\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)

b: \(=7\left(x-y\right)-\left(x-y\right)^2\)

\(=\left(x-y\right)\left(7-x+y\right)=0\)

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

\(=-\left(x^2-4x-1\right)\)

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

C=A+B

=>C=(x²-5xy+5y²-3x+18y)-(-x²+3xy-y²-x-7)

=>C=x²-5xy+5y²-3x+18y+x²-3xy+y²+x+7

=>C=(x²+x²)-(5xy+3xy)+(5y²+y²)-(3x-x)+18y+7

=>C=2x²+6y²-8xy-2x+18y+7

tính giá trị C khó quá nên mình làm có đc 1 nửa thôi, sorry nha

tham khảo

C=A+B

=>C=(x²-5xy+5y²-3x+18y)-(-x²+3xy-y²-x-7)

=>C=x²-5xy+5y²-3x+18y+x²-3xy+y²+x+7

=>C=(x²+x²)-(5xy+3xy)+(5y²+y²)-(3x-x)+18y+7

=>C=2x²+6y²-8xy-2x+18y+7

A = x^2 + 5y^2 + 4xy - 2y - 3 

= x^2 + 4xy + 4y^2 + y^2 - 2y + 1 - 4

= ( x + 2y )^2 + ( y - 1 )^2 - 4 >= -4 

Dấu ''='' xảy ra khi y = 1 ; x = -2 

Vậy GTNN A là -4 khi x = -2 ; y = 1

Ta có: \(A=x^2+5y^2+4xy-2y-3\)

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

\(=\left(x+2y\right)^2+\left(y-1\right)^2-4\ge-4\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

Ta có

M   =   3 x 2 ( x 2   +   y 2 )   +   3 y 2 ( x 2   +   y 2 )   –   5 ( y 2   +   x 2 )     =   ( x 2   +   y 2 ) ( 3 x 2   +   3 y 2   –   5 )     =   ( x 2   +   y 2 ) [ 3 ( x 2   +   y 2 )   –   5 ]

Mà x 2   +   y 2   =   1 nên M = 1.(3.1 – 5) = -2. Vậy M = -2

Đáp án cần chọn là: D

\(a,\) Vì x,y tlt nên \(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\dfrac{2x_1+5x_2}{2y_1+5y_2}=\dfrac{5}{10}=\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{1}{2}y_1\\x_2=\dfrac{1}{2}y_2\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{2}y\)

\(b,y=3\Leftrightarrow x=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\\ c,x=5\Leftrightarrow5=\dfrac{1}{2}y\Leftrightarrow y=10\)

Đáp án C

G T ⇔ x 2 + y − 3 x + y 2 − 4 y + 4 = 0 y 2 + x − 4 y + x 2 − 3 x + 4 = 0

có nghiệm  ⇔ Δ x ≥ 0 Δ y ≥ 0 ⇔ 0 ≤ x ≤ 4 3 1 ≤ y ≤ 7 3

Và:

x y = 3 x + 4 y − x 2 − y 2 − 4 ⇒ P = 3 x 3 + 18 x 2 + 45 x − 8 ⏟ f x + − 3 y 3 + 3 y 2 + 8 y ⏟ g y

 Xét hàm số f x = 3 x 3 + 18 x 2 + 45 x − 8 trên  0 ; 4 3 ⇒ max 0 ; 4 3 f x = f 4 3 = 820 9

Xét hàm số g x = − 3 y 3 + 3 y 2 + 8 y trên  1 ; 7 3 ⇒ max 1 ; 7 3 g x = f 4 3 = 80 9

Vật P ≤ max 0 ; 4 3 f x + max 1 ; 7 3 g x = 100

Dấu “=” xảy ra khi  x = y = 4 3

a: \(=3x^4+3x^2y^2+2x^2y^2+2y^4+y^2\)

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

\(=3x^2+3y^2=3\)

b: \(=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)

c: \(=\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(y-x\right)+3=3\)

d: \(=\left(x+y\right)^2-4\left(x+y\right)+1\)

=9-12+1

=-2