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ĐKXĐ : \(x\notin\left\{0;-1;-2;-3;-4\right\}\)
Ta có \(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)
\(\Leftrightarrow\dfrac{2x+4}{x.\left(x+4\right)}+\dfrac{2x+4}{\left(x+1\right).\left(x+3\right)}+\dfrac{1}{x+2}=0\)
\(\Leftrightarrow\dfrac{2x+4}{\left(x+2\right)^2-4}+\dfrac{2x+4}{\left(x+2\right)^2-1}+\dfrac{1}{x+2}=0\) (*)
Đặt x + 2 = a \(\left(a\ne0\right)\)
(*) \(\Leftrightarrow\dfrac{2a}{a^2-4}+\dfrac{2a}{a^2-1}+\dfrac{1}{a}=0\)
\(\Leftrightarrow\dfrac{2}{a-\dfrac{4}{a}}+\dfrac{2}{a-\dfrac{1}{a}}+\dfrac{1}{a}=0\) (**)
Đặt \(\dfrac{1}{a}=b\left(b\ne0\right)\) \(\Rightarrow ab=1\)
Ta được (**) \(\Leftrightarrow\dfrac{2}{a-4b}+\dfrac{2}{a-b}+b=0\)
\(\Leftrightarrow\dfrac{2b}{1-4b^2}+\dfrac{2b}{1-b^2}+b=0\)
\(\Leftrightarrow\dfrac{2}{1-4b^2}+\dfrac{2}{1-b^2}=-1\)
\(\Rightarrow4-10b^2=-4b^4+5b^2-1\)
\(\Leftrightarrow4b^4-15b^2+5=0\) (***)
Đặt b2 = t > 0
Ta có (***) <=> \(4t^2-15t+5=0\Leftrightarrow t=\dfrac{15\pm\sqrt{145}}{8}\) (tm)
\(\Leftrightarrow b=\pm\sqrt{\dfrac{15\pm\sqrt{145}}{8}}\)
mà x + 2 = a ; ab = 1
nên \(x=\pm\sqrt{\dfrac{8}{15\pm\sqrt{145}}}-2\)
Thử lại ta có phương trình có 4 nghiệm như trên
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)
b) ĐKXĐ : \(x\ne\pm1\)
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1+2x-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{2}{\sqrt{x}}\)
a) ĐKXĐ : \(x\ge0;x\ne16\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)
\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0\)
\(\Leftrightarrow\dfrac{5}{5\sqrt{x}+10}-\dfrac{\sqrt{x}+2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{5-\sqrt{x}-2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-3\right)}{5\sqrt{x}+10}>0\)
Mà: \(5\sqrt{x}+10\ge10>0\forall x\)
\(\Leftrightarrow\sqrt{x}>3\)
\(\Leftrightarrow x>9\)
_________
\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{4}{2\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)}{2\sqrt{x}+6}< 0\)
Mà: \(2\sqrt{x}+6\ge6>0\forall x\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
\(\Leftrightarrow0\le x\le1\)
\(\text{a) }\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ =\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\\ =\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
\(\text{b) }\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\\ =1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2017}-\dfrac{1}{2018}\\ =2016+\dfrac{1}{2}-\dfrac{1}{2018}\\ =\dfrac{2034698}{1009}\)
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
\(C=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{9-5}}\)
\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{6+2\sqrt{5}}{4}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)}{5-1}+\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{5}+1}{2}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{2}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)
\(D=\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-2x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(\sqrt{x}+1\right)}=-\dfrac{2}{\sqrt{x}+1}\)