tìm GTNN \(M=\dfrac{x+12}{\sqrt{x}-2}\)
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TD
1
26 tháng 11 2023
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(M=A\cdot B=\dfrac{x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
=>\(M=\dfrac{x}{\sqrt{x}+2}\)
=>\(M=\dfrac{x-4+4}{\sqrt{x}+2}=\sqrt{x}-2+\dfrac{4}{\sqrt{x}+2}\)
=>\(M=\sqrt{x}+2+\dfrac{4}{\sqrt{x}+2}-4\)
=>\(M>=2\cdot\sqrt{\left(\sqrt{x}+2\right)\cdot\dfrac{4}{\sqrt{x}+2}}-4=0\)
Dấu '=' xảy ra khi \(\sqrt{x}+2=\sqrt{4}=2\)
=>\(\sqrt{x}=0\)
=>x=0(nhận)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 3 2022
ĐKXĐ: \(x\ge0;x\ne1\)
\(M=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(M=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\ge1-\dfrac{2}{0+1}=-1\)
\(M_{min}=-1\) khi \(x=0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 3 2022
ĐKXĐ: \(x\ge-2;x\ne-1\)
\(M=\dfrac{x^2-2x}{x^3+1}+\dfrac{1}{2}\left(\dfrac{1-\sqrt{x+2}+1+\sqrt{x+2}}{1-\left(x+2\right)}\right)\)
\(=\dfrac{x^2-2x}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}=\dfrac{x^2-2x-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=-\dfrac{1}{x^2-x+1}\)
\(M=-\dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge-\dfrac{1}{\dfrac{3}{4}}=-\dfrac{4}{3}\)
\(M_{min}=-\dfrac{4}{3}\) khi \(x=\dfrac{1}{2}\)
NT
0
16 tháng 8 2018
\(a.P=\dfrac{x\sqrt{x}-47}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}+12}{\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{x\sqrt{x}-47-4\left(x-9\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}-3x+3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{x+3}{\sqrt{x}+1}\left(x\ne9;x\ge0\right)\)
\(b.P=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right).\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\)
\(\Leftrightarrow\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\ge4-2=2\)
\(\Rightarrow P_{Min}=2."="\Leftrightarrow x=1\left(TM\right)\)
NL
1
26 tháng 12 2021
\(\Leftrightarrow Mx^2=x^2-2x+\sqrt{2015}\\ \Leftrightarrow x^2\left(M-1\right)+2x-\sqrt{2015}=0\)
Ta có \(\Delta'\ge0\Leftrightarrow1+\sqrt{2015}\left(M-1\right)\ge0\)
\(\Leftrightarrow1+\sqrt{2015}M-\sqrt{2015}\ge0\\ \Leftrightarrow M\ge\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\)
Vậy \(M_{min}=\dfrac{\sqrt{2015}-1}{\sqrt{2015}}\Leftrightarrow x=-\dfrac{b'}{a}=-\dfrac{1}{M-1}=\dfrac{-\sqrt{2015}}{\sqrt{2015}-1}\)
26 tháng 7 2023
1:
a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
căn x+1>=1
=>2/căn x+1<=2
=>-2/căn x+1>=-2
=>A>=-2+1=-1
Dấu = xảy ra khi x=0
b: 
5 tháng 7 2022
\(A=\dfrac{x-4+16}{\sqrt{x}+2}=\sqrt{x}+2+\dfrac{16}{\sqrt{x}+2}-4\)
\(\Leftrightarrow A\ge2\cdot\sqrt{16}-4=2\cdot4-4=4\)
Dấu '=' xảy ra khi \(\sqrt{x}+2=4\)
hay x=4
24 tháng 8 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 5 2019
\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{2xy}+\frac{\left(x+y\right)^2}{2xy}\)
\(S\ge\frac{4\left(x+y\right)^2}{x^2+y^2+2xy}+\frac{\left(x+y\right)^2}{\frac{\left(x+y\right)^2}{2}}=\frac{4\left(x+y\right)^2}{\left(x+y\right)^2}+2=6\)
\(\Rightarrow S_{min}=6\) khi \(x=y\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)
\(M=\dfrac{x+12}{\sqrt{x}-2}=\dfrac{x-4+16}{\sqrt{x}-2}\)
\(=\sqrt{x}+2+\dfrac{16}{\sqrt{x}-2}\)
\(=\sqrt{x}-2+\dfrac{16}{\sqrt{x}-2}+4\)
=>\(M>=2\cdot\sqrt{\left(\sqrt{x}-2\right)\cdot\dfrac{16}{\sqrt{x}-2}}+4=2\cdot4+4=12\)
Dấu '=' xảy ra khi \(\left(\sqrt{x}-2\right)^2=16\)
=>\(\left[{}\begin{matrix}\sqrt{x}-2=-4\\\sqrt{x}-2=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(loại\right)\\\sqrt{x}=6\end{matrix}\right.\)
=>x=36(nhận)