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\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)
Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2
\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)
\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)
\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6
\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)
\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2
\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)
\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4
Bài 2 :
a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)
Dấu ''='' xảy ra khi x = 2
b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)
Dấu ''='' xảy ra khi x = -1
Bài 1 :
a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)
c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
Bài 1a)
\(P\left(x\right)=x^{2018}+4x^2+10\)
VÌ \(x^{2018}\ge0\forall x;4x^2\ge0\forall x\)
\(\Rightarrow x^{2018}+4x^2+10\ge10\forall x\)
Hay \(P\left(x\right)\ge10\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Bài 1b)
\(M\left(x\right)=x^2+x+1\)
\(M\left(x\right)=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(M\left(x\right)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-1}{2}\)
\(E=5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x+105\)
\(=\left(5x^7+10x^6-20x^5-35x^4+20x^3-5x^2+40x\right)+105\)
\(=5x\left(x^6+2x^5-4x^4-7x^3+4x^2-x+8\right)+105\)
Thay \(x^6+2x^5-4x^4-7x^3+4x^2-x+8=0\)vào đa thức ta được:
\(E=5x.0+105=105\)
