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21 tháng 8 2019

\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(:\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)

\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)

Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)

Nhưng \(2\sqrt{x}+1\ge1\)

\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

Vậy \(x\in\left\{0;4\right\}\)

17 tháng 5 2019

Hỏi đáp Toán

18 tháng 5 2019

chữ đẹp vcl

NV
15 tháng 5 2019

\(P=\frac{B}{A}=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}}{\sqrt{x}-2}\)

Để \(\left|P\right|>P\Rightarrow P< 0\)

\(\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-2}< 0\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

\(x\) nguyên \(\Rightarrow x=\left\{2;3\right\}\)

NV
12 tháng 2 2020

\(A=\frac{1}{6}\left(6-2x\right)\left(12-3y\right)\left(2x+3y\right)\)

\(A\le\frac{1}{6}\left(\frac{6-2x+12-3y+2x+3y}{3}\right)^3=36\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

\(A=\frac{\frac{ab}{\sqrt{2}}\sqrt{2\left(c-2\right)}+\frac{bc}{\sqrt{3}}\sqrt{3\left(a-3\right)}+\frac{ca}{2}\sqrt{4\left(b-4\right)}}{abc}\)

\(A\le\frac{\frac{abc}{2\sqrt{2}}+\frac{abc}{2\sqrt{3}}+\frac{abc}{4}}{abc}=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)