Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

a. \(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)

vì \(\left(x-1\right)^2\ge0\) với mọi x

=> (x-1)^2 +4 \(\ge\) vợi mọi x

Pmin=4 <=> x-1=0 <=>x=1

1.

b)\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu = xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\) và \(y+3=0\)

\(\Leftrightarrow x=\frac{1}{2}\) và \(y=-3\)

Vậy GTNN của M là \(\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)và \(y=-3\)

1.

a) \(P=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu = xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy GTNN của P là 4 

Câu 1.

P = x² - 2x + 5 

= ( x² - 2x + 1 ) + 4

= ( x - 1 )² + 4 ≥ 4 ∀ x

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinP = 4 <=> x = 1

Q = 2x² - 6x

= 2( x² - 3x + 9/4 ) - 9/2

= 2( x - 3/2 )² - 9/2 ≥ -9/2 ∀ x

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MinQ = -9/2 <=> x = 3/2

M = x² + y² - x + 6y + 10

= ( x² - x + 1/4 ) + ( y² + 6y + 9 ) + 3/4

= ( x - 1/2 )² + ( y + 3 )² + 3/4 ≥ 3/4 ∀ x

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

=> MinM = 3/4 <=> x = 1/2 ; y = -3

Câu 2.

A = 4x - x² + 3

= -( x² - 4x + 4 ) + 7

= -( x - 2 )² + 7 ≤ 7 ∀ x

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxA = 7 <=> x = 2

B = x - x²

= -( x² - x + 1/4 ) + 1/4

= -( x - 1/2 )² + 1/4 ≤ 1/4 ∀ x

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

=> MaxB = 1/4 <=> x = 1/2

N = 2x - 2x²

= -2( x² - x + 1/4 ) + 1/2

= -2( x - 1/2 )² + 1/2 ≤ 1/2 ∀ x

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

=> MaxB = 1/2 <=> x = 1/2

Làm gần xong thì lỡ bấm out ra TT

\(P=x^2-2x+5=\left(x-1\right)^2+4\ge4\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy minP = 4 <=> x = 1

\(Q=2x^2-6x=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

Vậy minQ = - 9/2 <=> x = 3/2

\(M=x^2+y^2-x+6y+10\)

\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\)

Vì \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

Vậy minM = 3/4 <=> x = 1/2 và y = - 3

a,\(P=x^2-2x+5\)

\(=\left(x^2-2x+1\right)+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu"="xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy \(Min_P=4\Leftrightarrow x=1\)

b,\(2x^2-6x\)

\(=2\left(x^2-3x\right)\)

\(=2\left(x^2-2x\frac{3}{2}+\frac{9}{4}\right)-\frac{9}{4}\)

\(=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\forall x\)

Dấu"="xảy ra khi \(2\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)

Vậy \(Min_Q=-\frac{9}{2}\Leftrightarrow x=\frac{3}{x}\)

c,\(M=x^2+y^2-x+6y+10\)

\(=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x,y\)

Dấu"="xảy ra khi \(\orbr{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}}\)

Vậy \(Min_M=\frac{3}{4}\Leftrightarrow x=\frac{1}{2};y=-3\)

x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1 
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x) 
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1] 
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0 
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4 
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4 
Vay gia tri nho nhat P=4 khi x=1 
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4] 
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2 
Vay gia tri nho nhat Q= -9/2 khi x= 3/2 
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4 
= ( x-1/2)^2 + (y+3)^2 +3/4 
M>= 3/4 
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3 
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7] 
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7 
Vay GTLN A=7 khi x=2 
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4] 
GTLN B= 1/4 khi x=1/2 
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4) 
= -2[(x-1/2)^2 +9/4] 
GTLN N= -9/2 khi x=1/2

2−2x+5

2+2x−5)

2+2x+1)+6

2+6

2≤0∀x

2+6≤6∀x

Dấu "=" xảy ra ⇔

$x=-1$

Vậy $MA{X}_M=6\iff x=-1$

2+3

22−4x−3)

2−4x+4−7)

2−7]

2+7

22+7≤7

2=0⇔x=2

Vậy MAXA=7 khi x = 2

mới chỉ có GTLN thôi XD

a) \(x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\)

MIN P = 4 khi \(x-1=0=>x=1\)

b) \(2x^2-6x\)

\(=2\left(x^2-3x\right)\)

\(=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)

\(=\frac{-18}{4}+2\left(x^2-\frac{3}{2}\right)^2\le\frac{-18}{4}\)

MIN Q = \(\frac{-18}{4}\)khi \(x^2-\frac{3}{2}=0\)

\(=>x^2=\frac{3}{2}\)

\(=>\orbr{\begin{cases}x=-\sqrt{\frac{3}{2}}\\x=\sqrt{\frac{3}{2}}\end{cases}}\)

Ủng hộ nha

a) P=x^2-2x+5

=x²-2x+1+4

=(x-1)²+4

Ta thấy;\(\left(x-1\right)^2+4\ge0+4=4\)

Dấu = <=>x-1=0 =>x=1

Vậy...