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\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}=\frac{x^2-2x.2007+2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}\)
\(=\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)
A min =\(\frac{2006}{2007}\)khi \(x-2007=0\)
\(\Leftrightarrow x=2007\)
\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}\)
\(A=\frac{x^2-2x.2007-2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}\)
\(A=\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)
\(\Rightarrow Amin=\frac{2006}{2007}\)khi \(x-2007=0\)
\(\Rightarrow x=2007\)
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự
thiếu đề : \(\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}.\)
Bài 2 :
a, Để \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4^2-4}{5}\)
\(\Rightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)
b,\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4x^2-4}{5}\)
\(B=\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\left[\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\left[\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{4}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(B=\frac{8}{5}\)
=> giá trị của B ko phụ thuộc vào biến x
bài 1
=\(^{\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x+1\right)^2}\)
=\(\left(2x+1+2x-1\right)^2\)
=\(\left(4x\right)^2\)
=\(16x^2\)
Tại x=100 thay vào biểu thức trên ta có:
16*100^2=1600000
\(A=\frac{x^2-2x+2007}{2007x^2}=\frac{2006}{2007^2}+\frac{x^2-4014x+2007^2}{2007^2x^2}=\frac{2006}{2007^2}+\frac{\left(x-2007\right)^2}{2007^2x^2}\ge\frac{2006}{2007^2}\)
Dấu ''='' xảy ra \(\Leftrightarrow\) x = 2007
\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}\)
\(=\frac{x^2-2x.2007+2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}\)
\(=\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)
A min =\(\frac{2006}{2007}\)khi \(x-2007=0\) hay \(x=2007\)