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b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
b, Ta có : \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y-4=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y=4\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y=x+2y\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y-x-2y=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+x-7y=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x^2+2xy+3y^2+1,5xy-4,5xy+x-7y=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\left(x+2y\right)+1,5y\left(x+2y\right)-4,5xy+x-7y=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}4x+6y-4,5xy+x-7y=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x-y-4,5xy=0\\x+2y=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5\left(4-2y\right)-y-4,5y\left(4-2y\right)=0\\x=4-2y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}20-10y-y-18y+9y^2=0\\x=4-2y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}20-29y+9y^2=0\\x=4-2y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}9y^2-9y-20y+20=0\\x=4-2y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(9y-20\right)\left(y-1\right)=0\\x=4-2y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=1\\y=\frac{20}{9}\end{matrix}\right.\\x=4-2y\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=1\\y=\frac{20}{9}\end{matrix}\right.\\\left[{}\begin{matrix}x=4-2.1=4-2=2\\x=4-\frac{2.20}{9}=-\frac{4}{9}\end{matrix}\right.\end{matrix}\right.\)
Vậy phương trình có 2 nghiệm ( x; y ) = \(\left(2;1\right)\), ( x; y ) = \(\left(-\frac{4}{9};\frac{20}{9}\right)\)
a, Ta có : \(\left\{{}\begin{matrix}2x-y=5\\x^2+xy+y^2=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2x-5\\x^2+x\left(2x-5\right)+\left(2x-5\right)^2=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2x-5\\x^2+2x^2-5x+4x^2-20x+25=7\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2x-5\\7x^2-25x+18=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2x-5\\7x^2-7x-18x+18=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2x-5\\\left(7x-18\right)\left(x-1\right)=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2x-5\\\left[{}\begin{matrix}x=1\\x=\frac{18}{7}\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=2.1-5=2-5=-3\\y=2.\left(\frac{18}{7}\right)-5=\frac{1}{7}\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\x=\frac{18}{7}\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ phương trình trên có 2 nghiệm là ( x; y ) = ( 1; -3 ) , ( x; y ) \(=\left(\frac{18}{7};\frac{1}{7}\right)\)
1/ \(\left\{{}\begin{matrix}x^3+y^3=1\left(1\right)\\x^2y+2xy^2+y^3=2\left(2\right)\end{matrix}\right.\)
Lấy (1). 2 - (2) ta được:
\(2x^3+y^3-x^2y-2xy^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(2x-y\right)=0\)
Đến đây dễ rồi nhé ^^
2/ Ta viết lại pt thứ 2 của hệ:
\(y^2-4\left(x+2\right)y+16+16x-5x^2=0\)
\(\Leftrightarrow y^2-4\left(x+2\right)y+4\left(x+2\right)^2-9x^2=0\)
\(\Leftrightarrow\left[y-2\left(x+2\right)\right]^2-\left(3x\right)^2=0\)
\(\Leftrightarrow\left(x+y-4\right)\left(y-5x-4\right)=0\)
Bạn làm tiếp nhé!
3/ Ta viết lại pt thứ nhất của hệ
\(x^2-x\left(2y-3\right)+y^2-3y-4=0\)
\(\Leftrightarrow x^2-x\left(2y-3\right)+\dfrac{4y^2-12y+9}{4}-\dfrac{25}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{2y+3}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-y-4\right)\left(x-y+1\right)=0\)
Bạn làm tiếp được chứ?
4/ Viết lại pt thứ 2 của hệ
\(\left(y+\sqrt{x}\right)^2-\left(y\sqrt{x}\right)^2=0\)
\(\Leftrightarrow\left(y-\sqrt{x}-y\sqrt{x}\right)\left(y-\sqrt{x}+y\sqrt{x}\right)=0\)
a. ĐKXĐ: ..
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}-\sqrt{2\left(x+y\right)}=4\\x+2y+\dfrac{2\sqrt{\left(x+y\right)\left(2x+5y\right)}}{3}=24\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2\left(2x+5y\right)}=a\ge0\\\sqrt{2\left(x+y\right)}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=4\\\dfrac{a^2+b^2}{6}+\dfrac{ab}{3}=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left(a+b\right)^2=144\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=4\\\left[{}\begin{matrix}a+b=12\\a+b=-12\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(a;b\right)=\left(8;4\right)\\\left(a;b\right)=\left(-4;-8\right)\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\left(2x+5y\right)=64\\2\left(x+y\right)=16\end{matrix}\right.\) \(\Leftrightarrow...\)
b.
Thế pt trên xuống dưới:
\(x^4+6y^4=\left(x+2y\right)\left(x^3+3y^3-2xy^2\right)\)
\(\Leftrightarrow2x^3y-2x^2y^2-xy^3=0\)
\(\Leftrightarrow xy\left(2x^2-2xy-y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-\left(1+\sqrt{3}\right)x\\y=\left(-1+\sqrt{3}\right)x\end{matrix}\right.\)
Thế vào pt đầu ...
Đề cho hơi xấu, nếu pt đầu là \(x^3+3y^3-2x^2y=1\) thì đẹp hơn nhiều
pt 1 là \(x^2\) hợp lí hơn chứ
thầy mk cho là x cơ . làm mà hoa hết mắt 2@@