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\(M=4x-x^2+3\\ =-(x^2-4x-3)\\ =-(x^2-4x+4)+7\\ =-(x+2)^2+7 \leq7,\forall x\in \mathbb{R}\quad (\mathrm{vì}-(x+2)^2\leq0)\)
Dấu bằng xảy ra khi và chỉ khi \(-(x+2)^2=0\Leftrightarrow x+2=0 \Leftrightarrow x=-2\).
Vậy \(\mathrm{Max}M=7\Leftrightarrow x=-2\).
\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
\(a,A=x^2-6x+11=\left(x-3\right)^2+2\)\(\Leftrightarrow Amin=2\)
Dấu = xảy ra \(\Leftrightarrow x=3\)
\(2x^2+10x-1=2\left(x^2+5x-\frac{1}{2}\right)=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{27}{4}\right)=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\)
\(\Rightarrow Bmin=\frac{-27}{2}.''=''\Leftrightarrow x=\frac{-5}{2}\)
\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)
a: Ta có: \(A=-x^2+4x+3\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu '=' xảy ra khi x=2
b: Ta có: \(B=-x^2+x\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)
\(\Leftrightarrow V\ge-1\forall x\)
Dấu '=' xảy ra khi x=1
a)
\(A=x^2-4x+1=x^2-2.2x+2^2-3\)
\(=(x-2)^2-3\)
Vì \((x-2)^2\geq 0, \forall x\Rightarrow A\geq 0-3=-3\)
Vậy GTNN của $A$ là $-3$ khi $x=2$
b) \(B=(x-2)(x-6)+7=x^2-6x-2x+12+7\)
\(=x^2-8x+19=(x^2-2.4x+4^2)+3\)
\(=(x-4)^2+3\)
Vì \((x-4)^2\geq 0, \forall x\Rightarrow B\geq 0+3=3\)
Vậy GTNN của $B$ là $3$ khi $x=4$
c)
\(C=4x-x^2=4-(x^2-4x+4)=4-(x-2)^2\)
Vì \((x-2)^2\geq 0\Rightarrow C\leq 4-0=4\)
Vậy GTLN của $C$ là $4$ khi $x=2$
d) \(D=x^2-2x+y^2-4y+16=(x^2-2x+1)+(y^2-4y+4)+11\)
\(=(x-1)^2+(y-2)^2+11\)
Vì \((x-1)^2\geq 0; (y-2)^2\geq 0, \forall x,y\)
\(\Rightarrow D\geq 0+0+11=11\)
Vậy GTNN của $D$ là $11$ khi \(\left\{\begin{matrix} x-1=0\\ y-2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)
a: Ta có: \(x^2+x+1\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)
b: Ta có: \(-x^2+x+2\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
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 \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
a) A=x2−6x+11A=x2−6x+11
⇒A=x2−6x+9+2⇒A=x2−6x+9+2
⇒A=(x−3)2+2⇒A=(x−3)2+2
Ta có: (x−3)2≥0∀x(x−3)2≥0∀x
⇒(x−3)2+2≥2∀x⇒(x−3)2+2≥2∀x
Dấu "=" xảy ra ⇔⇔ x = 3
Vậy MINMIN A=2⇔x=3A=2⇔x=3
b) B=2x2+10x−1B=2x2+10x−1
⇒B=2(x2+5)−1⇒B=2(x2+5)−1
⇒B=2(x2+2⋅52⋅x+254)−252−1⇒B=2(x2+2⋅52⋅x+254)−252−1
⇒B=2(x2+2⋅52⋅x+254)−232⇒B=2(x2+2⋅52⋅x+254)−232
Ta có: 2(x2+2⋅52⋅x+254)≥0∀x2(x2+2⋅52⋅x+254)≥0∀x
⇒2(x2+2⋅52⋅x+254)−232≥−232∀x⇒2(x2+2⋅52⋅x+254)−232≥−232∀x
Dấu "=" xảy ra ⇔⇔ x = −52−52
Vậy MINMIN B=−232⇔x=−52B=−232⇔x=−52
c) C=5x−x2C=5x−x2
⇒C=−(x2−5x)⇒C=−(x2−5x)
⇒C=−(x2−2⋅52⋅x+254)+254⇒C=−(x2−2⋅52⋅x+254)+254
⇒C=−(x−52)2+254⇒C=−(x−52)2+254
Ta có: −(x−52)2≤0∀x−(x−52)2≤0∀x
⇒−(x−52)2+254≤254∀x⇒−(x−52)2+254≤254∀x
Dấu "=" xảy ra ⇔⇔ x = 5252
Vậy MAXMAX C=254⇔x=52
Giá trị nhỏ nhất của hệ thức
\(A=x^2\)\(-6x+11\)
\(A=\left(x^2+6x+9\right)+2\)
\(A=\left(x-3\right)^2\)\(+2\)lớn hơn hoặc bằng \(2\)
\(A=2=>x=3\)
Giá trị nhỏ nhất
\(B=2x^2\)\(+10x-1\)
\(B=2\left(x^2+5x-\frac{1}{2}\right)\)
\(B=2\left(x+\frac{5}{2}\right)^2\)\(-\frac{27}{4}\))
\(B=2\left(x+\frac{5}{2}\right)^2\)\(-\frac{27}{2}\)
\(B\)≥ \(-\frac{27}{2}\)
\(=>2x^2\)\(+10x-1=-\frac{27}{2}\)\(=>\left(x+\frac{5}{2}\right)^2\)\(=0\)
\(x+\frac{5}{2}\)\(=0=>x=-\frac{5}{2}\)
Giá trị lớn nhất
\(C=5x-x^2\)
\(C=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}\)
\(C=\left(x-\frac{5}{2}\right)^2\)\(+\frac{25}{4}\)bé hơn hoặc bằng \(\frac{25}{4}\)
\(C=\frac{25}{4}\)\(=>x-\frac{5}{2}\)\(=0=>x=\frac{5}{2}\)
Giá trị lớn nhất
\(M=4x-x^2\)\(+3\)
\(M=-x^2\)\(+4x+3=-\left(x^2-4x-3\right)\)
\(M=\left(x-2\right)^2\)\(-7=-\left(x-2\right)^2\)\(+7\)
\(-\left(x-2\right)^2\)≤ \(0\)\(=>-\left(x-2\right)^2\)\(+7\)≤ \(7\)
Dấu " = " khi \(\left(x-2\right)^2\)\(=0\)
\(=>x-2=0\)
\(x=0+2=2\)
\(=>M=7=>x=2\)
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