GTNN=2011

giải ra jup mk V

GTNN = 2011

Vì \(\left(\left|x-3\right|+2\right)^2\ge0\left(\forall x\in Z\right)\)

       \(\left|y+3\right|\ge3\left(\forall y\in Z\right)\)

\(\Rightarrow P=\left(\left|x-3\right|+2\right)^2+\left|y+3\right|+2007\ge2007\)

Dấu "=" xảy ra khi \(\left(\left|x-3\right|+2\right)^2=0\Rightarrow\left|x-3\right|+2=0\Rightarrow\left|x-3\right|=-2\)

                                  \(\Rightarrow x\in\varnothing\) (Vì giá trị của GTTĐ không thể là một số âm)

         \(\left|y+3\right|=0\Rightarrow y+3=0\Rightarrow y=-3\)

Vậy \(P_{min}=2007\Leftrightarrow y=-3;x\in\varnothing\)

Vì | x - 3 | \(\ge\)0                                             ( 1 )

=> | x - 3 | + 2 \(\ge\)2

=> ( | x - 3 | + 2 )²  \(\ge\) 2² = 4

Vì | y + 3 | \(\ge\) 0                                          ( 2 )

Từ ( 1 ) và ( 2 ) => ( | x - 3 | + 2 )² + | y + 3 | + 2007 \(\ge\) 4 + 0 + 2007

                         => P \(\ge\) 2011

Dấu "=" xảy ra khi | x - 3 | = 0 và | y + 3 | = 0

                         => x - 3 = 0 và y + 3 = 0

                         => x = 3 và y = -3

Vậy GTNN của P là 2011 khi ( x ; y ) = ( 3 ; -3 )

\(A=\left(2x-y+1\right)^2+\left(x-3\right)^2-4y+2007\)

\(=4x^2+y^2+1-4xy+4x-2y+x^2-6x+9-4y+2007\)

\(=5x^2-4xy-2x-6y+y^2+2017\)

\(=\left[y^2-2y\left(2x+3\right)+\left(2x+3\right)^2\right]+\left(x^2-14x+49\right)+1959\)

\(=\left(y-2x-3\right)^2+\left(x-7\right)^2+1959\ge1959\)

\(minA=1959\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=17\end{matrix}\right.\)

BÀI 2 a, x²+x+1=(x²+1/2*2*x+1/4)-1/4+1=(x+1/2)² +3/4

MÀ (x+1/2)²>=0 với mọi giá trị của x .Dấu"=" xảy ra khi x+1/2=0 =>x=-1/2

    =>(x+1/2)²+3/4>=3/4 với mọi giá trị của x .Dấu "=" xảy ra khi x=-1/2

   =>x²+x+1 có giá trị nhỏ nhất là 3/4 khi x=-1/2

   b,A=y(y+1)(y+2)(y+3)

=>A =[y(y+3)] [(y+1)(y+2)]

  =>A=(y²+3y) (y²+3y+2)

Đặt X=y²+3y+1

=>A=(X+1)(X-1)

=>A=X²-1

=>A=(y²+3y+1)²-1

MÀ (y²+3y+1)²>=0 với mọi giá trị của y

=>(y²+3y+1)²-1>=-1

Vậy GTNN của Alà -1

c,B=x³+y³+z³-3xyz

=>B=(x³+y³)+z³-3xyz

=>B=(x+y)³-3xy(x+y)+z³-3xyz

=>B=[(x+y)³+z³]-3xy(x+y+z)

=>B=(x+y+z)(x²+2xy+y²-xz-yz+z²)-3xy(x+y+z)

=>B=(x+y+z)(x²+2xy+y²-xz-yz+z²-3xy)

=>B=(x+y+z)(x²+y²+z²-xy-xz-yz)

a)|x- 2006| -|2007- x|

Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x-2006\right|-\left|2007-x\right|\ge\left|x-2006-2007-x\right|=4013\)

Dấu = khi \(\left(x-2006\right)\left(2007-x\right)\ge0\)

\(\Rightarrow2006\le x\le2007\)

\(\Rightarrow\begin{cases}2006\le x\le2007\\\left(x-2006\right)\left(2007-x\right)=0\end{cases}\)\(\Rightarrow\begin{cases}x=2006\\x=2007\end{cases}\)

Vậy MinB=4013 khi x=2006 hoặc x=2007

b)Ta có:\(\begin{cases}y^2\\\left|x-16\right|\end{cases}\ge0\)

\(\Rightarrow y^2+\left|x-16\right|-9\ge0-9=-9\)

\(\Rightarrow C\ge-9\)

Dấu = khi \(\begin{cases}y^2=0\\\left|x-16\right|=0\end{cases}\)\(\Rightarrow\begin{cases}x=16\\y=0\end{cases}\)

Vậy MinC=-9 khi x=16 và y=0

hi sư huynh đệ mới lớp 6

\(A=\frac{x^2-2x+2007}{2007x^2},\left(x\ne0\right)\)

\(A=\frac{2007x^2-2x.2007+2007^2}{2007x^2}=\frac{x^2-2x.2007+2007^2}{2007x^2}+\frac{2006x^2}{2007x^2}=\) \(\frac{\left(x-2007\right)^2}{2007x^2}+\frac{2006}{2007}\ge\frac{2006}{2007}\)

\(A_{min}=\frac{2006}{2007}\) khi \(x-2007=0\) hay \(x=2007\)

Chúc bạn học tốt !!!

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<=> x = 2007

Vậy x = 2007 thì Amin

bài này từng có trên violimpic đấy bạn

a. B=|x- 2006| -|2007- x|

       Vì |x- 2006|\(\ge\)0

            |2007- x|\(\ge\)0

Suy ra:|x- 2006| -|2007- x|\(\ge\)0

   Dấu = xảy ra khi x-2006=0;x=2006

                               2007-x=0;x=2007

      Vậy Min B=0 khi x=2006;x=2007

b) C= y² +|x-16|-9

       Vì y²\(\ge\)0

           |x-16|\(\ge\)0

               Suy ra: y² +|x-16|-9\(\ge\)-9

   Dấu = xảy ra khi x-16=0;x=16

                               y²=0;y=0

Vậy Max C=-9 khi x=16;y=0