hông bít

bằng 63

\(C=4x^2+9y^2+4x-9y+3\)

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

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

Mà \(\left(2x+1\right)^2+\left(3y-\frac{3}{2}\right)^2\ge0\Rightarrow\left(2x+1\right)^2+\left(3y-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy \(C_{Min}=-\frac{1}{4}\)khi và chỉ khi \(\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-\frac{3}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+1=0\\3y-\frac{3}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}}\)

\(D=2x^2+y^2+2xy-10x+2y+2023\)

\(=\left(x^2+2xy+y^2\right)+x^2-10x+2y+2023\)

\(=\left(x+y\right)^2+2x+2y+\left(x^2-12x+36\right)+2023-36\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(x-6\right)^2+1987-1\)

\(=\left(x+y+1\right)^2+\left(x-6\right)^2+1986\)

Mà \(\left(x+y+1\right)^2+\left(x-6\right)^2\ge0\Rightarrow\left(x+y+1\right)^2+\left(x-6\right)^2+1986\ge1986\)

Vậy \(D_{Min}=1986\)khi và chỉ khi \(\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-6\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x-6=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=-7\end{cases}}}\)

Đặt `a+b=x, b+c=y, c+a=z`

`->x+y+z=2 (a+b+c)`

`(a+b)^3 +(b+c)^3 + (c+a)^3 - 8 (a+b+c)^3`

`= x^3 + y^3 + z^3 - 2^3 (a+b+c)^3`

`=x^3 +y^3 +z^3 - [2 (a+b+c)]^3`

`=x^3 +y^3+z^3 - (x+y+z)^3`

`= x^3 + y^3 +z^3 - [x^3 +y^3 +z^3 + 3 (x+y) (y+z) (x+z)]`

`= -3 (x+y)(y+z)(x+z)`

`= -3 (2b + a+c) (2c+a+b) (2a +b+c)`

Đặt : \(\hept{\begin{cases}a+b=x\\b+c=y\\c+a=z\end{cases}}\Rightarrow\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3-8\left(a+b+c\right)^3=x^3+y^3+z^3-\left(x+y+z\right)^3\)

\(=x^3+y^3+z^3-\left(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\right)=-3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(=-3\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)\)

Bài 2 : 

a, \(\left(2x+3y\right)^3=8x^3+3.4x^2.3y+3.2x.9y^2+27y^3\)

\(=8x^3+36x^2y+54xy^2+27y^3\)

b, \(\left(x+3y\right)^3=x^3+3x^2.3y+3x.9y^2+27y^3=x^3+9x^2y+27xy^2+27y^3\)

c, \(\left(2a-b\right)^3=8a^3-3.4a^2.b+3.2a.b^2-b^3=8a^3-12a^2b+6ab^2-b^3\)

d, \(\left(\frac{1}{2}x-2y\right)^3=\frac{1}{16}x^3-\frac{3}{4}x^2.2y+\frac{3}{2}x.4y^2-8y^3=\frac{1}{16}x^3-\frac{3}{2}x^2y+6xy^2-8y^3\)

ta có : hai tam giác ABD bằng CND ( do ABCD là hình bình hành nên )

\(S_{ABD}=S_{CBD}\Leftrightarrow\frac{1}{2}AH.BD=\frac{1}{2}CK.BD\Rightarrow AH=CK\)

mà AH song song với CK  (do cùng vuông góc với BD) 

nên AHCK là hình bình hành