\(x^2+y^2-xy-2x-2y+9=x^2+y^2+2xy-2x-2y+9-3xy\)

\(=\left(x+y\right)^2-2\left(x+y\right)+9-3xy=\left(x+y-2\right)\left(x+y\right)+9-3xy.\)

\(đếnđâytịt\)

b 

c, =3 dễ

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

Câu b bạn không làm à? Làm hộ mình với! Còn câu a thì còn -3xy thì?

26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

2/

a, \(A=2x^2+6x-5=2\left(x^2+3x-\frac{5}{2}\right)=2\left(x^2+2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{19}{4}\right)=2\left[\left(x+\frac{3}{2}\right)^2-\frac{19}{4}\right]=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow A=\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

Dấu "=" xảy ra khi x=-3/2

Vậy Amin=-19/2 khi x=-3/2

b,bài này phải tìm min 

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

Vì \(\left(x-2\right)^2\ge0\Rightarrow B=\left(x-2\right)^2+4\ge4\)

Dấu "=" xảy ra khi x = 2

Vậy Bmin=4 khi x=2

Bài 2)Ta có:

\(2x^2+6x-5\)

\(=2x^2+6x+\frac{9}{2}-\frac{19}{2}\)

\(=2\left(x^2+3x+\frac{9}{4}\right)-\frac{19}{2}\)

\(=2\left(x+\frac{3}{2}\right)^2-\frac{19}{2}\ge-\frac{19}{2}\)

a, A=3.(2/3)^3-2.(1/2)^3-6.(2/3)^2.(1/2)^2+(2/3).(1/2)

      =8/9-1/4-2/3+1/3=8/9-1/4-1/3=11/36

b,  B=-1+(-1/18)+1/12+2/3=-11/36

MK KO BT MK MỚI HO C LỚP 6

AI HỌC LỚP 6 CHO MK XIN

a)B=3x³ -2y³-6x²y²+xy

   B=(3x³-6x²y²)+(xy-2y³)

   B=3x²(x-2y²)+y(x-2y²)

    B=(x-2y²)(3x²+y)
tại x=\(\frac{2}{3}\)và y=\(\frac{1}{2}\)ta có B=(x-2y²)(3x²+y)=(\(\frac{2}{3}\)-2*^{\(\frac{1}{2}\)}^2 )(3*\(\frac{2}{3}\)^2+\(\frac{1}{2}\))=\(\frac{1}{6}\)*\(\frac{11}{6}\)=\(\frac{11}{36}\)

b)C= 2x+xy²-x²y-2y

   C=(2x-2y)+(xy²-x²y)

   C=2(x-y)-xy(x-y)

   C=(2-xy)(x-y)

tại x=\(-\frac{1}{2}\)và y=\(-\frac{1}{3}\)ta có C=(2-xy)(x-y)=(2-\(-\frac{1}{2}\)*\(-\frac{1}{3}\))(\(-\frac{1}{2}\)+\(\frac{1}{3}\))=\(\frac{-11}{36}\)