Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

\(\left|x-1\right|+\left(y+2\right)^{2016}=0\)

Ta thấy: \(\hept{\begin{cases}\left|x-1\right|\ge0\\\left(y+2\right)^{2016}\ge0\end{cases}}\)

\(\Rightarrow\left|x-1\right|+\left(y+2\right)^{2016}\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-1\right|=0\\\left(y+2\right)^{2016}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

\(\Rightarrow A=2x^5-5y^3+2017=2\cdot1^5-5\cdot\left(-2\right)^3+2017=2059\)

\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1=1\)

Vậy C=1

\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)

\(C=2\left(x+y\right)+13x^3y^2\left(x-y\right)+15xy\left(x-y\right)+1\)

Mà x - y = 0 (bài cho)

\(\Rightarrow C=2.0+13x^3y^2.0+15xy.0+1\)

\(C=1\)

Vậy C=1

biết x-y=0

B=13-5+2022=2030

\(\left|x-1\right|+\left(y+2\right)^{2022}=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left(y+2\right)^{2022}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \Rightarrow B=13.1-5\left(-8\right)+2022=13+40+2022=2075\)

|x-1|+(y+2)²⁰²²=0

Do |x-1| và (y+2)²⁰²² đều ≥0⇒\(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

⇒B=13.(1)⁷-5.(-2)³+2022=13+40+2022=2075

\(C = 2.(x-y)+13x^3y^2.(x-y)+15.xy.\)

\((y-x) +1\)

\(C = 2.( x- y )+13x^3y^2.(x-y)-15.xy.\)

\(( x - y )+1\)

\(C = (x - y)(2 + 13x^3y^2 - 15 ) +1\)

\(C =(x- y)(13x^3y^2 - 13 )+ 1\)

Ta có:

\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1\)

=\(0+0+0+1=1\)

\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)

\(=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)\)

\(=0+0+1=1\)

~^~

bài đó nhân liên hợp là ra

Bạn tham khảo cách làm của bạn Thắng Nguyễn ở đây nhé

Câu hỏi của Băng Mikage - Toán lớp 9 - Học toán với OnlineMath

Ta có:\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)+15xy\left(y-x\right)+1\)Thế \(x-y=0\) vào C ta được:

\(C=0+0+0+1\)

C = 0