cho x2+y2+2x+2y+2=0
tinh Q =(x+2)2017+(y+2) 2018
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13 tháng 1 2023
3x^2+3y^2+4xy-2x+2y+2=0
=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0
=>x=1 và y=-1
M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1
LN
29 tháng 12 2019
Theo đề bài : 2x2 + 2y2 + 2xy - 2x + 2y + 2 = 0
\(\Rightarrow\) ( x2 + 2xy + y2 ) + ( x2 - 2x + 1 ) + ( y2 + 2y + 1 ) = 0
( x + y )2 + ( x - 1 )2 + ( y + 1 )2 = 0
Ta thấy : \(\left(x+y\right)^2\ge0;\forall x,y\in R\)
\(\left(x-1\right)\ge0;\forall x\in R\)
\(\left(y+1\right)^2\ge0;\forall y\in R\)
\(\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0;\forall x,y\in R\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\left(\text{Thỏa mãn}\right)\)
Thay \(x=1\) và \(y=-1\) vào \(A=\left(x-2\right)^{2017}+\left(y+1\right)^{2018}\) , ta được :
\(A=\left(x-2\right)^{2017}+\left(y+1\right)^{2018}\)
\(A=\left(1-2\right)^{2017}+\left(-1+1\right)^{2018}\)
\(A=-1+0\)
\(A=-1\)
Vậy \(A=-1\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2+2xy-2x+2y+2=0\\x=1\\y=-1\end{matrix}\right.\)
NN
0
NN
0
9 tháng 1 2023
a: \(\dfrac{\left(x+1\right)}{x^2+2x-3}=\dfrac{\left(x+1\right)}{\left(x+3\right)\cdot\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x+2\right)\left(x+5\right)}{\left(x+3\right)\left(x-1\right)\left(x+2\right)\left(x+5\right)}\)
\(\dfrac{-2x}{x^2+7x+10}=\dfrac{-2x}{\left(x+2\right)\left(x+5\right)}=\dfrac{-2x\left(x+3\right)\left(x-1\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x-1\right)}\)
b: \(\dfrac{x-y}{x^2+xy}=\dfrac{x-y}{x\left(x+y\right)}=\dfrac{y^2\left(x-y\right)}{xy^2\left(x+y\right)}\)
\(\dfrac{2x-3y}{xy^2}=\dfrac{\left(2x-3y\right)\left(x+y\right)}{xy^2\left(x+y\right)}\)
c: \(\dfrac{x-2y}{2}=\dfrac{\left(x-2y\right)\left(x-xy\right)}{2\left(x-xy\right)}\)
\(\dfrac{x^2+y^2}{2x-2xy}=\dfrac{x^2+y^2}{2\left(x-xy\right)}\)
KD
1
15 tháng 12 2017
2x2+y2+9=6x+2xy
=>2x2+y2+9-6x-2xy=0
=>(x2-2xy+y2)+(x2-6x+9)=0
=>(x-y)2+(x-3)2=0
do (x-y)2 ≥ 0 ∀ x,y
(x-3)2 ≥ 0 ∀x
=>(x-y)2+(x-3)2 =0 khi
=>\(\left[{}\begin{matrix}x-y=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=x=3\\x=3\end{matrix}\right.\)
thay x=3 và y=3
Q=32017.32018-32018. 32017+\(\dfrac{1}{9}.3.3\)
Q=1
21 tháng 10 2021
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
\(x^2+y^2+2x+2y+2=0\)
<=> \(\left(x+1\right)^2+\left(y+1\right)^2=0\)
<=> \(\hept{\begin{cases}x+1=0\\y+1=0\end{cases}}\)
<=> \(x=y=-1\)
\(Q=\left(-1+2\right)^{2017}+\left(-1+2\right)^{2018}=2\)
Ta có: \(x^2+y^2+2x+2y+2=0\)
\(\left(x^2+2.x.1+1^2\right)+\left(y^2+2.y.1+1^2\right)=0\)
\(\left(x+1\right)^2+\left(y+1\right)^2=0\)
Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Mà \(\left(x+1\right)^2+\left(y+1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}\)
\(Q=\left(x+2\right)^{2017}+\left(y+2\right)^{2018}\)
\(Q=\left(-1+2\right)^{2017}+\left(-1+2\right)^{2018}\)
\(Q=1^{2017}+1^{2018}\)
\(Q=1+1\)
\(Q=2\)
Vậy \(Q=2\)
Tham khảo nhé~