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1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
\(a,x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=a^2-2\)
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)
....................................
Ta có \(x^4+x^2+1=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(\Rightarrow\frac{x^2}{x^4+x^2+1}=\frac{x^2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=a\cdot\frac{x}{x^2+x+1}=\frac{ax}{x^2-x+1+2x}=\frac{ax}{a+2x}\)
(ko biết có đúng ko nữa..)
\(a,\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x^2-6x}\)
\(=\frac{7}{x}-\frac{x}{x+6}+\frac{36}{x\left(x-6\right)}\)
\(=\frac{7\left(x-6\right)\left(x+6\right)-x\left(x-6\right)+36\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)}\)
\(=\frac{7\left(x^2-6\right)-x^2+6x+36x+216}{x\left(x^2-6\right)}\)
\(=\frac{7x^2-42-x^2+6x+36x+216}{x\left(x^2-6\right)}\)
\(=\frac{6x^2+42x+216}{x\left(x^2-6\right)}\)
\(=\frac{6\left(x^2+7x+36\right)}{x\left(x^2-6\right)}\)
Đề sai nhé, phải là như này nè :
\(b,\frac{1}{x^2-x+1}-\frac{1}{x^2+x+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{x^2+x+1-\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{x^2+x+1-x^2+x-1}{x^4+x^2+1}\)\(-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x}{x^4+x^2+1}-\frac{2x}{x^4-x^2+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x\left(x^4-x^2+1\right)-2x\left(x^4+x^2+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)}+\frac{4x^3}{x^8-x^4+1}\)
\(=\frac{2x^5-2x^3+2x-2x^5-2x^3-2x}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}\)
\(=-\frac{4x^3}{x^8-x^4+1}+\frac{4x^3}{x^8-x^4+1}=0\)
\(M=\frac{x^2}{x^4+x^2+1}=\frac{x^2}{x^4+2x^2+1-x^2}=\frac{x^2}{\left(x^2+1\right)^2-x^2}=\frac{x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(\frac{x}{x^2-x+1}=a\Rightarrow\frac{1}{a}=\frac{x^2-x+1}{x}=\frac{x^2+x-2x+1}{x}=\frac{x^2+x+1}{x}-2\)
\(\Rightarrow\frac{x}{x^2+x+1}=\frac{a}{2a+1}\)
Suy ra \(M=a.\frac{a}{2a+1}=\frac{a^2}{2a+1}\).