\(Q=\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)
Tìm x \(\in\)Z để Q\(\in\)Z
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30 tháng 8 2020
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne\frac{9}{4}\end{matrix}\right.\)
Ta có: \(Q=\frac{\sqrt{x}+2}{-\sqrt{x}+2}+\frac{3\sqrt{x}-4}{2\sqrt{x}-3}+\frac{-7\sqrt{x}+10}{-2x+7\sqrt{x}-6}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(2\sqrt{x}-3\right)}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}+\frac{\left(3\sqrt{x}-4\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}+\frac{-7\sqrt{x}+10}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{2x+\sqrt{x}-6-3x+10\sqrt{x}-8-7\sqrt{x}+10}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{-x+4\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{-\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}-2}{2\sqrt{x}-3}\)
b) Để Q<-4 thì Q+4<0
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}-3}+\frac{4\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
Trường hợp 1: \(\left\{{}\begin{matrix}9\sqrt{x}-14>0\\2\sqrt{x}-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}>14\\2\sqrt{x}< 3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>\frac{14}{9}\\\sqrt{x}< \frac{3}{2}\end{matrix}\right.\)
⇔Loại vì \(\frac{14}{9}>\frac{3}{2}\)
Trường hợp 2: \(\left\{{}\begin{matrix}9\sqrt{x}-14< 0\\2\sqrt{x}-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}< 14\\2\sqrt{x}>3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}< \frac{14}{9}\\\sqrt{x}>\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< \frac{196}{81}\\x>\frac{9}{4}\end{matrix}\right.\Leftrightarrow\frac{9}{4}< x< \frac{196}{81}\)
Kết hợp ĐKXĐ, ta được:
\(\frac{9}{4}< x< \frac{196}{81}\)
Vậy: Để Q<-4 thì \(\frac{9}{4}< x< \frac{196}{81}\)
6 tháng 7 2016
điều kiện \(x\ge0\)và x khác 1/4
Q= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{2x+7\sqrt{x}-4}\)
=\(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)
đề Q>1/2 thì \(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}>\frac{1}{2}\)
<=> \(8x+34\sqrt{x}+8>2x+7\sqrt{x}-4\)<=> \(6x+27\sqrt{x}+12>0\) với mọi x>=0
vậy Q>1/2 khi x>=0 và x khác 1/4
12 tháng 9 2020
Đk: x \(\ge\)0; x \(\ne\)1; x \(\ne\)9
1) \(B=\left(\frac{2x+3}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(B=\frac{2x+3-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\)
\(B=\frac{-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(B=\frac{-\left(x+2\sqrt{x}-\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+2}{3-\sqrt{x}}\)
2. \(B=\frac{\sqrt{x}+2}{3-\sqrt{x}}=\frac{-\left(3-\sqrt{x}\right)+5}{3-\sqrt{x}}=-1+\frac{5}{3-\sqrt{x}}\)
Để B \(\in\)Z <=> 5 \(⋮\)\(3-\sqrt{x}\)
<=> \(3-\sqrt{x}\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Do \(3-\sqrt{x}\le\)3 => 3 - \(\sqrt{x}\)\(\in\){1; -1; -5}
Lập bảng:
| \(3-\sqrt{x}\) | 1 | -1 | -5 |
| x | 4 | 16 | 64 |
Vậy ...
Rút gọn : \(Q=\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\) (ĐK : \(0\le x\ne\frac{1}{4}\))
\(=\frac{\left(\sqrt{x}+4\right)\left(4\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+4\right)}=\frac{4\sqrt{x}+1}{2\sqrt{x}-1}\)
\(Q\in Z\Leftrightarrow\frac{4\sqrt{x}+1}{2\sqrt{x}-1}\in Z\Leftrightarrow2+\frac{3}{2\sqrt{x}-1}\in Z\Leftrightarrow\frac{3}{2\sqrt{x}-1}\in Z\Rightarrow\left(2\sqrt{x}-1\right)\inƯ\left(3\right)\)
Do \(x\ge0\)nên \(2\sqrt{x}-1\ge-1\Rightarrow\left(2\sqrt{x}-1\right)\in\left\{-1;1;3\right\}\Leftrightarrow x\in\left\{0;1;4\right\}\)