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Tự tìm ĐKXĐ nhé
\(P=\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
c, \(P=\frac{\sqrt{x}+4}{\sqrt{x}+2}=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow1+\frac{2}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1;0\right\}\)
\(\Rightarrow x=\left\{0\right\}\)
Kết hợp với ĐKXĐ =>...
\(2ab+3bc+4ca=5abc\)
Do a,b,c lần lượt là độ dài 3 cạnh của tam giác
\(\Rightarrow\frac{2ab}{abc}+\frac{3bc}{abc}+\frac{4ca}{abc}=\frac{5abc}{abc}\Rightarrow\frac{2}{c}+\frac{3}{a}+\frac{4}{b}=5\)
Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với x,y >0 (Dấu "=" xảy ra khi x=y)
Ta có: \(P=\frac{7}{a+b-c}+\frac{6}{b+c-a}+\frac{5}{a+c-b}\)
\(=\left(\frac{2}{b+c-a}+\frac{2}{c+a-b}\right)+\left(\frac{3}{c+a-b}+\frac{3}{a+b-c}\right)+\left(\frac{4}{a+b-c}+\frac{4}{b+c-a}\right)\)
\(=2\left(\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)+3\left(\frac{1}{c+a-b}+\frac{1}{a+b-c}\right)+4\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}\right)\)
\(\ge\frac{8}{2c}+\frac{12}{2a}+\frac{16}{2b}=2\left(\frac{2}{c}+\frac{3}{a}+\frac{4}{b}\right)=10\)
Vậy ...
ĐKXĐ tất cả các câu bạn tự tìm
\(C=\frac{4\left(\sqrt{x}+3\right)+3}{\sqrt{x}+3}=4+\frac{3}{\sqrt{x}+3}\le4+\frac{3}{3}=5\)
\(C_{max}=5\) khi \(x=0\)
\(A=\frac{2\left(\sqrt{x}+2\right)-17}{\sqrt{x}+2}=2-\frac{17}{\sqrt{x}+2}\ge2-\frac{17}{2}=-\frac{13}{2}\)
\(A_{min}=-\frac{13}{2}\) khi \(x=0\)
\(B=\frac{x+2\sqrt{x}+1+9}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)^2+9}{\sqrt{x}+1}=\sqrt{x}+1+\frac{9}{\sqrt{x}+1}\)
\(B\ge2\sqrt{\frac{9\left(\sqrt{x}+1\right)}{\sqrt{x}+1}}=6\Rightarrow B_{min}=6\) khi \(\sqrt{x}+1=3\Leftrightarrow x=4\)
\(A=\frac{2\left(\sqrt{x}+2\right)+1}{\sqrt{x}+2}=2+\frac{1}{\sqrt{x}+2}\)
Để A nguyên \(\Rightarrow\sqrt{x}+2=Ư\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+2=-1\\\sqrt{x}+2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-3\left(l\right)\\\sqrt{x}=-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x nguyên để A nguyên
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}< 1\)
Mặt khác \(A+2=\frac{\sqrt{x}-2}{\sqrt{x}+1}+2=\frac{\sqrt{x}-2+2\sqrt{x}+2}{\sqrt{x}+1}=\frac{3\sqrt{x}}{\sqrt{x}+1}\ge0\)
\(\Rightarrow A\ge-2\Rightarrow-2\le A< 1\)
Mà A nguyên \(\Rightarrow A=\left\{-2;-1;0\right\}\)
- Với \(A=-2\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}=-2\Rightarrow\sqrt{x}-2=-2\sqrt{x}-2\)
\(\Rightarrow3\sqrt{x}=0\Rightarrow x=0\)
- Với \(A=-1\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}=-1\Rightarrow\sqrt{x}-2=-\sqrt{x}-1\)
\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)
- Với \(A=0\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}=0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)
Vậy \(x=\left\{0;\frac{1}{4};4\right\}\)