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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 \(MAX\) \(C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
a) \(A=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\)
\(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+2\ge2\)
Đẳng thức xảy ra <=> x - 3 = 0 => x = 3
Vậy AMin = 2 , đạt được khi x = 3
b) \(B=5x-x^2=-x^2+5x=-x^2+5x-\frac{25}{4}+\frac{25}{4}=-\left(x^2-5x+\frac{25}{4}\right)+\frac{25}{4}=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
\(-\left(x-\frac{5}{2}\right)^2\le0\forall x\Rightarrow-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Đẳng thức xảy ra <=> x - 5/2 = 0 => x = 5/2
Vậy BMax = 25/4 , đạt được khi x = 5/2
c) \(2x-2x^2-5=-2x^2+2x-5=-2\left(x^2-x+\frac{1}{4}\right)-\frac{9}{2}=-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\)
\(-2\left(x-\frac{1}{2}\right)^2\le0\forall x\Rightarrow-2\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)
Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2
Vậy CMax = -9/2 , đạt được khi x = 1/2
a)\(A=4x^2+4x+11\)
\(=4x^2+4x+1+10\)
\(=\left(2x+1\right)^2+10\ge10\)
Dấu = khi \(x=\frac{-1}{2}\)
Vậy MinA=10 khi \(x=\frac{-1}{2}\)
b)\(B=3x^2-6x+1\)
\(=3x^2-6x+3-2\)
\(=3\left(x^2-2x+1\right)-2\)
\(=3\left(x-1\right)^2-2\ge-2\)
Dấu = khi \(x=1\)
Vậy MinB=-2 khi \(x=1\)
c)\(C=x^2-2x+y^2-4y+6\)
\(=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy MinC=1 khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Ta có : A = x2 - 6x + 15
= x2 - 6x + 9 + 6
= (x - 3)2 + 6 \(\ge6\forall x\in R\)
Vậy Amin = 6 khi x = 3.
a, Ta có: \(B=2x^2+10x-1=2x^2+10x+\dfrac{25}{2}-\dfrac{27}{2}\)
\(=2\left(x^2+2.x.\dfrac{5}{2}+\dfrac{25}{4}\right)-\dfrac{27}{2}\)
\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge\dfrac{-27}{2}\)
Dấu " = " khi \(2\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{-5}{2}\)
Vậy \(MIN_B=\dfrac{-27}{2}\) khi \(x=\dfrac{-5}{2}\)
b, Ta có: \(C=5x-x^2=-\left(x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{25}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
Dấu " = " khi \(-\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(MAX_C=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)
\(A=x^2+5x+7\)
\(A=\left(x^2+5x+\frac{25}{4}\right)+\frac{3}{4}\)
\(A=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x+\frac{5}{2}\right)^2=0\)
\(\Leftrightarrow\)\(x+\frac{5}{2}=0\)
\(\Leftrightarrow\)\(x=\frac{-5}{2}\)
Vậy GTNN của \(A\) là \(\frac{3}{4}\) khi \(x=\frac{-5}{2}\)
Chúc bạn học tốt ~
\(B=6x-x^2-5\)
\(-B=x^2-6x+5\)
\(-B=\left(x^2-6x+9\right)-4\)
\(-B=\left(x-3\right)^2-4\ge-4\)
\(B=-\left(x-3\right)^2+4\le4\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(x-3\right)^2=0\)
\(\Leftrightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Vậy GTLN của \(B\) là \(4\) khi \(x=3\)
Chúc bạn học tốt ~
\(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra <=> x = 3
Vậy MinA = 1
\(B=5x^2-10x+3=5\left(x^2-2x+1\right)-2=5\left(x-1\right)^2-2\ge-2\forall x\)
Dấu "=" xảy ra <=> x = 1
Vậy MinB = -2
\(C=2x^2+8x+y^2-10y+43=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)
Dấu "=" xảy ra <=> x = -2 ; y = 5
Vậy MinC = 10
\(A=x^2-6x+10\)
\(=\left(x^2-6x+9\right)+1\)
\(=\left(x-3\right)^2+1\ge1\forall x\)
Dấu"=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)
Vậy \(Min_A=1\Leftrightarrow x=3\)
b,\(B=5x^2-10x+3\)
\(=5\left(x^2-2x+1\right)-2\)
\(=5\left(x-1\right)^2-2\ge-2\forall x\)
Dấu"=" xảy ra khi \(x-1=0\Leftrightarrow x=1\)
Vậy \(Min_B=-2\Leftrightarrow x=1\)
c,\(C=2x^3+8x+y^2-10+43\)
\(=2x^2+8x+8+y^2-10y+25+10\)
\(=2\left(x^2+4x+4\right)+\left(y^2-10y+25\right)+10\)
\(=2\left(x+2\right)^2+\left(y-5\right)^2+10\ge10\forall x,y\)
Dấu"=" xảy ra khi \(\orbr{\begin{cases}x+2=0\\y-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\y=5\end{cases}}}\)
Vậy \(Min_C=10\Leftrightarrow x=-2;y=5\)
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 \(\left(x-2\right)^2\)\(=0\)
\(=>x-2=0\)
\(x=0+2=2\)
\(=>M=7=>x=2\)
Em đóng góp ít ý kiến thế này thôi ạ mong anh thông cảm