\(A=x^2-4x^2+2-1=\left(x-2\right)^2-1\)
suy ra Amin=-1
\(B=4x^2+4x+11=4\left(x^2+x+\frac{11}{4}\right)=4\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{10}{4}\right)=4\left(x+\frac{1}{2}\right)^2+10\) Suy ra Bmin = 10
phần B có bạn làm rồi nha mình không làm nữa
A=x2-4x+1=x2-4x+4-3=(x-2)2-3
Vì (x-2)2\(\ge\)0\(\forall\)x \(\Rightarrow\)(x-2)2-3\(\ge\)-3\(\forall\)x
Vậy minA = -3
C=(x-1)(x+3)(x+2)(x+6)
C=(x-1)(x+6)(x+3)(x+2)
C=(x2+5x-6)(x2+5x+6)
Đặt x2+5x+6=t . Ta có:
C= (t-12).t=t2-12t=t2-12+36-36=(t-6)2-36
C= (x2+5x+6-6)2-36=(x2+5x)2-36
Vì (x2+5x)2\(\ge\)0\(\forall\)x \(\Rightarrow\)(x2+5x)2-36\(\ge\)-36\(\forall\)x
Vậy minC= -36
D=5-8x-x2=-(x2+8x-5)=-(x2+8x+16-21)=-\(\left[\left(x+4\right)^2-21\right]\)
D=-(x+4)2+21=21-(x+4)2
Vì (x+4)2\(\ge\)0\(\forall\)x\(\Rightarrow\)21-(x+4)2\(\le\)21\(\forall\)x
Vậy maxD=21
E=4x-x2+1=-(x2-4x-1)=-(x2-4x+4-5)=-\(\left[\left(x-2\right)^2-5\right]\)=-(x-2)2+5=5-(x-2)2
Vì (x-2)2\(\ge0\forall x\)\(\Rightarrow\)5-(x-2)2\(\le5\forall x\)
Vậy maxE=5
C,D,E dùng phương pháp đó tách tiếp ra
A=x2−4x+1A=x2−4x+1
⇔A=x2−4x+4−3⇔A=x2−4x+4−3
⇔A=(x2−4x+4)−3⇔A=(x2−4x+4)−3
⇔A=(x−2)2−3≥−3;∀x⇔A=(x−2)2−3≥−3;∀x
⇔AMin=−3⇔AMin=−3
"="⇔x−2=0⇔x=2"="⇔x−2=0⇔x=2
Vậy ...
GTLN:
D=5−8x−x2D=5−8x−x2
⇔D=21−16−8x−x2⇔D=21−16−8x−x2
⇔D=21−(16+8x+x2)⇔D=21−(16+8x+x2)
⇔D=21−(4+x)2≤21;∀x⇔D=21−(4+x)2≤21;∀x
⇔DMax=21⇔DMax=21
"="⇔4+x=0⇔x=−4