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Bài 1 :
a) \(A=x^2-6x+11\)
\(A=x^2-2\cdot x\cdot3+3^2+2\)
\(A=\left(x-3\right)^2+2\ge2\forall x\)
Dấu "=' xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
b) \(B=2x^2+10x-1\)
\(B=2\left(x^2+5x-\frac{1}{2}\right)\)
\(B=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)
\(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
c) \(C=5x-x^2\)
\(C=-\left(x^2-5x\right)\)
\(C=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]\)
\(C=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)
\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Bài 2 :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[x+\left(y+z\right)\right]^3-x^3-y^3-z^3\)
\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3-x^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+y^3+3y^2z+3yz^2+z^3-y^3-z^3\)
\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+3yz\left(y+z\right)\)
\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)+yz\right]\)
\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
1,A=(x2-6x+9)+2
=(x-3)2+2
ta thấy (x-3)2>=0 với mọi x
=>(x-3)2+2>=2 với mọi x
hay A>=2
dấu "="xảy ra x-3=0<=>x=3
vậy MinA=2 khi x=3
ý b sai đầu bài bạn nhé
C=-(x2-5x)
=-(x2-5x+25/4)+25/4
=-(x-5/2)2+25/4
ta thấy -(x-5/2)2<=0 với mọi x
=>-(x-5/2)2+25/4 <=25/4 với mọi x
hay C<=25/4
dấu "=" xảy ra khi x-5/2=0<=>x=5/2
vậy MaxC=25/4 khi x=5/2
k mk nha
a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)
c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)
\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)
d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)
\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)
e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)
\(minE=-20\Leftrightarrow x=-3\)
f) \(G=\dfrac{1}{4}x^2+x-\dfrac{1}{3}=\left(\dfrac{1}{2}x+1\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)
\(A=x^2+3x+4=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Do \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Mấy câu còn lại làm tương tự nhé em^^
a)4x2-4x+3
=[(2x)2-4x+1]+2
=(2x+1)2+2 \(\ge\)2 với mọi x
Vậy GTNN của 4x2-4x+3 là 2 tại
(2x+1)2+2=2
<=>(2x+1)2 =0
<=>2x+1 =0
<=>x =\(\frac{-1}{2}\)
b)-x2+2x-3
=(-x2+2x-1)-2
= -(x2-2x+1)-2
=-(x-1)2-2 \(\le\)-2
Vậy GTLN của -x2+2x-3 là -2 tại :
-(x-1)2-2=-2
<=>-(x-1)2 =0
<=>x-1 =0
<=>x =1
\(\dfrac{3x^2-1}{x^2+2}=\dfrac{6x^2-2}{2\left(x^2+2\right)}=\dfrac{7x^2-\left(x^2+2\right)}{2\left(x^2+2\right)}=\dfrac{7x^2}{2\left(x^2+2\right)}-\dfrac{1}{2}\ge=-\dfrac{1}{2}\)
GTNN của biểu thức là \(-\dfrac{1}{2}\), xảy ra khi \(x=0\)
Biểu thức ko tồn tại GTLN
Bài 1:
1: \(=x^2-6x+9+2=\left(x-3\right)^2+2>=2\)
Dấu = xảy ra khi x=3
2: \(=2\left(x^2+5x-\dfrac{1}{2}\right)\)
\(=2\left(x^2+5x+\dfrac{25}{4}-\dfrac{27}{4}\right)\)
\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}>=-\dfrac{27}{2}\)
Dấu = xảy ra khi x=-5/2
3: =-(x^2-5x)
=-(x^2-5x+25/4-25/4)
=-(x-5/2)^2+25/4<=25/4
Dấu = xảy ra khi x=5/2