đk: \(x,y\ge0\)

Đặt \(\hept{\begin{cases}\sqrt{x}+\sqrt{y}=a\\\sqrt{xy}=b\end{cases}}\) với \(a,b\ge0\)

\(\Rightarrow x+y=\left(\sqrt{x}+\sqrt{y}\right)^2-2\sqrt{xy}=a^2-2b\)

Khi đó \(HPT\Leftrightarrow\hept{\begin{cases}a+4b=16\\a^2-2b=10\end{cases}}\)

Đến đây thì dễ dàng rồi: \(HPT\Leftrightarrow\hept{\begin{cases}b=\frac{16-a}{4}\\a^2-2b=10\end{cases}}\)

\(\Leftrightarrow a^2-\frac{16-a}{2}=10\)

\(\Leftrightarrow2a^2+a-36=0\)

\(\Leftrightarrow\left(2a^2-8a\right)+\left(9a-36\right)=0\)

\(\Leftrightarrow\left(a-4\right)\left(2a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=4\\a=-\frac{9}{2}\left(ktm\right)\end{cases}}\Rightarrow\hept{\begin{cases}a=4\\b=\frac{16-4}{4}=3\end{cases}}\)

Gọi \(\sqrt{x},\sqrt{y}\) là 2 nghiệm của PT \(t^2-4t+3=0\)

\(\Leftrightarrow\left(t-1\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=1\\t=3\end{cases}}\Leftrightarrow\left(\sqrt{x};\sqrt{y}\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;9\right);\left(9;1\right)\right\}\)

Công sức ngồi gõ hết câu c xong đi chơi về mở máy bấm gửi thì bảo "Lỗi xảy ra khi cập nhật dữ liệu:"

May thiệt ấy nhỉ? :<

understand

me too

a) \(\hept{\begin{cases}\left(x-1\right)\left(2x+y\right)=0\\\left(y+1\right)\left(2y-x\right)=0\end{cases}}\)
\(\cdot x=1\Rightarrow\hept{\begin{cases}0=0\\\left(y+1\right)\left(2y-1\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\y=-1;y=\frac{1}{2}\end{cases}}\)
\(\cdot y=-1\Rightarrow\hept{\begin{cases}\left(x-1\right)\left(2x-1\right)=0\\0=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1;x=\frac{1}{2}\\0=0\end{cases}}\)
\(\cdot x=2y\Rightarrow\hept{\begin{cases}\left(2y-1\right)5y=0\\0=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=0\\y=\frac{1}{2}\Rightarrow x=1\end{cases}}\)
\(y=-2x\Rightarrow\hept{\begin{cases}0=0\\\left(1-2x\right)5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\Rightarrow y=-1\\x=0\Rightarrow y=0\end{cases}}\)

b) \(\hept{\begin{cases}x+y=\frac{21}{8}\\\frac{x}{y}+\frac{y}{x}=\frac{37}{6}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\\left(\frac{21}{8}-y\right)^2+y^2=\frac{37}{6}y\left(\frac{21}{8}-y\right)\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\2y^2-\frac{21}{4}y+\frac{441}{64}=-\frac{37}{6}y^2+\frac{259}{16}y\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{21}{8}-y\\1568y^2-4116y+1323=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{8}\\y=\frac{9}{4}\end{cases}}hay\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{3}{8}\end{cases}}\)

c) \(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\\\frac{2}{xy}-\frac{1}{z^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x-y\right)^2=-4x^2y^2+2xy\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}8x^2y^2-4x^2y-4xy^2+x^2+y^2-2xy+2xy=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4x^2y^2-4x^2y+x^2+4x^2y^2-4xy^2+y^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(2xy-x\right)^2+\left(2xy-y\right)^2=0\\\frac{1}{z^2}=\frac{2}{xy}-4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=\frac{-1}{2}\end{cases}}\)
d) \(\hept{\begin{cases}xy+x+y=71\\x^2y+xy^2=880\end{cases}}\). Đặt \(\hept{\begin{cases}x+y=S\\xy=P\end{cases}}\), ta có: \(\hept{\begin{cases}S+P=71\\SP=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P\left(71-P\right)=880\end{cases}}\Leftrightarrow\hept{\begin{cases}S=71-P\\P^2-71P+880=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S=16\\P=55\end{cases}}hay\hept{\begin{cases}S=55\\P=16\end{cases}}\)
\(\cdot\hept{\begin{cases}S=16\\P=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=16\\xy=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y\left(16-y\right)=55\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16-y\\y^2-16y+55=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=5\\y=11\end{cases}}hay\hept{\begin{cases}x=11\\y=5\end{cases}}\)

\(\cdot\hept{\begin{cases}S=55\\P=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=55\\xy=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y\left(55-y\right)=16\end{cases}}\Leftrightarrow\hept{\begin{cases}x=55-y\\y^2-55y+16=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{55-3\sqrt{329}}{2}\\y=\frac{55+3\sqrt{329}}{2}\end{cases}}hay\hept{\begin{cases}x=\frac{55+3\sqrt{329}}{2}\\y=\frac{55-3\sqrt{329}}{2}\end{cases}}\)

e) \(\hept{\begin{cases}x\sqrt{y}+y\sqrt{x}=12\\x\sqrt{x}+y\sqrt{y}=28\end{cases}}\). Đặt \(\hept{\begin{cases}S=\sqrt{x}+\sqrt{y}\\P=\sqrt{xy}\end{cases}}\), ta có \(\hept{\begin{cases}SP=12\\P\left(S^2-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\P\left(\frac{144}{P^2}-2P\right)=28\end{cases}}\Leftrightarrow\hept{\begin{cases}S=\frac{12}{P}\\2P^4+28P^2-144P=0\end{cases}}\)
Tự làm tiếp nhá! Đuối lắm luôn

Đặt:

\(\hept{\begin{cases}\sqrt{x+1}=2+t\\\sqrt{y+1}=2-t\end{cases}\Rightarrow t=0}\)

\(ĐK:\hept{\begin{cases}x\ge-1\\y\ge-1\\xy\ge0\end{cases}}\)

Hệ tương đương \(\hept{\begin{cases}x+y-\sqrt{xy}=3\\2\sqrt{xy+x+y+1}=14-\left(x+y\right)\end{cases}}\)

Đặt S=x+y;P=\(\sqrt{xy}\)(\(P\ge0\))

\(\Rightarrow\hept{\begin{cases}S-P=3\left(3\right)\\2\sqrt{P^2+S+1}=14-S\left(4\right)\end{cases}}\)

Thay (3) \(S=3+P\)vào (4) ta được:

\(2\sqrt{P^2+P+4}=11-P\Leftrightarrow\hept{\begin{cases}P\le11\\3P^2+26P-105=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}P\le11\\\orbr{\begin{cases}P=3\left(n\right)\\P=\frac{-35}{3}\left(L\right)\end{cases}}\end{cases}}\)đến đây tự xét

\(\Rightarrow P=3\Rightarrow S=3\Rightarrow\hept{\begin{cases}x+y=6\\xy=9\end{cases}}\Rightarrow x=y=3\)

\(\Leftrightarrow\hept{\begin{cases}P\le11\\\orbr{\begin{cases}P=3\left(n\right)\\P=\frac{-35}{3}\left(L\right)\end{cases}}\end{cases}}\)

pt(1)<=>\(\left(\sqrt{x-1}+\sqrt{y}\right)^2=4\)