hì

mik lp6

nên k bít

xin lỗi ha

\(PT\Leftrightarrow\left(x^2-4xy+4y^2\right)+4x-8y+4+y^2-16=0\)

\(\Leftrightarrow\left(x-2y\right)^2+4\left(x-2y\right)+4+y^2=16\)

\(\Leftrightarrow\left(x-2y+2\right)^2+y^2=16\)

Vì \(\left(x+2y+2\right)^2+y^2\) là tổng hai số chính phương 

nên \(\left(\left(x+2y+2\right)^2;y^2\right)\in\left\{0;16\right\}\)xét 2 TH là ra

\(4x^2+4y-4xy+5y^2+1=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{4}\\y=-\frac{1}{2}\end{cases}}\)

https://tuhoc365.vn/qa/1-tim-nghiem-nguyen-cua-phuong-trinh-x2-5y2-4xy-4x-4y-3-0-2-tim-tat-ca-cac-so-ng/

\(x^2-4xy+5y^2-16=0\)

\(\Leftrightarrow\left(x-2y\right)^2+y^2=16\)

Ta xét các TH:

TH1: \(\left\{{}\begin{matrix}x-2y=0\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x-2y=4\\y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)

Vậy ta tìm được cặp số (x; y) là \(\left(8;4\right);\left(4;0\right)\)

\(4x^2+5y^2-4xy+4y+1=0\)

\(\Leftrightarrow4x^2-4xy+y^2+4y^2+4y+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-y=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{4}\\y=\frac{-1}{2}\end{matrix}\right.\)

4x² +5y² - 4xy+ 4y+1=0

(=) (4x²-4xy+ y²)+ (4y²+4y+1)=0

(=) ( 2x-y)² + ( 2y+1)²=0

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

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

(=)\(\left\{{}\begin{matrix}x=\frac{-1}{4}\\y=\frac{-1}{2}\end{matrix}\right.\)

\(4x^2+5y^2-4xy+4y+1=0\\ \Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4y^2+4y+1\right)=0\\ \Leftrightarrow\left(2x-y\right)^2+\left(2y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\2y+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{4}\\y=-\frac{1}{2}\end{matrix}\right.\)

c) Đặt \(a=\sqrt{x-4},b=\sqrt{y-4}\)với \(a,b\ge0\)thì pt đã cho trở thành:

\(2\left(a^2+4\right)b+2\left(b^2+4\right)a=\left(a^2+4\right)\left(b^2+4\right)\). chia 2 vế cho \(\left(a^2+4\right)\left(b^2+4\right)\)thì pt trở thành : 

\(\frac{2b}{b^2+4}+\frac{2a}{a^2+4}=1\). Để ý rằng a=0 hoặc b=0 không thỏa mãn pt.

Xét \(a,b>0\). Theo BĐT  AM-GM ta có: \(b^2+4\ge2\sqrt{4b^2}=4b,a^2+4\ge4a\)

\(\Rightarrow VT\le\frac{2a}{4a}+\frac{2b}{4b}=1\), dấu đẳng thức xảy ra khi và chỉ khi \(\hept{\begin{cases}a^2=4\\b^2=4\end{cases}\Leftrightarrow a=b=2\Leftrightarrow x=y=8}\)

Vậy x=8,y=8 là nghiệm của pt