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\(...=A=x^3-3x^2+3x-1+1013\)
\(A=\left(x-1\right)^3+1013=\left(11-1\right)^3+1013=1000+1013=2013\)
\(...B=x^3-6x^2+12x-8-100\)
\(B=\left(x-2\right)^3-100=\left(12-2\right)^3-100=1000-100=900\)
\(...C=\left(x-2y\right)^3=\left(-2y-2y\right)^3=\left(-4y\right)^3=-64y^3\)
\(...D=x^3+9x^2+27x+9+2018\)
\(D=\left(x+3\right)^3+2018=\left(-23+3\right)^3+2018=-8000+2018=-5982\)
a) \(A=x^3-3x^2+3x+1012\)
\(A=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1+1013\)
\(A=\left(x-1\right)^3+1013\)
Thay x=11 vào A ta có:
\(A=\left(11-1\right)^3+1013=10^3+1013=1000+1013=2013\)
b) \(B=x^3-6x^2+12x-108\)
\(B=x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-8-100\)
\(B=\left(x-2\right)^3-100\)
Thay x=12 vào B ta có:
\(B=\left(12-2\right)^3-100=10^3-100=1000-100=900\)
c) \(C=x^3+6x^2y+12xy^2+8y^3\)
\(C=x^3+3\cdot2y\cdot x^2+3\cdot\left(2y\right)^2\cdot x+\left(2y\right)^3\)
\(C=\left(x+2y\right)^3\)
Thay x=-2y vào C ta được:
\(C=\left(-2y+2y\right)^3=0^3=0\)
d) \(D=x^3+9x^2+27x+2027\)
\(D=x^3+3\cdot3\cdot x^2+3\cdot3^2\cdot x+27+2000\)
\(D=\left(x+3\right)^3+2000\)
Thay x=-23 vào D ta có:
\(D=\left(-23+3\right)^3+2000=\left(-20\right)^3+2000=-8000+2000=-6000\)
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 \(x=\dfrac{1}{2}\)
Bài 1: \(A=2x^2-8x=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4\right)-8=2\left(x-2\right)^2-8\ge-8\)
Vậy MinA= -8 \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
\(B=3x^2-3x=3\left(x^2-x\right)=3\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{3}{4}\)
\(=3\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\ge-\dfrac{3}{4}\)
Vậy \(Min_B=-\dfrac{3}{4}\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
\(C=x^2+y^2-2x+4y+7=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+2\ge2\)
Vậy \(Min_C=2\Leftrightarrow x=1;y=-2\)
\(D=x^2+4y^2+x+4y+2=\left(x^2+x+\dfrac{1}{4}\right)+\left(4y^2+4y+1\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\left(2y+1\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy \(Min_D=\dfrac{3}{4}\Leftrightarrow x=y=-\dfrac{1}{2}\)
Bài 2: \(A=x-x^2=-\left(x^2-x\right)=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Vậy \(Max_A=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(B=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)\)
\(=-2\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)
\(=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)
Vậy \(Max_B=\dfrac{9}{8}\Leftrightarrow x=\dfrac{3}{4}\)
\(C=2x-2x^2-3=-2\left(x^2-x+\dfrac{3}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{5}{4}\right)=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{2}\le-\dfrac{5}{2}\)
Vậy \(Max_C=-\dfrac{5}{2}\Leftrightarrow x=\dfrac{1}{2}\)
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
\(a,x^2+x+1=\left(x^2+2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì: \(\left(x+\frac{1}{2}\right)^2\ge0,\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4},\forall x\)
Dấu '' = '' xảy ra khi : \(x+\frac{1}{2}=0\Rightarrow x=\frac{-1}{2}\)
Vậy GTLN của biểu thức = 3/4 khi x=-1/2
\(b,2+x-x^2=-x^2+x+2\)
\(=-\left(x^2-x-2\right)=-\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{9}{4}\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\)
Vì: \(-\left(x-\frac{1}{2}\right)^2\le0,\forall x\)
\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4},\forall x\)
Dấu '' = '' xảy ra khi: x-1/2=0 => x=1/2
Vậy GTNN của biểu thức = 9/4 khi x=1/2
\(c,x^2-4x+1=\left(x^2-2.x.2+4\right)-3=\left(x-2\right)^2-3\)
Vì \(\left(x-2\right)^2\ge0,\forall x\Rightarrow\left(x-2\right)^2-3\ge-3,\forall x\)
Dấu ''='' xảy ra khi x-2=0 => x=2
Vậy GTLN của biểu thức = -3 khi x=2
Các câu khác tương tự
\(d,4x^2+4x+11=\left[\left(2x\right)^2+2.2x.1+1\right]+10=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0,\forall x\Rightarrow\left(2x+1\right)^2+10\ge10,\forall x\)
Dấu ''='' xảy ra khi 2x+1=0 => x=-1/2
Vậy GTNN của biểu thức =10 khi x=-1/2
\(e,3x^2-6x+1=3\left(x^2-2x+1\right)-2=3\left(x-1\right)^2-2\)
Vì \(3\left(x-1\right)^2\ge0,\forall x\Rightarrow3\left(x-1\right)^2-2\ge-2,\forall x\)
Dấu ''='' xảy ra khi x-1=0 => x=1
Vậy GTNN của biểu thức =-2 khi x=1
\(f,x^2-2x+y^2-4y+6=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(=\left(x-1\right)^2+\left(y-2\right)^2+1\)
Vì \(\left(x-1\right)^2\ge0,\forall x;\left(y-2\right)^2\ge0,\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1,\forall x,y\)
Dấu ''='' xảy ra khi \(\orbr{\begin{cases}x-1=0\\y-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\y=2\end{cases}}}\)
Vậy GTNN của biểu thức =1 khi x=1 và y=2
a) A = x2 + 12x + 39
= ( x2 + 12x + 36 ) + 3
= ( x + 6 )2 + 3 ≥ 3 ∀ x
Đẳng thức xảy ra ⇔ x + 6 = 0 => x = -6
=> MinA = 3 ⇔ x = -6
B = 9x2 - 12x
= 9( x2 - 4/3x + 4/9 ) - 4
= 9( x - 2/3 )2 - 4 ≥ -4 ∀ x
Đẳng thức xảy ra ⇔ x - 2/3 = 0 => x = 2/3
=> MinB = -4 ⇔ x = 2/3
b) C = 4x - x2 + 1
= -( x2 - 4x + 4 ) + 5
= -( x - 2 )2 + 5 ≤ 5 ∀ x
Đẳng thức xảy ra ⇔ x - 2 = 0 => x = 2
=> MaxC = 5 ⇔ x = 2
D = -4x2 + 4x - 3
= -( 4x2 - 4x + 1 ) - 2
= -( 2x - 1 )2 - 2 ≤ -2 ∀ x
Đẳng thức xảy ra ⇔ 2x - 1 = 0 => x = 1/2
=> MaxD = -2 ⇔ x = 1/2
Ta có A = x2 + 12x + 39 = (x2 + 12x + 36) + 3 = (x + 6)2 + 3 \(\ge\)3
Dấu "=" xảy ra <=> x + 6 = 0
=> x = -6
Vậy Min A = 3 <=> x = -6
Ta có B = 9x2 - 12x = [(3x)2 - 12x + 4] - 4 =(3x - 2)2 - 4 \(\ge\)-4
Dấu "=" xảy ra <=> 3x - 2 =0
=> x = 2/3
Vậy Min B = -4 <=> x = 2/3
b) Ta có C = 4x - x2 + 1 = -(x2 - 4x - 1) = -(x2 - 4x + 4) + 5 = -(x - 2)2 + 5 \(\le\)5
Dấu "=" xảy ra <=> x - 2 = 0
=> x = 2
Vậy Max C = 5 <=> x = 2
Ta có D = -4x2 + 4x - 3 = -(4x2 - 4x + 1) - 2 = -(2x - 1)2 - 2 \(\le\)-2
Dấu "=" xảy ra <=> 2x - 1 = 0
=> x = 0,5
Vậy Max D = -2 <=> x = 0,5
\(A=x^2-x+3=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}+3=\left(x-2\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\left(\left(x-2\right)^2\ge0\right)\)
\(\Rightarrow Min\left(A\right)=\dfrac{11}{4}\)
\(B=x^2-4x+1=x^2-4x+4-4+1=\left(x-2\right)^2-3\ge-3\left(\left(x-2\right)^2\ge0\right)\)
\(\Rightarrow Min\left(B\right)=-3\)
Câu C bạn xem lại đề
\(D=3-4x-x^2=3+4-4-4x-x^2=7-\left(x^2+4x+4\right)=7-\left(x+2\right)^2\le7\left(-\left(x+2\right)^2\le0\right)\)
\(\Rightarrow Max\left(D\right)=7\)
\(A=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ =\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\in R\)
Vậy GTNN của A là 11/4 khi x=1/2