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

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

Ta có : P = x² - 2x + 5 = x² - 2x + 1 + 4 = (x - 1)² + 4

Vì \(\left(x-1\right)^2\ge0\forall x\)

Suy ra : \(P=\left(x-1\right)^2+4\ge4\forall x\)

Nên : P_min = 4 khi x = 1

b) Ta có Q = 2x² - 6x = 2(x² - 3x) = 2(x² - 3x + \(\frac{9}{4}-\frac{9}{4}\) ) = \(2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)

Vì \(2\left(x-\frac{3}{2}\right)^2\ge0\forall x\) 

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

Vậy \(Q_{min}=-\frac{9}{2}\) khi \(x=\frac{3}{2}\)

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

b.\(B=x-x^2=\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\)

c.\(C=2x-2x^2-5=-\frac{9}{2}-2\left(x^2-x+\frac{1}{4}\right)=-\frac{9}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{9}{2}\)

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

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 

Biểu thức nào em?

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1