\(\sqrt{x^2+y^2-2xy+2x-2y+5}+2y^2-8y+2015\)

\(=\sqrt{\left(x^2+y^2-2xy\right)+2\left(x-y\right)+1+4}+2\left(y^2-4y+4\right)+2007\)\(=\sqrt{\left(x-y+1\right)^2+4}+2\left(y-2\right)^2+2007\ge2007\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

TA có :

\(H=x^2+2xy+y^2-2x-2y=\left(x^2+y^2+1+2xy-2x-2y\right)-1=\left(x+y-1\right)^2-1\)

Vì  \(\left(x+y-1\right)^2\ge0\) nên \(\left(x+y-1\right)^2-1\ge-1\)

Vậy GTNN của H là -1 khi x+y-1=0 => x+y = 1

BẢO HÙNG HÓM HỈNH LỚP TAO LÀM CHO CÒN TAO CHO Ý H

H=\(X^2+2XY+Y^2-2X-2Y\)

H=\(\left(X+Y\right)^2-2\left(X+Y\right)\)

H=\(\left(X+Y\right)^2\)\(-2.\left(X+Y\right).1+1\))-1

H=\(\left(X+Y-1\right)^2-1\)

VẬY GTNN LÀ -1

Bạn ơi , có sai đề ko z ?

Ta co :

\(B=y^2-2y\left(1-y\right)+1-2y+y^2+y^2-8y+16+x^2+2x+1+2002\)

B=\(\left(y-1+y\right)^2+\left(y-4\right)^2+(x+1)^2+2002\)

Vi \(\left(2y-1\right)^2;\left(y-4\right)^2;\left(x+1\right)^2\) luon lon hon hoac bang 0 nen

ta co : minB=2002

\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)

\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)

⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)

\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)

\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)

\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)

b)B=27y^3-27y^2x+9yx^2-x^3 
= 27 . (1/3x)^3 - 27.(1/3x)².x + 9.1/3.x.x^2 - x^3 
= x^3 - 3x^3 + 3x^3 - x^3 
= 0

d) D=50y^2+x(x-2y)+14y(x-y) 

=50y^2 +x^2 -2xy +14xy -14y^2 

=36y^2 +x^2 +12xy 

=(6y + x)^2 

=81