\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}+\dfrac{4\sqrt{x}-6}{9-x}\)(x\(\ge\)0;x\(\ne\)9)
tìm x để A nhận giá trị nguyên
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16 tháng 5 2021
`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
16 tháng 5 2021
`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
28 tháng 8 2023
a:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
7 tháng 11 2021
\(a,A=\dfrac{2\cdot2-4}{2-1}=0\\ b,B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ c,AB=\dfrac{2\sqrt{x}-4}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-4}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\\ AB=5-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)
Vì \(\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+1}>0\) nên \(AB< 5\)
7 tháng 11 2021
a. \(x=4\Rightarrow A=\dfrac{2.\sqrt{4}-4}{\sqrt{4}-1}=0\)
b. \(\Rightarrow B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow B=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
AT
12 tháng 7 2021
\(A=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(A< \dfrac{3}{5}\Rightarrow\dfrac{3}{5}-A>0\Rightarrow\dfrac{3}{5}-\dfrac{\sqrt{x}-3}{\sqrt{x}-1}>0\)
\(\Rightarrow\dfrac{3\left(\sqrt{x}-1\right)-5\left(\sqrt{x}-3\right)}{5\left(\sqrt{x}-1\right)}>0\Rightarrow\dfrac{12-2\sqrt{x}}{5\left(\sqrt{x}-1\right)}>0\)
\(\Rightarrow\dfrac{2}{5}.\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\Rightarrow\dfrac{6-\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-\sqrt{x}>0\\\sqrt{x}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}6-\sqrt{x}< 0\\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1< x< 36\\\left\{{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\left(l\right)\end{matrix}\right.\)
\(\Rightarrow1< x< 36\)
12 tháng 7 2021
\(=>A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
để \(A< \dfrac{3}{5}< =>\dfrac{\sqrt{x}-3}{\sqrt{x}-1}< \dfrac{3}{5}\)
\(< =>\dfrac{5\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-1\right)}{5\left(\sqrt{x}-1\right)}< 0\)
\(< =>\dfrac{2\sqrt{x}-12}{5\left(\sqrt{x}-1\right)}< 0\)
\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}2\sqrt{x}-12>0\\5\left(\sqrt{x}-1\right)< 0\end{matrix}\right.\\\left[{}\begin{matrix}2\sqrt{x}-12< 0\\5\left(\sqrt{x}-1\right)>0\end{matrix}\right.\end{matrix}\right.\)\(=>\left\{{}\begin{matrix}\left[{}\begin{matrix}x>36\\x< 1\end{matrix}\right.\\\left[{}\begin{matrix}x< 36\\x>1\end{matrix}\right.\end{matrix}\right.=>1< x< 36\left(tm\right)\)
AH
Akai Haruma
Giáo viên
11 tháng 4 2023
Lời giải:
$A=\frac{10\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+4)}-\frac{(2\sqrt{x}-3)(\sqrt{x}-1)}{(\sqrt{x}+4)(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}-1)(\sqrt{x}+4)}$
$=\frac{10\sqrt{x}-(2\sqrt{x}-3)(\sqrt{x}-1)-(\sqrt{x}+1)(\sqrt{x}+4)}{(\sqrt{x}+4)(\sqrt{x}-1)}$
$=\frac{-3x+10\sqrt{x}-7}{(\sqrt{x}+4)(\sqrt{x}-1)}$
$=\frac{-(\sqrt{x}-1)(3\sqrt{x}-7)}{(\sqrt{x}+4)(\sqrt{x}-1)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}$
9 tháng 7 2022
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
15 tháng 8 2021
1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)
Lời giải:
\(A=\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\frac{\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{4\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}\\ =\frac{\sqrt{x}(\sqrt{x}-3)+\sqrt{x}+3-(4\sqrt{x}-6)}{(\sqrt{x}-3)(\sqrt{x}+3)}\\ =\frac{x-6\sqrt{x}+9}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{(\sqrt{x}-3)^2}{(\sqrt{x}-3)(\sqrt{x}+3)}\\ =\frac{\sqrt{x}-3}{\sqrt{x}+3}\)