\(GTNN\)
\(F=\frac{x+5}{\sqrt{x}+2}\)
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NM
Nguyễn Minh Quang
Giáo viên
27 tháng 8 2021
\(F=\frac{x-1+16}{\sqrt{x}+1}=\sqrt{x}-1+\frac{16}{\sqrt{x}+1}=\sqrt{x}+1+\frac{16}{\sqrt{x}+1}-2\)
\(\ge2\sqrt{\left(\left(\sqrt{x}+1\right).\frac{16}{\sqrt{x}+1}\right)}-2=8-2=6\) vậy GTNN của F=6 khi \(\sqrt{x}+1=\frac{16}{\sqrt{x}+1}\Leftrightarrow x=9\)
\(G=\frac{x-9+4}{\sqrt{x}+3}=\sqrt{x}-3+\frac{4}{\sqrt{x}+3}=\sqrt{x}+3+\frac{4}{\sqrt{x}+3}-6=\frac{5}{9}\left(\sqrt{x}+3\right)+\frac{4}{9}\left(\sqrt{x}+3\right)+\frac{4}{\sqrt{x}+3}-6\)
\(\ge\frac{5}{9}\left(\sqrt{x}+3\right)+2\sqrt{\left(\frac{4}{9}\left(\sqrt{x}+3\right).\frac{4}{\sqrt{x}+3}\right)}-6\ge\frac{5}{3}+\frac{8}{3}-6=-\frac{5}{3}\) vậy GTNN G =- 5/3 khi x=0
27 tháng 8 2021
\(F=\frac{x-1+16}{\sqrt{x}+1}=\frac{x-1}{\sqrt{x}+1}+\frac{16}{\sqrt{x}+1}=\sqrt{x}-1+\frac{16}{\sqrt{x}+1}\)
\(=\left[\left(\sqrt{x}+1\right)+\frac{16}{\sqrt{x}+1}\right]-2\ge2\sqrt{\left(\sqrt{x}+1\right)\cdot\frac{16}{\sqrt{x}+1}}-2=6\)
Dấu "=" xảy ra <=> x = 9
11 tháng 2 2020
A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
11 tháng 2 2020
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
RD
5 tháng 5 2019
Ta có:\(F=\frac{x^2}{\sqrt{y}}+\frac{y^2}{\sqrt{x}}\)
\(F=\frac{x^2}{\sqrt{y}}+x^2\sqrt{y}+\frac{y^2}{\sqrt{x}}+y^2\sqrt{x}-x^2\sqrt{y}-y^2\sqrt{x}\)
\(F\ge2x^2+2y^2-x^2\sqrt{y}-y^2\sqrt{x}=4-x^2\sqrt{y}-y^2\sqrt{x}\)
Đặt \(A=x^2\sqrt{y}+y^2\sqrt{x}\)
\(A\le\frac{x^2\left(y+1\right)+y^2\left(x+1\right)}{2}=\frac{x^2y+y^2x+2}{2}\)
Ta có:\(x^2y+y^2x=xy\left(x+y\right)\le\frac{x^2+y^2}{2}.\sqrt{2\left(x^2+y^2\right)}=1.2=2\)
\(\Rightarrow A\le\frac{2+2}{2}=2\)
\(\Rightarrow F\ge4-2=2\)
"="<=>x=y=1
AN
12 tháng 12 2019
a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0
\(F=\frac{x+5}{\sqrt{x}+2}=\frac{\left(\sqrt{x}+2\right)\sqrt{x}-\left(\sqrt{x}+2\right)2+9}{\sqrt{x}+2}=\sqrt{x}-2+\frac{9}{\sqrt{x}+2}=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\)
theo bdt co si ta co:\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge2\sqrt{\sqrt{x}+2.\frac{9}{\sqrt{x}+2}}=6\)
\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\ge6-4=2\)
=>Fmin=2 xay ra dau = khi\(\sqrt{x}+2=\frac{9}{\sqrt{x}+2}\)