Question

tìm giá trị nhỏ nhất của biểu thức 

a, A=x^2-6x+11

b, B=x^2-20x+101

c, C= x^2-6x+11

d, D= (x-1)(x+2)(x+3)(x+6)

e,E= x^2-2x+y^2+4y+8

f, x^2-4x+y^2-8y+6

g, G=x^2-4xy+5y^2+10x-22y+28

Answer 1

a/ Ta có:

\(A=x^2-6x+11\)

\(A=x\cdot x-3x-3x+3\cdot3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

Vì \(\left(x-3\right)^2\ge0\)

Nên GTNN của \(\left(x-3\right)^2\)là 0

=> \(A_{min}=0+2=2\)

Answer 2

mình chỉ biết a. thôi

a) ta có : \(A=x^2-6x+11\)

\(A=x.x-3x-3x+3.3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

vì \(\left(x-3\right)^2\ge0\)

nên GTNN của \(\left(x-3\right)^2\)là \(0\)

\(\Rightarrow\)\(A_{min}\)\(=0+2=2\)

Answer 3

oOo Không đủ can đảm để oOo copy mà nói nhưu mk tự làm

Answer 4

a/ Ta có:

\(A=x^2-6x+11\)

\(A=x\cdot x-3x-3x+3\cdot3+2\)

\(A=x\left(x-3\right)-3\left(x-3\right)+2\)

\(A=\left(x-3\right)\left(x-3\right)+2\)

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

Mà \(\left(x-3\right)^2\ge0\)

Nên GTNN của \(\left(x-3\right)^2\)là 0

Vậy \(A_{min}=0+2=2\)

b/Ta có:

\(B=x^2-20x+101\)

\(B=x\cdot x-10x-10x+10\cdot10+1\)

\(B=x\left(x-10\right)-10\left(x-10\right)+1\)

\(B=\left(x-10\right)\left(x-10\right)+1\)

\(B=\left(x-10\right)^2+1\)

Mà \(\left(x-10\right)^2\ge0\)

=> GTNN của \(\left(x-10\right)^2\)là 0

Vậy \(B_{min}=0+1=1\)

c/Ta có:

\(C=x^2-6x+11\)

\(C=x\cdot x-3x-3x+3\cdot3+2\)

\(C=x\left(x-3\right)-3\left(x-3\right)+2\)

\(C=\left(x-3\right)\left(x-3\right)+2\)

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

Mà \(\left(x-3\right)^2\ge0\)

nên GTNN của \(\left(x-3\right)^2\)là 0

=> \(C_{min}=0+2=2\)

Answer 5

a, Ta có \(A=x^2-6x+11\)

\(A=x^2-2.x.3+3^2+2\)

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

Vì \(\left(x-3\right)^2\ge0\) dấu = khi \(x-3=0\Leftrightarrow x=3\)

      \(2>0\)

\(\Rightarrow\left(x-3\right)^2+2\ge2\) dấu = khi \(x=3\) 

\(\Rightarrow x^2-6x+11\ge2\) dấu = khi \(x=3\)

Vậy \(A_{min}=2\) khi \(x=3\)

b, Ta có \(B=x^2-20x+101\)

\(B=x^2-2.x.10+10^2+1\)

\(B=\left(x-10\right)^2+1\)

Vì \(\left(x-10\right)^2\ge0\) dấu = khi \(x-10=0\Leftrightarrow x=10\)

     \(1>0\)

\(\Rightarrow\left(x-10\right)^2+1\ge1\)dấu = khi \(x=10\)

\(\Rightarrow x^2-20x+101\ge1\)dấu = khi \(x=10\)

Vậy \(A_{min}=1\) khi \(x=10\)

d,Ta có  \(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(D=\text{[}\left(x-1\right)\left(x+6\right)\text{]}.\text{[}\left(x+2\right)\left(x+3\right)\text{]}\)

\(D=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(D=\left(x^2+5x\right)^2-6^2\)

\(D=\left(x^2+5x\right)^2-36\)

Vì \(\left(x^2+5x\right)^2\ge0\) dấu = khi \(x^2+5x=0\Leftrightarrow\left(x+5\right)x=0\Leftrightarrow\hept{\begin{cases}x=0\\x+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-5\end{cases}}}\)

\(\Rightarrow\left(x^2+5x\right)-36\ge-36\) dấu = khi \(\hept{\begin{cases}x=-5\\x=0\end{cases}}\)

Vậy \(D_{min}=-36\) khi \(\hept{\begin{cases}x=0\\x=-5\end{cases}}\)

e, Ta có \(E=x^2-2x+y^2+4y+8\)

\(E=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)

\(E=\left(x-1\right)^2+\left(y+2\right)^2+3\)

Vì \(\left(x-1\right)^2\ge0\) dấu = khi x=1

\(\left(y+2\right)^2\ge0\) dấu = khi \(y=-2\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\) dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

Vậy \(E_{min}=3\) dấu = khi \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

\(f,F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Vì \(\left(x-2\right)^2\ge0\) dấu = khi \(x=2\)

\(\left(y-4\right)^2\ge0\) dấu = khi \(y=4\)

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) dấu = khi \(\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(F_{min}=-14\) dấu = khi \(\hept{\begin{cases}x=2\\y=4\end{cases}}\)