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a)
Rút gọn :
\(P=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{\left(x+1\right)\left(x-1\right)+x+\left(2-x^2\right)\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{x^2-1+x+2x-2-x^3+x^2}{x\left(x-1\right)}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\frac{-x^3+2x^2+3x-3}{x\left(x-1\right)}\right)\)
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 \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)}=\frac{4}{x-2}\Rightarrow x^2+4x+4=4\)
\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)
a)Ta có : \(4x^2=1\)
\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
mà \(x\ne-\frac{1}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào B , ta được:
\(B=\frac{\left(\frac{1}{2}\right)^2-\frac{1}{2}}{2.\frac{1}{2}+1}=\frac{\frac{1}{4}-\frac{1}{2}}{1+1}=\frac{-\frac{1}{4}}{2}=-\frac{1}{8}\)
Vậy \(B=-\frac{1}{8}\)khi \(4x^2=1\)
b)Ta có : \(A=\frac{1}{x-1}-\frac{x}{1-x^2}\)
\(=\frac{1}{x-1}+\frac{x}{x^2-1}\)
\(=\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow M=A.B=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x^2-x}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x\left(x-1\right)}{2x+1}\)
\(=\frac{x}{x+1}\)
Vậy \(M=\frac{x}{x+1}\)
c)Ta có: \(x< x+1\forall x\)
\(\Rightarrow M=\frac{x}{x+1}< \frac{x+1}{x+1}=1\forall x\ne-1\)
Vậy với mọi \(x\ne-1\)thì \(M< 1\)
a xác định khi và chỉ khi x^2 -1 khác 0 suy ra x^2 khác 1 suy ra x khác 1
\(\frac{x^2-9}{x^2+2x+1}\)khác 0 suy ra x^2-9 khác 0 suy ra x^2 khác 9 suy ra x khác 3
1-x khác 0 suy ra x khác 1
vậy xác định khi x khác 1 và 3
b A = \(\frac{x+3}{x^2-1}\cdot\frac{x^2+2x+1}{x^2-9}-\frac{x}{1-x}\)
= \(\frac{\left(x+3\right)\cdot\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}-\frac{x}{1-x}\)
= \(\frac{x+1}{\left(x-1\right)\left(x-3\right)}+\frac{x}{x-1}\)
= \(\frac{x+1+x\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x+1+x^2-3x}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-2x+1}{\left(x-1\right)\left(x-3\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x-3\right)}=\frac{x-1}{x-3}\)
\(ĐKXĐ:x\ne\pm1\)
a) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)
\(\Leftrightarrow P=\frac{\left(2x+3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+3x+5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{2x^2+x-3-x^2-3x-2+3x+5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x^2+x}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x}{x-1}\)
b) Để \(P\inℤ\)
\(\Leftrightarrow x⋮x-1\)
\(\Leftrightarrow x-1+1⋮x-1\)
\(\Leftrightarrow1⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)