tìm gtnn B = x^2 – 3x + 10
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TM
2
27 tháng 8 2016
A=3x2-x+4
\(=3\left(x^2-\frac{x}{3}+\frac{4}{3}\right)\)
\(=3\left(x-\frac{1}{6}\right)^2+\frac{47}{12}\ge0+\frac{47}{12}=\frac{47}{12}\)
Dấu = khi \(x=\frac{1}{6}\)
Vậy MinA=\(\frac{47}{12}\Leftrightarrow x=\frac{1}{6}\)
27 tháng 8 2016
B=(x-2)(x-5)(x2-7x-10)
=(x2-7x+10)(x2-7x-10)
Đặt t=x2-7x+10 đc:
B=t(t-20)=t2-20t
=t2-20t+100-100
=(t-10)2-100
Thay t=x2-7x+10 ta đc:
\(B=\left(x^2-7x+10-10\right)-100\ge0-100=-100\)
\(\Rightarrow B\ge-100\)
Dấu = khi \(\left[\begin{array}{nghiempt}x=0\\x=7\end{array}\right.\)
Vậy MinB=-100 khi \(\left[\begin{array}{nghiempt}x=0\\x=7\end{array}\right.\)
10 tháng 9 2020
\(A=x^2-6x+10=\left(x-3\right)^2+1\ge1\)
\(\Rightarrow A_{min}=1\Leftrightarrow x=3\)
\(B=4x^2-4x+25=\left(2x-1\right)^2+24\ge24\)
\(\Rightarrow B_{min}=24\Leftrightarrow x=\frac{1}{2}\)
\(C=3x^2+9x+12=3\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\ge\frac{21}{4}\)
\(\Rightarrow C_{min}=\frac{21}{4}\Leftrightarrow x=\frac{-3}{2}\)
TP
1
24 tháng 10 2021
\(B=2x\left(x-4\right)-10=2x^2-8x-10\)
\(=2\left(x^2-4x+4\right)-18=2\left(x-2\right)^2-18\ge-18\)
\(minB=-18\Leftrightarrow x=2\)
18 tháng 11 2018
\(A=x^2-6x+10\)
\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)
\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\) \(\forall x\in z\)
\(\Leftrightarrow A_{min}=1khix=3\)
\(B=3x^2-12x+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)
\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\) \(\forall x\in z\)
\(\Leftrightarrow B_{min}=-11khix=2\)
TN
1
7 tháng 9 2015
A = x2+ 3x+ 7
=x2 + 2*x*3/2+9/4 + 19/4
=(x+3/2)2 +19/4
ta có (x+3/2)2>0 nên (x+3/2)2+ 19/4>hoặc=19/4
=> AMin khi x+3/2=0
=>x=-3/2
5 tháng 10 2021
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
5 tháng 10 2021
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
13 tháng 9 2021
\(6,\\ a,\\ 1,A=x^2+3x+7=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
\(2,B=\left(x-2\right)\left(x-5\right)\left(x^2-7x+10\right)=\left(x-2\right)^2\left(x-5\right)^2\ge0\)
Dấu \("="\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(b,\\ 1,A=11-10x-x^2=-\left(x+5\right)^2+36\le36\)
Dấu \("="\Leftrightarrow x=-5\)
6 tháng 12 2018
\(A=2x^2+3x-10\)
\(A=2\left(x^2+\frac{3}{2}x-5\right)\)
\(A=2\left[x^2+2\cdot x\cdot\frac{3}{4}+\left(\frac{3}{4}\right)^2-\frac{89}{16}\right]\)
\(A=2\left[\left(x+\frac{3}{4}\right)^2-\frac{89}{16}\right]\)
\(A=2\left(x+\frac{3}{4}\right)^2-\frac{89}{8}\ge\frac{-89}{8}\forall x\)vì \(2\left(x+\frac{3}{4}\right)^2\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{3}{4}=0\Leftrightarrow x=\frac{-3}{4}\)
T
6 tháng 12 2018
Hình như lớp 8 chưa học BĐT cô si nhỉ?
ĐK: \(x\ne0;\).Không mất tính tổng quát,giả sử \(x\ge1\).Đặt \(x=\frac{1+m}{1}\left(m\ge0\right)\)
Ta có:
\(B=\frac{1+m}{1}+\frac{1}{1+m}\ge\frac{1+m}{1+m}+\frac{1}{1+m}=\frac{2+m}{1+m}=\frac{2+m}{1}:\frac{1+m}{1}\ge2:1=2\) (Do \(m\ge0\))
1) A = x^2 + 3x + 10
=x^2 + 2x.3/2 + 9/4 + 31/4
=(x+3/2)^2 + 31/4
mà (x+ 3/2 )^2 >=0 với mọi x
=> Min (x+3/2)^2 = 0
=> Min (x+3/2)^2 + 31/4 = 31/4
Vậy GTNN của A là 31/4 khi:
x+ 3/2 = 0 <=> x= - 3/2
\(B=x^2-3x+10\)
\(=x^2-3x+\frac{9}{4}+\frac{31}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)