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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: \(\hept{\begin{cases}\left|x-1\right|\ge0\\\left(y+2\right)^{2016}\ge0\end{cases}}\)
\(\Rightarrow\left|x-1\right|+\left(y+2\right)^{2016}\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x-1\right|=0\\\left(y+2\right)^{2016}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow A=2x^5-5y^3+2017=2\cdot1^5-5\cdot\left(-2\right)^3+2017=2059\)
Ta có:
\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1\)
=\(0+0+0+1=1\)
\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)
\(=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)\)
\(=0+0+1=1\)
~^~
Ta có:
\(A=\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^4}-\frac{1}{y^4}\right)=\frac{1}{\left(x+y\right)^3}.\frac{\left(y^2+x^2\right)\left(x+y\right)\left(y-x\right)}{x^4y^4}=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}\)
\(B=\frac{1}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x+y\right)^4x^3y^3}\)
\(C=\frac{1}{\left(x+y\right)^5}\left(\frac{1}{x^2}-\frac{1}{y^2}\right)=\frac{y-x}{\left(x+y\right)^4x^2y^2}\)
\(\Rightarrow A+B+C=\frac{\left(x^2+y^2\right)\left(y-x\right)}{\left(x+y\right)^2x^4y^4}+\frac{\left(y-x\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)^4x^3y^3}+\frac{\left(y-x\right)}{\left(x+y\right)^4x^2y^2}\)
\(=\frac{y^3-x^3}{x^4y^4\left(x+y\right)^2}\)
b/ Thế vô rồi tính nhé
Đoạn gần cuối thay y-x= 1 luôn
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2x^4y^4}+\left(\frac{\left(x+y\right)^2}{\left(x+y\right)^4\left(xy\right)^3}\right)\\ \)
\(A+B+C=\frac{x^2+y^2}{\left(x+y\right)^2\left(xy\right)^4}+\frac{1}{\left(x+y\right)^2\left(xy\right)^3}\)
\(A+B+C=\frac{x^2+y^2+xy}{\left[\left(x+y\right)xy\right]^2\left(xy\right)^2}\) giờ mới thay không biết đã tối giản chưa
vì (x-2016)2 và (y+2017)2 đều lớn hơn hoặc = 0
=> (x-2016)2 = 0
=> x = 2016
=> (y+2017)2 = 0
=> y = -2017
mik nha chế
( x - 2016 )2 + ( y + 2017 )2 = 0
=> ( x - 2016 )2 = 0 và ( y + 2017 )2 = 0
+) ( x - 2016 )2 = 0
=> x - 2016 = 0
=> x = 2016
+) ( y + 2017 )2 = 0
=> y + 2017 = 0
=> y = -2017
Vậy x = 2016, y = -2017
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Ta thấy \(VT\ge VP\forall x;y\) để đấu "=" xảy ra \(\Leftrightarrow x=1;y=-1\) thay vào M :
\(M=\left(-1+1\right)^{2015}+\left(1-2\right)^{2016}+\left(-1+1\right)^{2017}=1\)
\(\left|x-1\right|+\left(y+2\right)^{2016}=0\)
=>x-1=0 và y+2=0
=>x=1 và y=-2
\(C=13\cdot1^5-3\cdot\left(-2\right)^3+2017=13+2017-3\cdot\left(-8\right)=2030+24=2054\)