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DT
0
LA
1
22 tháng 7 2018
a,\(\sqrt{x^2-3}\le x^2-3\)
\(\Leftrightarrow x^2-3\le x^4-6x^2+9\)
\(\Leftrightarrow x^4-6x^2-x^2+12\ge0\)
\(\Leftrightarrow x^4-7x^2+12\ge0\)
\(\Leftrightarrow x^4-\frac{2.7}{2}.x^2+\frac{49}{4}-\frac{1}{4}\ge0\)
\(\Leftrightarrow\left(x^2-\frac{7}{2}\right)^2\ge\frac{1}{4}\)
\(\Leftrightarrow x^2-\frac{7}{2}\ge\frac{1}{2}\Leftrightarrow x^2\ge4\)
\(\Leftrightarrow x\le-2\)và \(x\ge2\)
KL:
b,\(\sqrt{x^2-6x+9}>x-6\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}>x-6\)
\(\Leftrightarrow|x-3|>x-6\)
Với x\(\ge\)3 phương trình <=>x-3>x-6 (luôn đúng)
Với x<3 phương trình <=> 3-x>x-6 <=>x<9/2 <=>x<4,5
KL:
VP
0
VP
1
21 tháng 8 2019
\(\text{a) ĐKXĐ: }x\ge\sqrt{3}\)
\(\sqrt{x^2-3}\le x^2-3\)
\(\Leftrightarrow\left(\sqrt{x^2-3}\right)^2\le\left(x^2-3\right)^2\)
\(\Leftrightarrow x^2-3\le x^4-6x^2+9\)
\(\Leftrightarrow x^2-3-x^4+6x^2-9\le0\)
\(\Leftrightarrow-x^4+7x^2-12\le0\)
\(\Leftrightarrow-x^2+4x^2+3x^2-12\le0\)
\(\Leftrightarrow\left(-x^4+4x^2\right)+\left(3x^2-12\right)\le0\)
\(\Leftrightarrow-x^2\left(x^2-4\right)+3\left(x^2-4\right)\le0\)
\(\Leftrightarrow\left(x^2-4\right)\left(3-x^2\right)\le0\)
\(\text{Đến đây EZ rồi}\)
30 tháng 7 2021
a, đk: \(x\ge0,x\ne9,x\ne4\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4-x+3\sqrt{x}-\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2-\sqrt{x}}{-\left(\sqrt{x}-3\right)\left(2-\sqrt{x}\right)}=\dfrac{-1}{\sqrt{x}-3}\)
b,\(Q< -1=>\dfrac{-1}{\sqrt{x}-3}+1< 0< =>\dfrac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(< =>\dfrac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)
\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left[{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\)\(< =>\left[{}\begin{matrix}\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\end{matrix}\right.\)\(< =>9< x< 16\)
c, \(=>2Q=\dfrac{-2}{\sqrt{x}-3}=1+\dfrac{1}{\sqrt{x}-3}\in Z\)
\(< =>\sqrt{x}-3\inƯ\left(1\right)=\left\{\pm1\right\}\)\(=>x\in\left\{16;4\right\}\)(loại 4)
=>x=16
NT
30 tháng 7 2021
a) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-3\dfrac{\sqrt{x}-1}{x-5\sqrt{x}+6}\)
Ta có \(x-5\sqrt{x}+6=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3>0\\\sqrt{x}-2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x>9\\x>2\end{matrix}\right.\) \(\Leftrightarrow x>9\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-3\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(x-4\right)-\left(x-2\sqrt{x}-3\right)-\left(3\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\) \(=\dfrac{-1}{\left(\sqrt{x}-3\right)}=\dfrac{1}{3-\sqrt{x}}\)
b) \(Q< -1\Leftrightarrow\dfrac{1}{3-\sqrt{x}}< -1\) \(\Leftrightarrow\dfrac{1}{3-\sqrt{x}}+1< 0\) \(\Leftrightarrow\dfrac{4-\sqrt{x}}{3-\sqrt{x}}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-\sqrt{x}>0\\3-\sqrt{x}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4-\sqrt{x}< 0\\3-\sqrt{x}>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x>16\\x< 9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow9< x< 16\)
Vậy để \(Q< -1\) thì \(S=\left\{x/9< x< 16\right\}\)
c) \(2Q\in Z\Leftrightarrow\dfrac{2}{3-\sqrt{x}}\in Z\)
\(\Rightarrow3-\sqrt{x}\inƯ\left(2\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}3-\sqrt{x}=2\\3-\sqrt{x}=-2\\3-\sqrt{x}=1\\3-\sqrt{x}=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=25\\x=4\\x=16\end{matrix}\right.\)
Kết hợp với ĐKXĐ,ta có để \(2Q\in Z\) thì \(x\in\left\{16;25\right\}\)
7 tháng 3 2021
a) Ta có: \(M=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{-\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3}{\sqrt{x}-2}\)
a) ĐKXĐ : \(\orbr{\begin{cases}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{cases}}\)
\(\sqrt{x^2-3}=x^2-3\)
\(\Leftrightarrow\sqrt{x^2-3}=\sqrt{x^2-3}\cdot\sqrt{x^2-3}\)
\(\Leftrightarrow\sqrt{x^2-3}-\sqrt{x^2-3}\cdot\sqrt{x^2-3}=0\)
\(\Leftrightarrow\sqrt{x^2-3}\left(1-\sqrt{x^2-3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-3}=0\\\sqrt{x^2-3}=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x^2-3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\in\left\{\pm\sqrt{3}\right\}\\x\in\left\{\pm2\right\}\end{cases}}\)( thỏa mãn )
b) ĐKXĐ : \(x\le6\)
\(\sqrt{x^2-6x+9}=6-x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)
\(\Leftrightarrow\left|x-3\right|=6-x\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=6-x\\x-3=x-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=9\\0x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x\in\varnothing\end{cases}}\)( thỏa mãn )