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2
22 tháng 10 2021
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
6 tháng 8 2017
a) \(A=x^2-6x+11\)
\(\Rightarrow A=x^2-6x+9+2\)
\(\Rightarrow A=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = 3
Vậy \(MIN\) \(A=2\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(\Rightarrow B=2\left(x^2+5\right)-1\)
\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{25}{2}-1\)
\(\Rightarrow B=2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\)
Ta có: \(2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)\ge0\forall x\)
\(\Rightarrow2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)-\dfrac{23}{2}\ge-\dfrac{23}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{-5}{2}\)
Vậy \(MIN\) \(B=\dfrac{-23}{2}\Leftrightarrow x=\dfrac{-5}{2}\)
c) \(C=5x-x^2\)
\(\Rightarrow C=-\left(x^2-5x\right)\)
\(\Rightarrow C=-\left(x^2-2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(\Rightarrow C=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Ta có: \(-\left(x-\dfrac{5}{2}\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)
Vậy \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
18 tháng 10 2019
2. \(Q=\left(x-3\right)\left(4x+5\right)+2019\)
\(Q=4x^2+5x-12x-15+2019\)
\(Q=4x^2-7x+2004\)
\(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\)
\(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)
\(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)
\(\Rightarrow Q\ge\frac{32255}{16}\)
\(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)
3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\) (do a+b=1)
\(T=4a^2-4ab+4a^2-6a^2-6b^2\)
\(T=-2a^2-4ab-2b^2\)
\(T=-2\left(a^2+2ab+b^2\right)\)
\(T=-2\left(a+b\right)^2\)
\(T=-2.1^2=-2.1=-2\) (do a+b=1)
MT
1
TT
17 tháng 12 2017
\(2x^2+10x-1=\\ 2\left(x^2+5x-\dfrac{1}{2}\right)\\ =2\left(x^2+2\cdot\dfrac{5}{2}\cdot x+\dfrac{25}{4}-\dfrac{27}{4}\right)\\ =2\left(\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{4}\right)\\ =2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\)
vì \(2\left(x+\dfrac{5}{2}\right)^2\ge0\Rightarrow2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge-\dfrac{27}{2}\)vậy Min \(2x^2+10x-1\) \(=-\dfrac{27}{2}\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=0\)
\(\Rightarrow x+\dfrac{5}{2}=0\Rightarrow x=-\dfrac{5}{2}\)
N
8 tháng 12 2018
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). dấu = xảy ra khi x2=0
=> x=0
Vậy \(B_{min}\Leftrightarrow x=0\)
ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)
dấu = xảy ra khi \(x+1=0\)
\(\Rightarrow x=-1\)
Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)