h) Ta có: \(\left\{{}\begin{matrix}\left|x-7\right|=\left|7-x\right|\ge7-x\\\left|x+5\right|\ge x+5\end{matrix}\right.\)
\(\Rightarrow\left|7-x\right|+\left|x+5\right|\ge\left(7-x\right)+\left(x+5\right)\)
\(\Rightarrow\left|x-7\right|+\left|x+5\right|\ge12\)
\(\Rightarrow H\ge12\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}7-x\ge0\\x+5\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le7\\x\ge-5\end{matrix}\right.\)
\(\Leftrightarrow-5\le x\le7\)
Vậy, MinH = 12 \(\Leftrightarrow-5\le x\le7\)
a) Ta có: \(A=2x^2-8x+10\)
\(=2\left(x^2-4x+5\right)\)
\(=2\left(x^2-4x+2^2+1\right)\)
\(2\left[\left(x-2\right)^2+1\right]\)
Ta lại có: \(\left(x-2\right)^2\ge0\)
\(\Rightarrow2\left[\left(x-2\right)^2+1\right]\ge2\)
\(\Rightarrow A\ge2\)
Dấu bằng xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy MinA = 2 \(\Leftrightarrow x=2\)
\(a.2x^2-8x+10=2\left(x^2-4x+4\right)+2=2\left(x-2\right)^2+2\ge2\)
\(\Rightarrow A_{Min}=2."="\Leftrightarrow x=2\)
\(b.B=x\left(x+1\right)\left(x^2+x-4\right)=\left(x^2+x\right)\left(x^2+x-4\right)\)
Đặt : \(x^2+x=t\) , ta có :
\(B=t\left(t-4\right)=t^2-4t+4-4=\left(t-2\right)^2-4\ge-4\)
\(\Rightarrow B_{MIN}=-4."="\Leftrightarrow t=2\Leftrightarrow x^2+x=2\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(d.D=9x^2-6x+5=9x^2-6x+1+4=\left(3x-1\right)^2+4\ge4\)
\(\Rightarrow D_{Min}=4."="\Leftrightarrow x=\dfrac{1}{3}\)
\(e.E=x^2+3x-1=x^2+2.\dfrac{3}{2}x+\dfrac{9}{4}-1-\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{13}{4}\)
\(\Rightarrow E_{Min}=-\dfrac{13}{4}."="\Leftrightarrow x=-\dfrac{3}{2}\)
\(f.F=\left(x^2+5x+4\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
Đặt : \(x^2+5x+5=t\) , ta có :
\(F=\left(t-1\right)\left(t+1\right)=t^2-1\ge-1\)
\(\Rightarrow F_{Min}=-1."="\Leftrightarrow t=0\Leftrightarrow x^2+5x+5=0\Leftrightarrow x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+5-\dfrac{25}{4}=0\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=\dfrac{5}{4}\left(Tự-giải-ra-nhé\right)\)
\(g.G=\left(x-3\right)\left(x-4\right)\left(x^2-7x+8\right)=\left(x^2-7x+12\right)\left(x^2-7x+8\right)\)
Tự đặt và làm như phần f nhé .
\(h.H=|x-7|+|x+5|=|7-x|+|x+5|\ge|7-x+x-5|=2\)
\(\Rightarrow H_{Min}=2."="\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\7-x\ge0\end{matrix}\right.\)\(\Leftrightarrow-5\le x\le7\)
\(i.\) \(I=\left(2x-1\right)^2-3\left|2x-1\right|+2=\left|2x-1\right|^2-2.\dfrac{3}{2}\left|2x-1\right|+\dfrac{9}{4}+2-\dfrac{9}{4}=\left(\left|2x-1\right|-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow I_{Min}=-\dfrac{1}{4}."="\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(k.\) Tương tự phần h nhé bạn .
E = x2 + 3x - 1
= x2 + 2 . \(\dfrac{3}{2}\)x + \(\dfrac{9}{4}\) - 1 - \(\dfrac{9}{4}\)
= \(\left(x+\dfrac{3}{2}\right)^2\) - \(\dfrac{13}{4}\)
Vì \(\left(x+\dfrac{3}{2}\right)^2\) ≥ 0 => \(\left(x+\dfrac{3}{2}\right)^2\) - \(\dfrac{13}{4}\) ≥ 0 - \(\dfrac{13}{4}\)
<=> E ≥ \(\dfrac{-13}{4}\)
Dấu = xảy ra khi: \(x+\dfrac{3}{2}\) = 0 => x = \(\dfrac{-3}{2}\)
Vậy GTNN của E = \(\dfrac{-13}{4}\) khi x = \(\dfrac{-3}{2}\)
Tìm GTNN của biểu thức:
a) A = 2x2 - 8x + 10
= 2x2 - 2.x.4 + 42 - 6
= (2x - 4)2 - 6
Vì (2x - 4)2 ≥ 0 => (2x - 4)2 - 6 ≥ 0 - 6 (với mọi x)
<=> A ≥ -6
Dấu = xảy ra khi: 2x - 4 = 0 => 2x = 4 => x = 2
Vậy GTNN của A = -6 khi x = 2
D = 9x2 - 6x + 5
= (3x)2 - 2.3x.1 + 12 + 4
= (3x - 1)2 + 4
Vì (3x - 1)2 ≥ 0 => (3x - 1)2 + 4 ≥ 0 + 4 (với mọi x)
<=> D ≥ 4
Dấu = xảy ra khi: 3x - 1 = 0 => 3x = 1 => x = \(\dfrac{1}{3}\)
Vậy GTNN của D = 4 khi x = \(\dfrac{1}{3}\)