11: \(\dfrac{1}{3}x^2y^2\left(6x+\dfrac{2}{3}x^2-y\right)\)

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

12: \(\dfrac{3}{4}x^3y^2\left(4x^2y-x+y^5\right)\)

\(=3x^5y^3-\dfrac{3}{4}x^4y^2+\dfrac{3}{4}x^3y^7\)

13: \(-5x^2y^4\left(3x^2y^3-2x^3y^2-xy\right)\)

\(=-15x^4y^7+10x^5y^6+5x^3y^5\)

1 x − x y = x 2 + x y − 2 y 2 ( 1 ) x + 3 − y 1 + x 2 + 3 x = 3 ( 2 )

Điều kiện:  x > 0 y > 0 x + 3 ≥ 0 x 2 + 3 x ≥ 0 ⇔ x > 0 y > 0

( 1 ) ⇔ y − x y x = ( x − y ) ( x + 2 y ) ⇔ ( x − y ) x + 2 y + 1 y x = 0 ⇔ x = y do  x + 2 y + 1 y x > 0 , ∀ x , y > 0

Thay y = x vào phương trình (2) ta được:

( x + 3 − x ) ( 1 + x 2 + 3 x ) = 3 ⇔ 1 + x 2 + 3 x = 3 x + 3 − x ⇔ 1 + x 2 + 3 x = x + 3 + x ⇔ x + 3 . x − x + 3 − x + 1 = 0 ⇔ ( x + 1 − 1 ) ( x − 1 ) = 0 ⇔ x + 3 = 1 x = 1 ⇔ x = − 2 ( L ) x = 1 ( t m ) ⇒ x = y = 1

Vậy hệ có nghiệm duy nhất (1;1)

2,

M + N = 3xyz - 3x² + 5xy - 1 + 5x² + xyz - 5xy + 3 - y

= -3x² + 5x² + 3xyz + xyz + 5xy - 5xy - y - 1 + 3

= 2x² + 4xyz - y +2.

M - N = (3xyz - 3x² + 5xy - 1) - (5x² + xyz - 5xy + 3 - y)

= 3xyz - 3x² + 5xy - 1 - 5x² - xyz + 5xy - 3 + y

= -3x² - 5x² + 3xyz - xyz + 5xy + 5xy + y - 1 - 3

= -8x² + 2xyz + 10xy + y - 4.

N - M = (5x² + xyz - 5xy + 3 - y) - (3xyz - 3x² + 5xy - 1)

= 5x² + xyz - 5xy + 3 - y - 3xyz + 3x² - 5xy + 1

= 5x² + 3x² + xyz - 3xyz - 5xy - 5xy - y + 3 + 1

= 8x² - 2xyz - 10xy - y + 4.

3,

a) P + (x² – 2y²) = x² – y² + 3y² – 1

P = (x² – y² + 3y² – 1) - (x² – 2y²)

P = x² – y² + 3y² – 1 - x² + 2y²

P = x² – x² – y² + 3y² + 2y² – 1

P = 4y² – 1.

Vậy P = 4y² – 1.

b) Q – (5x² – xyz) = xy + 2x² – 3xyz + 5

Q = (xy + 2x² – 3xyz + 5) + (5x² – xyz)

Q = xy + 2x² – 3xyz + 5 + 5x² – xyz

Q = 7x² – 4xyz + xy + 5

Vậy Q = 7x² – 4xyz + xy + 5.

4,

a, Thu gọn : x2+2xy-3x3+2y3+3x3-y3

= x2+2xy+(-3x3+3x3)+2y3-y3

=x2+2xy+2y3-y3

Thay x=5,y=4 vào đa thức x2+2xy+2y3-y3 Ta có:

5² + 2.5.4 + 4³ = 25 + 40 + 64 = 129.

Vậy giá trị của đa thức x2+2xy+2y3-y3 tại x=5,y=4 là 129

b,

Thay x = -1; y = -1 vào biểu thức xy-x2y2+x4y4-x6y6+x8y8 Ta Có

M = (-1)(-1) - (-1)².(-1)² + (-1)^4. (-1)⁴-(-1)⁶.(-1)⁶ + (-1)⁸.(-1)⁸

= 1 -1 + 1 - 1+ 1 = 1.

Vậy giá trị của biểu thức xy-x2y2+x4y4-x6y6+x8y8 tại x=-1, y=-1 là 1

5,

a, C=A+B

C = x² – 2y + xy + 1 + x² + y - x²y² - 1

C = 2x² – y + xy - x²y²

b) C + A = B => C = B - A

C = (x² + y - x²y² - 1) - (x² – 2y + xy + 1)

C = x² + y - x²y² - 1 - x² + 2y - xy - 1

C = - x²y² - xy + 3y - 2.

dễ mà , có khó đâu bạn

ây dà,tự túc đi bạn

\(\left(xy+1\right)\left(2y-x\right)=2x^3y^2\Leftrightarrow2xy^2+2y-x^2y-x=2x^3y^2\)

\(\Leftrightarrow2xy^2+2y-x^2y+x=2x\left(x^2y^2+1\right)=2y^2.2x=4xy^2\)

\(\Leftrightarrow2y-x^2y+x-2xy^2=0\Leftrightarrow\left(2y+x\right)\left(1-xy\right)=0\Rightarrow\orbr{\begin{cases}x=-2y\\xy=1\end{cases}.}\)

Đến đây thì dễ rồi

Có 1 ý tưởng nhưng mà khùng v ler ấy :))

Từ \(x^2y^2+1=2y^2\Rightarrow x^2y^2-2y^2=-1\)

\(\Rightarrow y^2\left(x^2-2\right)=-1\Rightarrow y^2=\frac{1}{2-x^2}\Rightarrow y=\frac{1}{\sqrt{2-x^2}}\)

\(pt\left(1\right)\Rightarrow\left(x\sqrt{\frac{1}{\: 2-x^2}}+1\right)\left(2\sqrt{\frac{1}{\: 2-x^2}}-x\right)=2x^3\left(\sqrt{\frac{1}{\: 2-x^2}}\right)^2\)

\(\Leftrightarrow\frac{x^2\sqrt{2-x^2}}{x^2-2}-\frac{2\sqrt{2-x^2}}{x^2-2}-\frac{x^3}{x^2-2}=\frac{2x^3}{2-x^2}\)

\(\Leftrightarrow\frac{x^2\sqrt{2-x^2}}{x^2-2}-\frac{2\sqrt{2-x^2}}{x^2-2}+\frac{x^3}{x^2-2}=0\)

\(\Leftrightarrow\frac{x^2\sqrt{2-x^2}}{x^2-2}+1-\frac{2\sqrt{2-x^2}}{x^2-2}-2+\frac{x^3}{x^2-2}+1=0\)

\(\Leftrightarrow\frac{x^2\sqrt{2-x^2}+x^2-2}{x^2-2}-\frac{2\sqrt{2-x^2}-\left(2x^2-4\right)}{x^2-2}+\frac{x^3+x^2-2}{x^2-2}=0\)

\(\Leftrightarrow\frac{\frac{x^4\left(2-x^2\right)-\left(x^2-2\right)^2}{x^2\sqrt{2-x^2}-x^2+2}}{x^2-2}-\frac{\frac{4\left(2-x^2\right)-\left(2x^2-4\right)^2}{2\sqrt{2-x^2}+\left(2x^2-4\right)}}{x^2-2}+\frac{\left(x-1\right)\left(x^2+2x+2\right)}{x^2-2}=0\)

\(\Leftrightarrow\frac{\frac{-\left(x-1\right)\left(x+1\right)\left(x^2-2\right)\left(x^2+2\right)}{x^2\sqrt{2-x^2}-x^2+2}}{x^2-2}-\frac{\frac{-4\left(x-1\right)\left(x+1\right)\left(x^2-2\right)}{2\sqrt{2-x^2}+\left(2x^2-4\right)}}{x^2-2}+\frac{\left(x-1\right)\left(x^2+2x+2\right)}{x^2-2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{\frac{-\left(x+1\right)\left(x^2-2\right)\left(x^2+2\right)}{x^2\sqrt{2-x^2}-x^2+2}}{x^2-2}-\frac{\frac{-4\left(x+1\right)\left(x^2-2\right)}{2\sqrt{2-x^2}+\left(2x^2-4\right)}}{x^2-2}+\frac{x^2+2x+2}{x^2-2}\right)=0\)

\(\Rightarrow x=1\Rightarrow y=\frac{1}{\sqrt{2-x^2}}=1\)

Ôi chúa :)) nhầm dấu thảo nào ngồi từ chiều tới giờ ko ra :))

\(y^2=\frac{1}{2-x^2}\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{\sqrt{2-x^2}}\\y=-\frac{1}{\sqrt{2-x^2}}\end{cases}}\) 

a)2x^2+xy-y^2-x+2y-1

=2x^2+xy-x-(y-1)^2

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

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

=(2a-b)(a+b)

=(2x-y+1)(x+y-1)

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+2y^2-4y+3=0\\2x^2+2x^2y^2-4y=0\left(1\right)\end{matrix}\right.\Rightarrow}x^3+2y^2-4y-2x^2-2x^2y^2+4y=0\Rightarrow x^3+1-2x^2y^2+2y^2-2x^2+2=0\Rightarrow\left(x+1\right)\left(x^2-x+1\right)-2y^2\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(x^2-x+1-2xy^2+2y^2-2x+2\right)=0\Rightarrow x=-1\)Thay x=-1 vào (1) ta được y²-2y+1=0⇒ (y-1)²=0⇒y-1=0⇒y=1

Do đó Q=x²+y²=(-1)²+1²=2

a) Ta có: ( x2 -1 )( x2 + 2x )

= x2( x2 + 2x ) - ( x2 + 2x )

= x4 + 2x3 - x2 - 2x

b) Ta có ( x + 3 )( x2 + 3x -5 )

= x( x2 + 3x -5 ) + 3( x2 + 3x -5 )

= x3 + 3x2 - 5x + 3x2 + 9x - 15

= x3 + 6x2 + 4x - 15

c) Ta có ( x -2y )( x2y2 - xy + 2y )

= x( x2y2 - xy + 2y ) - 2y( x2y2 - xy + 2y )

= x3y2 - x2y + 2xy - 2x2y3 + 2xy2 - 4y2

d) Ta có ( 1/2xy -1 )( x3 -2x -6 )

= 1/2xy( x3 -2x -6 ) - ( x3 -2x -6 )

= 1/2x4y - x2y - 3xy - x3 + 2x + 6