1/HPT\(\Leftrightarrow\hept{\begin{cases}x^2+y^2=6-\left(x+y\right)=3\\\left(x+y\right)^2=9\end{cases}}\Rightarrow2xy=\left(x+y\right)^2-\left(x^2+y^2\right)=9-3=6\Rightarrow xy=3\)

Kết hợp đề bài có được: \(\hept{\begin{cases}x+y=3\\xy=3\end{cases}}\). Dùng hệ thức Viet đảo là xong.

ĐKXĐ: \(-1\le x,y\le1\)

\(\hept{\begin{cases}\sqrt{1-x}+\sqrt{1-y}=\sqrt{2}\left(3\right)\\\sqrt{1+x}+\sqrt{1+y}=\sqrt{6}\end{cases}}\)

<=> \(\hept{\begin{cases}1-x+1-y+2\sqrt{\left(1-x\right)\left(1-y\right)}=2\\1+x+1+y+2\sqrt{\left(1+x\right)\left(1+y\right)}=6\end{cases}}\)

<=> \(\hept{\begin{cases}2\sqrt{1-x-y+xy}=x+y\left(1\right)\\2\sqrt{xy+x+y+1}=4-x-y\left(2\right)\end{cases}}\)

Từ (1) và (2) cộng vế theo vế:

\(2\sqrt{xy-x-y+1}+2\sqrt{xy+x+y+1}=4\)

<=>\(\sqrt{xy-x-y+1}+\sqrt{xy+x+y+1}=2\)(đk: - 1 < = x,y < = 1)

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

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

<=> \(\sqrt{x^2y^2-x^2-y^2+1}=1-xy\) (đk: xy < = 1)

<=> \(x^2y^2-x^2-y^2+1=x^2y^2-2xy+1\)

<=> \(x^2+y^2-2xy=0\)

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

Thay x = y vào pt (3) => \(2\sqrt{1-x}=\sqrt{2}\) (đk: -1 < = x < = 1)

<=> 4(1 - x) = 2 <=> 4 - 4x = 2 <=> 2 = 4x <=> x = 1/2

=> x = y = 1/2 (tm)

giúp câu 2

\(4\left(x^2+xy+y^2\right)=3\left(x+y\right)^2+\left(x-y\right)^2.\)
Đặt (x+y)=a ; (x-y)=b là ok nhé !!!!

em ko biết làm :">

\(\hept{\begin{cases}2\sqrt{x-2}+3\sqrt{y-3}=14\\\sqrt{x-2}+\sqrt{y-3}=5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2\sqrt{x-2}+3\sqrt{y-3}=14\\2\sqrt{x-2}+2\sqrt{y-3}=10\end{cases}}\)

\(\Leftrightarrow2\sqrt{x-2}+3\sqrt{y-3}-2\sqrt{x-2}-2\sqrt{y-3}=14-10\)

\(\Leftrightarrow\sqrt{y-3}=4\Leftrightarrow y-3=16\Leftrightarrow y=19\)

\(\Rightarrow\sqrt{x-2}+\sqrt{19-3}=5\)

\(\Leftrightarrow x-2=\left(5-4\right)^2\Leftrightarrow x-2=1\Leftrightarrow x=3\)

\(\hept{\begin{cases}3\left(x+1\right)-y=6-2y\\2x-y=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x+3-y=6-2y\\2x-y=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x+y=3\\2x-y=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}6x+2y=6\\6x-3y=21\end{cases}}\)

\(\Leftrightarrow6x+2y-6x+3y=6-21\)

\(\Leftrightarrow5y=-15\Leftrightarrow y=-3\)

\(\Rightarrow x=\frac{7-3}{2}=2\)

\(\hept{\begin{cases}\sqrt{2}x+\left(\sqrt{2}+1\right)y=3\\x+\sqrt{2}y=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{2}x+\sqrt{2}y+y=3\\\sqrt{2}x+y=2\sqrt{2}\end{cases}}\)

\(\Leftrightarrow\sqrt{2}x+\sqrt{2y}+y-\sqrt{2}x-y=3-2\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}y=3-2\sqrt{2}\)

\(\Rightarrow y=\frac{3-2\sqrt{2}}{\sqrt{2}}=\frac{3}{\sqrt{2}}-2\)( em ko biết rút gọn sao :vv)

\(\Rightarrow x+\sqrt{2}\left(\frac{3}{\sqrt{2}}-2\right)=2\)

\(\Leftrightarrow x+3-2\sqrt{2}=2\)

\(\Leftrightarrow x=2\sqrt{2}-1\)

\(ĐKXĐ:x;y\ge0\)

\(\hept{\begin{cases}\sqrt{x}+\sqrt{y}=4\left(1\right)\\\sqrt{x+5}+\sqrt{y+5}=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+2\sqrt{xy}+y=16\\x+5+2\sqrt{\left(x+5\right)\left(y+5\right)}+y+5=36\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=16-2\sqrt{xy}\\x+y=26-2\sqrt{\left(x+5\right)\left(y+5\right)}\end{cases}}\)

\(\Rightarrow16-2\sqrt{xy}=26-2\sqrt{\left(x+5\right)\left(y+5\right)}\)

\(\Leftrightarrow-2\sqrt{xy}=10-2\sqrt{\left(x+5\right)\left(y+5\right)}\)

\(\Leftrightarrow\sqrt{xy}=\sqrt{\left(x+5\right)\left(y+5\right)}-5\)

\(\Leftrightarrow\sqrt{xy}+5=\sqrt{\left(x+5\right)\left(y+5\right)}\)

\(\Leftrightarrow xy+10\sqrt{xy}+25=xy+5\left(x+y\right)+25\)

\(\Leftrightarrow2\sqrt{xy}=x+y\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2=0\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y}\)

\(\Leftrightarrow x=y\)

Thế vô pt (1) được \(2\sqrt{x}=4\)

                          \(\Leftrightarrow\sqrt{x}=2\)

                         \(\Leftrightarrow x=y=4\)(Thỏa mãn ĐKXĐ)

Vậy hệ pt có nghiệm duy nhất \(\hept{\begin{cases}x=4\\y=4\end{cases}}\)

ai k mình k lại nhưng phải lên điểm mình tích gấp đôi

ĐKXĐ: \(x\ge0;x+y-4\ge0\) 

\(PT_{\left(1\right)}\Leftrightarrow\left(\sqrt{x^2+3}-2\right)+\left(\sqrt{x}-1\right)+\left(\sqrt{x+3}-2\right)=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{x-1}{\sqrt{x}+1}+\frac{x-1}{\sqrt{x+3}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{\left(x+1\right)}{\sqrt{x^2+3}+2}+\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x+3}+2}\right]=0\)

Cái ngoặc to vô nghiệm. Vậy x = 1.

Thay xuống PT (2) \(\Leftrightarrow3+\sqrt{y-3}=5\left(Đ\text{K:}y\ge3\right)\Leftrightarrow\sqrt{y-3}=2\Leftrightarrow y=7\)

Vậy x = 1; y = 7

P/s: Em ko chắc.