Question

[gaarakazekage]

Tìm GTNN
A= x² + 3x - 7
B= x -5\(\sqrt{x}\) -1
C=\(\frac{-4}{\sqrt{x}+7}\)
D= \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
E= \(\frac{x+7}{\sqrt{x}+3}\)
F= \(\frac{x^2+3x+5}{x^2}\)
G= \(\frac{4x+1}{x^2+3}\)
H= \(\sqrt{x^2+2x+5}\)
Tìm GTLN
A = -x² + 4x+3
B = -x² + x + 1
C = 5 - 3x +\(\sqrt{x}\)
D = \(\frac{7}{\sqrt{x}+3}\)
E = \(\frac{\sqrt{x}+6}{\sqrt{x}+1}\)
F = \(\frac{11}{x+3\sqrt{x}+7}\)

Answer 1

[Phạm Minh Quang]

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 = \(\frac{1}{3}\Leftrightarrow x=0\)

Answer 2

[Phạm Minh Quang]

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)