Tìm x
a) \(\sqrt{x}+\sqrt{x-5}\le\sqrt{5}\)
b) \(\frac{x+3}{x+2}< \frac{x+4}{x+5}\)
c) \(3^{x^2-x-6}< 1\)
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1.
a) \(x-4\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=0+4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
Vậy \(x\in\left\{0;16\right\}.\)
b) \(\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|-\frac{3}{4}=\frac{1}{5}\)
\(\Rightarrow\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|=\frac{1}{5}+\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|=\frac{19}{20}.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{5}\sqrt{x}-\frac{1}{20}=\frac{19}{20}\\\frac{3}{5}\sqrt{x}-\frac{1}{20}=-\frac{19}{20}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{3}{5}\sqrt{x}=1\\\frac{3}{5}\sqrt{x}=-\frac{9}{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1:\frac{3}{5}\\\sqrt{x}=\left(-\frac{9}{10}\right):\frac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{5}{3}\\\sqrt{x}=-\frac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{25}{9}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\frac{25}{9}.\)
Câu c) làm tương tự như câu b).
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Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
\(a,\sqrt{x}+\sqrt{x-5}\le\sqrt{5}\)
ĐKXĐ: \(\sqrt{x}\ge0;\sqrt{x-5}\ge0=>x\ge5\)
\(=>\left(\sqrt{x}+\sqrt{x-5}\right)^2\le\left(\sqrt{5}\right)^2\)
\(=>\left(\sqrt{x}\right)^2+2.\sqrt{x}.\sqrt{x-5}+\left(\sqrt{x-5}\right)^2\le5\)
\(=>x+2.\sqrt{x.\left(x-5\right)}+x-5\le5\)
\(=>2x+2\sqrt{x^2-5x}-5\le5=>2x+2\sqrt{x^2-5x}-10\le0\)
\(=>2\left(x+\sqrt{x^2-5x}\right)\le10=>x+\sqrt{x^2-5x}\le5\)
\(=>\sqrt{x^2-5x}\le5-x=>\left(\sqrt{x^2-5x}\right)^2\le\left(5-x\right)^2\)
\(=>x^2-5x\le25-10x+x^2=>25-10x+x^2-x^2+5x\ge0\)
\(=>25-5x\ge0=>5x\le25=>x\le5\)
Mà theo ĐKXĐ: \(x\ge5\) nên x chỉ có thể bằng 5
Vậy x=5
\(b,\frac{x+3}{x+2}<\frac{x+4}{x+5}=>\frac{\left(x+3\right)\left(x+5\right)}{\left(x+2\right)\left(x+5\right)}<\frac{\left(x+4\right)\left(x+2\right)}{\left(x+5\right)\left(x+2\right)}\) (ĐKXĐ: \(x\notin\left\{-5;-2\right\}\))
\(=>\left(x+3\right)\left(x+5\right)<\left(x+4\right)\left(x+2\right)=>x^2+8x+15\)\(<\)\(x^2+6x\)\(+8\)
\(=>x^2+6x+8-x^2-8x-15>0=>-2x-7>0=>-2x>7=>x>-\frac{7}{2}\)
\(c,3^{x^2-x-6}<1=3^0=>x^2-x-6<0\)
\(=>x^2+2x-3x-6<0=>x\left(x+2\right)-3\left(x+2\right)<0=>\left(x+2\right)\left(x-3\right)<0\)
Vì x+2 > x-3
=>x+2 > 0 và x-3 < 0
=>x > -2 và x < 3
=>-2 < x < 3
Vậy.............
- Oa, Phúc giỏi vãi đái ~~~