......................?

mik ko biết

mong bn thông cảm 

nha ................

a) x²+2y²+2xy-2y+1=0

\(\Leftrightarrow\)(x²+2xy+y²)+(y²-2y+1)=0

\(\Leftrightarrow\)(x+y)²+(y-1)²=0

\(\Leftrightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

Vậy x=-1, y=1

a/ \(x^2+2y^2+2xy-2y+1=0\)

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

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

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

<=> \(\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-y\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

b/ \(x^2+2y^2+2xy-2x+2=0\)

<=> \(\left(x^2+2xy+y^2\right)+\left(2y-2x+2\right)=0\)

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

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

<=> \(\hept{\begin{cases}x+y=0\\y-x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x+y=0\\y-x=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x+y=0\left(1\right)\\x-y=1\left(2\right)\end{cases}}\)

Trừ (1) và (2)

=> \(2y=-1\)

<=> \(y=-\frac{1}{2}\)

<=> \(x=\frac{1}{2}\)(vì \(x+y=0\)<=> \(x=-y\))

bạn viết lại đi sai đề hay sao ý

vãi 4 năm mà ko mtj thằng nào rep đã thế còn gấp

\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)

Ta có: \(\left(ax+by\right)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(\Leftrightarrow a^2x^2+2abxy+b^2y^2=a^2x^2+a^2y^2+x^2b^2+b^2y^2\)

\(\Leftrightarrow2abxy=a^2y^2+x^2b^2\)

\(\Leftrightarrow\left(ay-xb\right)^2=0\)

\(\Leftrightarrow ay=xb\)

hay \(\dfrac{a}{x}=\dfrac{b}{y}\)

sssss