Question

cho B = \(\frac{4}{x^2-2x+1}-\left(\frac{x}{x^2-1}-\frac{1}{x^3-x}\right):\frac{x^2-2x+1}{x^3+x}\)

a) Tìm ĐKXĐ và rút gọn B

b) tính giá trị B khi /x-1/ = 2

c) tìm x để B = -1

d) so sánh B với -2

e) GTNN của B

 

Answer 1

a.ĐKXĐ \(x\ne0,x\ne1\),\(x\ne-1\)

B=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2-1}{x^3-x}.\frac{x^3+x}{\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x.\left(x^2+1\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x-1\right)^2}\)=\(\frac{4}{\left(x-1\right)^2}-\frac{x^2+1}{\left(x-1\right)^2}\)

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

b.TH1 x=3\(\Rightarrow\)B=\(\frac{3-3^2}{2^2}=\frac{-3}{2}\)

TH2 x=-1\(\Rightarrow\)B=\(\frac{3-\left(-1\right)^2}{4}=\frac{1}{2}\)

c.B=-1\(\Leftrightarrow\frac{3-x^2}{\left(x-1\right)^2}=-1\)\(\Leftrightarrow x^2-3=x^2-2x+1\)\(\Leftrightarrow2x=4\Leftrightarrow x=2\)

d.B+2=\(\frac{3-x^2}{\left(x-1\right)^2}+2=\frac{x^2-4x+5}{\left(x-1\right)^2}=\frac{\left(x-2\right)^2+1}{\left(x-1\right)^2}\ge0\)với mọi x\(\Rightarrow B\)>-2