các bn ơi giúp mình với

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

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

Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)

\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

2) \(\sqrt{x^2-5x+4}+2\sqrt{x+5}=2\sqrt{x-4}+\sqrt{x^2+4x-5}\)

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Bài 1:

ĐKĐB suy ra $x(x+1)+y(y+1)=3x^2+xy-4x+2y+2$

$\Leftrightarrow 2x^2+x(y-5)+(y-y^2+2)=0$

Coi đây là PT bậc 2 ẩn $x$

$\Delta=(y-5)^2-4(y-y^2+2)=(3y-3)^2$Do đó:

$x=\frac{y+1}{2}$ hoặc $x=2-y$. Thay vào một trong 2 phương trình ban đầu ta thu được:

$(x,y)=(\frac{-4}{5}, \frac{-13}{5}); (1,1)$

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x^2+xy-4x+2y=2\\x\left(x+1\right)+y\left(y+1\right)=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3x^2+xy-4x+2y=2\left(1\right)\\x^2+x+y^2+y=4\left(2\right)\end{matrix}\right.\)

Trừ vế theo vế của (1) cho (2)\(\Leftrightarrow3x^2+xy-4x+2y-x^2-x-y^2-y=-2\Leftrightarrow2x^2-y^2-5x+y+xy+2=0\Leftrightarrow\left(2x-y-1\right)\left(x+y-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}y=2x-1\\y=2-x\end{matrix}\right.\)

TH1: y=2x-1

thay vào (2)\(\Leftrightarrow x^2+x+\left(2x-1\right)^2+2x-1=4\Leftrightarrow x^2+x+4x^2-4x+1+2x-5=0\Leftrightarrow5x^2-x-4=0\Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-\frac{4}{5}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}y=1\\y=-\frac{8}{5}\end{matrix}\right.\)

TH2: y=2-x

Thay vào (2)\(\Leftrightarrow x^2+x+\left(2-x\right)^2+2-x=4\Leftrightarrow x^2+x+4-4x+x^2+2-x=4\Leftrightarrow2x^2-4x+2=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)\(\Leftrightarrow y=1\)

Vậy (x;y)={(1;1);(\(-\frac{4}{5};-\frac{8}{5}\))}

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)