a: ĐKXĐ: x<>1; x<>2; x<>-2; x<>-1

\(P=\dfrac{2017x+2017-2016x+2016-2014x-2016}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-2015x+2017}{x^2-4}\)

Ta có: \(G=\left(\dfrac{x-\sqrt{x}+2}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x+2\sqrt{x}+1}{2x-2\sqrt{x}}\)

\(=\dfrac{x-\sqrt{x}+2-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

Lời giải:

Trong TH này ta thêm điều kiện $x$ là số nguyên dương.

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)

\(=1-\frac{1}{x+1}=\frac{x}{x+1}\)

Vậy \(\frac{x}{x+1}=\frac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)

\(\Rightarrow x\sqrt{2016-x}+2017x=(x+1)\sqrt{2017-x}+2016(x+1)\)

\(\Leftrightarrow x\sqrt{2016-x}=(x+1)\sqrt{2017-x}+2016-x\)

\(\Leftrightarrow x(\sqrt{2017-x}-\sqrt{2016-x})+\sqrt{2017-x}+2016-x=0\)

\(\Leftrightarrow \frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}+\sqrt{2017-x}+(2016-x)=0\)

Hiển nhiên ta thấy:

\(\frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}>0\)

\(\sqrt{2017-x}\geq 0\)

\(2016-x\geq 0\)

Do đó pt trên vô nghiệm

Tức là không tìm đc $x$ thỏa mãn.

\(=\dfrac{x+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

D. Tìm x thuộc Z biết 

x+(x+1)+(x+2)+....+2016+2017=2017 

=> ( x + x + x + ..+ x ) + ( 1 + 2 + 3+...+2016 + 2017 ) = 2017 

<=> 2017x + 2035153 = 2017 

=> 2017x = -2033136

=> x = -1008

Vậy ...

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