Sửa đề: Các dấu bằng ở yêu cầu là dấu cộng.

1. Có: \(x+y=3\)

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

\(\Leftrightarrow x^2+2xy+y^2=9\)

\(\Leftrightarrow x^2+y^2=9-2\cdot1=7\) (do \(xy=1\))

\(------\)

Lại có: \(x+y=3\)

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

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=27\)

\(\Leftrightarrow x^3+y^3+3\cdot1\cdot3=27\) (do x + y = 3; xy = 1)

\(\Leftrightarrow x^3+y^3=18\)

Ta có: \(x^2+y^2=7\)

\(\Leftrightarrow\left(x^2+y^2\right)^2=7^2\)

\(\Leftrightarrow x^4+y^4+2\cdot\left(xy\right)^2=49\)

\(\Leftrightarrow x^4+y^4=49-2\cdot1=47\) (do xy = 1)

\(x-y=-30\Rightarrow\dfrac{x}{-30}=\dfrac{1}{y}\\ y.z=-42\\ \Rightarrow\dfrac{z}{-42}=\dfrac{1}{y}\\ \Rightarrow\dfrac{x}{-30}=\dfrac{z}{-42}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{-30}=\dfrac{z}{-42}=\dfrac{z-x}{-42-\left(-30\right)}=\dfrac{-12}{-12}=1\)

\(\dfrac{x}{-30}=1\Rightarrow x=-30\\ \dfrac{z}{-42}=1\Rightarrow z=-42\)

\(x.y=-30\Rightarrow-30.y=-30\Rightarrow y=1\)

\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)

\(\Leftrightarrow x^{2010}+x^{2012}-2x^{2011}+y^{2010}+y^{2012}-2y^{2011}=0\)

\(\Leftrightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)

\(x^{2010};y^{2010}>0\Leftrightarrow x=y=1.\Rightarrow x^{2016}+y^{2016}=2\)

\(x^{2010}+y^{2010}=x^{2011}+y^{2011}=x^{2012}+y^{2012}\)

\(\Leftrightarrow x^{2010}+x^{2012}-2x^{2011}+y^{2010}+y^{2012}-2y^{2011}=0\)

\(\Leftrightarrow x^{2010}\left(x^2-2x+1\right)+y^{2010}\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow x^{2010}\left(x-1\right)^2+y^{2010}\left(y-1\right)^2=0\)

\(x^{2010};y^{2010}>0\Leftrightarrow x=y=1.\Rightarrow x^{2016}+y^{2016}=2\)

...................................................................................................................

ta có : 

c = (x+y) * (xy +x+y+2)

c = 3 * ( -5 ) + 3  + 2 

c= -10

Vì x+y+1=0=>x+y=-1

ta có : A=(x+y).(y+1).(x+1)

=(x+y).[y.(x+1)+1(x+1)]

=(-1).(xy+(y+x)+1)

=(-1).(z+(-1)+1)

=(-1).(z+0)=(-1).z=-z 

Vậy A=-z

nho tik nhé

Chọn đáp án A.

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

\(=1^2-3xy\)

=1+3=4

câu b tương tự

\(\text{a) Ta có:}xy=1\Rightarrow\hept{\begin{cases}2xy=2\\-2xy=-2\end{cases}}\)

\(\text{Ta lại có: }x^2+y^2=2\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=2+2=4\\x^2+y^2-2xy=2-2=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=4\\\left(x-y\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x+y=\pm2\\x-y=0\end{cases}}}\)

\(\text{b) Ta có: }x+y=5\)

\(\Rightarrow\left(x+y\right)^2=25\)

\(\Rightarrow x^2+2xy+y^2=25\)

\(\Rightarrow x^2+4+y^2=25\)

\(\Rightarrow x^2+y^2=21\)

\(\text{b) Ta có: }x^2+y^2=21\)

\(\Rightarrow x^2-2xy+y^2=21-2xy\)

\(\Rightarrow\left(x-y\right)^2=21-4\)

\(\Rightarrow\left(x-y\right)^2=17\)

\(\Rightarrow x-y=\pm\sqrt{17}\)