Ói , hoa mắt chóng mặt nhức đầu ,

đa thức lớp 5 hả bạm

đa thức lớp 5 à

mik ko bít

I don't now

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1:

ĐKXĐ: y>=0; x>=1

\(\begin{cases} xy + x + y = x^2 - 2y^2 \quad (1) \\ x\sqrt{2y} - y\sqrt{x-1} = 2x - 2y \quad (2) \end{cases}\)

(1)=>\(x^2 - xy - 2y^2 - x - y = 0\)

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

=>(x-2y)(x+y)-(x+y)=0

=>(x+y)(x-2y-1)=0

mà x+y>=0

nên x-2y-1=0

=>x=2y+1

Thay x=2y+1 vào (2), ta được:

\((2y + 1)\sqrt{2y} - y\sqrt{2y} = 2(2y + 1) - 2y\)

=>\(y\sqrt{2y} + \sqrt{2y} = 2y + 2\)

=>\(\sqrt{2y}(y + 1) = 2(y + 1)\)

=>\(\sqrt{2y}=2\)

=>2y=4

=>y=2(nhận)

=>x=2y+1=5(nhận)

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)