Tìm GTNN của
A = \(|x+2|+|3x-5|+|4x+7|\)
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T
1
30 tháng 10 2021
\(a,A=\left|2-4x\right|-6\ge-6\\ A_{min}=-6\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\\ b,x^2+1\ge1\Leftrightarrow B=1-\dfrac{4}{x^2+1}\ge1-\dfrac{4}{1}=-3\\ B_{min}=-3\Leftrightarrow x=0\)
S
1
3 tháng 12 2021
a, \(A=\left|x+2\right|+3\ge3\)
dấu "=" xảy ra\(\Leftrightarrow x=-2\)
Vậy \(A_{min}=3\Leftrightarrow x=-2\)
b,\(B=5+\left|2x-7\right|\ge5\)
dấu "=" xảy ra\(\Leftrightarrow x=\dfrac{7}{2}\)
Vậy \(B_{min}=5\Leftrightarrow x=\dfrac{7}{2}\)
c, \(-\left|4x+5\right|+1\le1\)
dấu "=" xảy ra\(\Leftrightarrow x=-\dfrac{5}{4}\)
Vậy \(C_{max}=1\Leftrightarrow x=-\dfrac{5}{4}\)
d, \(D=3-\left|x+3\right|\le3\)
dấu "=" xảy ra\(\Leftrightarrow x=-3\)
Vậy \(D_{max}=3\Leftrightarrow x=-3\)
NT
2
2 tháng 9 2021
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) \(B=3x^2+4x-13=3\left(x^2+\dfrac{4}{3}x+\dfrac{4}{9}\right)-\dfrac{35}{3}=3\left(x+\dfrac{2}{3}\right)^2-\dfrac{35}{3}\ge-\dfrac{35}{3}\)
\(minB=-\dfrac{35}{3}\Leftrightarrow x=-\dfrac{2}{3}\)
2 tháng 9 2021
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 \(x=-\dfrac{2}{3}\)
TP
1
24 tháng 10 2021
\(B=2x\left(x-4\right)-10=2x^2-8x-10\)
\(=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)
\(minB=-18\Leftrightarrow x=2\)
C
3
25 tháng 6 2021
`A=x^2-2x+5`
`=x^2-2x+1+4`
`=(x-1)^2+4>=4`
Dấu "=" `<=>x=1`
`B=4x^2+4x+3`
`=4x^2+4x+1+2`
`=(2x+1)^2+2>=2`
Dấu "=" xảy ra khi `x=-1/2`
`C=9x^2-6x+7`
`=9x^2-6x+1+6`
`=(3x-1)^2+6>=6`
Dấu '=' xảy ra khi `x=1/3`
`D=5x^2+3x+8`
`=5(x^2+3/5x)+8`
`=5(x^2+3/5x+9/100-9/100)+8`
`=5(x+3/10)^2+151/20>=151/20`
Dấu "=" xảy ra khi `x=-3/10`
AT
25 tháng 6 2021
\(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow A_{min}=4\) khi \(x=1\)
\(B=4x^2+4x+3=4x^2+4x+1+2=\left(2x+1\right)^2+2\)
Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+2\ge2\Rightarrow B_{min}=2\) khi \(x=-\dfrac{1}{2}\)
\(C=9x^2-6x+7=9x^2-6x+1+6=\left(3x-1\right)^2+6\)
Ta có: \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+6\ge6\Rightarrow C_{min}=6\) khi \(x=\dfrac{1}{3}\)
\(D=5x^2+3x+8\Rightarrow5\left(x^2+2.x.\dfrac{3}{10}+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\)
Ta có: \(5\left(x+\dfrac{3}{10}\right)^2\ge0\Rightarrow5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)
\(\Rightarrow D_{min}=\dfrac{151}{20}\) khi \(x=-\dfrac{3}{10}\)
N
1
20 tháng 10 2021
Bài 2:
a: Ta có: \(x^2+4x+7\)
\(=x^2+4x+4+3\)
\(=\left(x+2\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi x=-2
MH
1 tháng 9 2021
a)A=4(x+11/8)^2 -153/16
Min A=-153/16 khi x=-11/8
b)B=3(x-1/3)^2 -4/3
Min B=-4/3 khi x=1/3
1 tháng 9 2021
Bài 1:
a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)
\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)
b) \(B=3x^2-2x-1=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)
Bài 2:
a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)
b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)
\(maxB=11\Leftrightarrow x=-2\)
Ta có:
\(\hept{\begin{cases}\left|x+2\right|\ge0\forall x\\\left|3x-5\right|\ge0\forall x\\\left|4x+7\right|\ge0\forall x\end{cases}}\\ \Rightarrow\left|x+2\right|+\left|3x-5\right|+\left|4x+7\right|\ge0\\ A\ge0\\Min_A=0 \)