4}{x^2+y^2}+\frac{1}{xy}+\frac{1}{xy}\right)+\left(\frac{1}{xy}+16xy\right)+\frac{5}{xy}\)
\(\ge\frac{\left(1+1+2\right)^2}{\left(x+y\right)^2}+2\sqrt{\frac{1}{xy}.16xy}+\frac{5}{\frac{\left(x+y\right)^2}{4}}\)
\(=16+8+20=44\)
\(\Rightarrow B\ge11\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
a) \(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{x^2-1}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2-x-3}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{\left(x^2-x\right)+\left(2x^2+2x-3x-3\right)-\left(2x^2-x-3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x}{x+1}\)
MÌnh nghĩ đề câu b là với x>-4 mới đúng chứ
\(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{\left(x^2-1\right)}.\)
\(=\frac{x\left(x-1\right)+\left(2x-3\right)\left(x+1\right)-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)
\(\Rightarrow A.B=\frac{x}{\left(x+1\right)}.\frac{x\left(x+1\right)}{\left(x-2\right)}=\frac{x^2}{\left(x-2\right)}=\frac{x^2-4+4}{\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+4}{\left(x-2\right)}=x+2+\frac{4}{x-2}=x-2+\frac{4}{x-2}+4\)
Áp dụng BĐT Cô - Si cho 2 số dương \(x-2;\frac{4}{x-2}\)ta có :
\(x-2+\frac{4}{x-2}\ge2\sqrt{\frac{\left(x-2\right).4}{x-2}}=2\sqrt{4}=4\)
\(\Rightarrow x-2+\frac{4}{x-2}\ge4\Rightarrow x-2+\frac{4}{x-2}+4\ge8\)
Hay \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)}=\frac{4}{x-2}\Rightarrow x^2+4x+4=4\)
\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)
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.
1: ĐKXĐ: x \(\ne\) 0; 1
\(\frac{1}{x}+\frac{1}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{2x-1}{x\left(x-1\right)}=\frac{3}{2}\)
\(\Leftrightarrow3x^2-3x=4x-2\)
\(\Leftrightarrow3x^2-7x=-2\)
\(\Leftrightarrow9x^2-21x+12,25=6,25\)
\(\Leftrightarrow\left(3x-3,5\right)^2=6,25\)
\(\Leftrightarrow...\left(tl\right)\)