Tìm GTLN của h(x)=3x2-4x+3
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a) \(N=-1-x-x^2=-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{3}{4}=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\)
\(maxN=-\dfrac{3}{4}\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)
\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)
a: Ta có: \(N=-x^2-x-1\)
\(=-\left(x^2+x+1\right)\)
\(=-\left(x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
b: ta có: \(B=3x^2+4x-13\)
\(=3\left(x^2+\dfrac{4}{3}x-\dfrac{13}{3}\right)\)
\(=3\left(x^2+2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{43}{9}\right)\)
\(=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{43}{3}\ge-\dfrac{43}{3}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{2}{3}\)
\(A=\dfrac{3x^2+12x+17}{x^2+4x+5}=\dfrac{3\left(x^2+4x+5\right)+2}{x^2+4x+5}=3+\dfrac{2}{x^2+4x+5}\)
Ta có: \(x^2+4x+5=x^2+4x+4+1=\left(x+2\right)^2+1\ge1\)
\(\Rightarrow\dfrac{2}{x^2+4x+5}\le2\Rightarrow A\le3+2=5\)
\(\Rightarrow A_{max}=5\) khi \(x=-2\)
A=3(x^2+2/3x-1)
=3(x^2+2*x*1/3+1/9-10/9)
=3(x+1/3)^2-10/3>=-10/3
Dấu = xảy ra khi x=-1/3
\(B=1+\dfrac{15}{x^2+x+5}=1+\dfrac{15}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}}< =1+15:\dfrac{19}{4}=1+\dfrac{60}{19}=\dfrac{79}{19}\)
Dấu = xảy ra khi x=-1/2
Ta có: \(-x^2+4x+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\forall x\in R\)
⇒ max y = 7 tại x = 2
Sửa đề: Tìm GTNN....
\(h\left(x\right)=3x^2-4x+3=3\left(x^2-\dfrac{4}{3}x+1\right)\)
\(=3\left[\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}\right)+\dfrac{5}{9}\right]\)
\(=3\left[\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{9}\right]\)
Vì \(\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{9}\ge\dfrac{5}{9}\)
=> \(3\left[\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{9}\right]\ge3\cdot\dfrac{5}{9}=\dfrac{15}{9}\)
Dấu ''='' xảy ra khi \(x=\dfrac{2}{3}\)
Vậy GTNN của h(x) là \(\dfrac{15}{9}\Leftrightarrow x=\dfrac{2}{3}\)