Question

Giúp mik với mình cần gấp ạ:

Cho các số thực \(x\), \(y\), \(z\) thỏa mãn \(x+y+z=3\). Tìm giá trị nhỏ nhất của biểu thức: \(P=x(2x+z)+y(6y+z)\).

Answer 1

Ko

Answer 2

Ko

Answer 3

\(P=2x^2+xz+6y^2+yz=2x^2+6y^2+\left(x+y\right)z\)

\(=2x^2+6y^2+\left(x+y\right)\left(3-x-y\right)\)

\(=x^2+5y^2-2xy+3x+3y\)

\(=\left(x^2-2xy+y^2\right)+3\left(x-y\right)+4y^2+6y\)

\(=\left(x-y\right)^2+3\left(x-y\right)+\dfrac{9}{4}+\left(4y^2+6y+\dfrac{9}{4}\right)-\dfrac{9}{2}\)

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

Do \(\left\{{}\begin{matrix}\left(x-y+\dfrac{3}{2}\right)^2\ge0\\\left(2y+\dfrac{3}{2}\right)^2\ge0\end{matrix}\right.\) ; \(\forall x;y\)

\(\Rightarrow P\ge-\dfrac{9}{2}\)

Vậy \(P_{min}=-\dfrac{9}{2}\) khi \(\left\{{}\begin{matrix}x+y+z=3\\x-y+\dfrac{3}{2}=0\\2y+\dfrac{3}{2}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{4}\\y=-\dfrac{3}{4}\\z=6\end{matrix}\right.\)

Answer 4

Giải thích các bước giải:

Ta có: x+y+z=3→z=3−x−yx+y+z=3→z=3−x−y

→P=x(2x+3−x−y)+y(6y+3−x−y)→P=x(2x+3−x−y)+y(6y+3−x−y) 

→P=x2−2xy+3x+5y2+3y→P=x2−2xy+3x+5y2+3y

→P=x2−x(2y−3)+5y2+3y→P=x2−x(2y−3)+5y2+3y

→P=x2−2x⋅12(2y−3)+(12(2y−3))2+5y2+3y−(12(2y−3))2→P=x2−2x⋅12(2y−3)+(12(2y−3))2+5y2+3y−(12(2y−3))2

→P=(x−12(2y−3))2+14(16y2+24y−9)→P=(x−12(2y−3))2+14(16y2+24y−9)

→P=(x−12(2y−3))2+14(4y+3)2−92≥−92→P=(x−12(2y−3))2+14(4y+3)2−92≥−92

→GTNNP=−92→GTNNP=−92

→x−12(2y−3)=0→x−12(2y−3)=0 và 4y+3=0→x=−34,y=−944y+3=0→x=−34,y=−94

→z=3−(−34)−(−94)=6