cho \(P=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
Tìm x để P nguyên
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26 tháng 12 2020
\(P=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)
\(=\left[\frac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}-\frac{\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right]:\left[\frac{2\left(\sqrt{x}-1\right)^2}{x-1}\right]\)
Phương trình tương đương :
\(=\frac{2x^2-2x}{x^2-x}:\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2:\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\frac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
19 tháng 8 2020
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(P=\frac{x-3\sqrt{x}-x-9}{x-9}.\frac{x\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(P=\frac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{x\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(P=\frac{-3x}{2\left(\sqrt{x}+2\right)}\)
25 tháng 6 2023
a: \(P=\dfrac{x-\sqrt{x}-1-\sqrt{x}+1}{x-1}\cdot\dfrac{4\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)^2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)\cdot4\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{4}{x-1}\)
Để P nguyên dương thì x-1 thuộc {1;4;2}
=>x thuộc {2;5;3}
b: x+y+z=0
=>x=-y-z; y=-x-z; z=-x-y
\(P=\dfrac{x^2}{y^2+z^2-\left(y+z\right)^2}+\dfrac{y^2}{z^2+x^2-\left(x+z\right)^2}+\dfrac{z^2}{x^2+y^2-\left(x+y\right)^2}\)
\(=\dfrac{x^2}{-2yz}+\dfrac{y^2}{-2xz}+\dfrac{z^2}{-2xy}\)
\(=\dfrac{x^3+y^3+z^3}{2xyz}\cdot\left(-1\right)\)
\(=-\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)}{2xyz}\)
\(=-\dfrac{\left(-z\right)^3+z^3-3xy\cdot\left(-z\right)}{2xyz}=-\dfrac{3}{2}\)
3 tháng 8 2017
ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
a. Ta có \(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right):\frac{2x}{\sqrt{x}-1}\)
\(=2\sqrt{x}.\frac{\sqrt{x}-1}{2x}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b. Để \(A< 0\Rightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< 0\Rightarrow\sqrt{x}-1< 0\Rightarrow0\le x< 1\)
Vậy \(0\le x< 1\)thì \(A< 0\)
c. Ta có \(A=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)
Để A nguyên thì \(\sqrt{x}\inƯ\left(1\right)\Rightarrow x=1\)
Vậy với x=1 thì A nguyên
xét \(b^2+2a^2=\frac{b^2}{1}+\frac{a^2}{1}+\frac{a^2}{1}\ge\frac{\left(b+a+a\right)^2}{1+1+1}=\frac{\left(b+2a\right)^2}{3}\)
\(\Leftrightarrow\sqrt{b^2+2a^2}\ge\sqrt{\frac{\left(b+2a\right)^2}{3}}=\frac{b+2a}{\sqrt{3}}\)\(\Leftrightarrow\frac{\sqrt{b^2+2a^2}}{ab}\ge\frac{b+2a}{ab\sqrt{3}}\)
Tương tự \(\frac{\sqrt{c^2+2b^2}}{bc}\ge\frac{c+2b}{bc\sqrt{3}}\)và \(\frac{\sqrt{a^2+2c^2}}{ca}\ge\frac{a+2c}{ca\sqrt{3}}\)
\(\Rightarrow\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\frac{b+2a}{ab\sqrt{3}}+\frac{c+2b}{bc\sqrt{3}}+\frac{a+2c}{ca\sqrt{3}}\)\(=\frac{b}{ab\sqrt{3}}+\frac{2a}{ab\sqrt{3}}+\frac{c}{bc\sqrt{3}}+\frac{2b}{bc\sqrt{3}}+\frac{a}{ca\sqrt{3}}+\frac{2c}{ca\sqrt{3}}\)\(=\frac{1}{a\sqrt{3}}+\frac{2}{b\sqrt{3}}+\frac{1}{b\sqrt{3}}+\frac{2}{c\sqrt{3}}+\frac{1}{c\sqrt{3}}+\frac{2}{a\sqrt{3}}\)\(=\frac{3}{\sqrt{3}}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\sqrt{3}\cdot1980\)(đpcm)
Dấu "=" xăy ra khi a=b=c=3/1980