1.

\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)

\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)

\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)

\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)

\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)

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

ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

\(\Leftrightarrow a^2-3a+2+ab-b=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

mong mọi người giải giúp em vs 

bài 1:

b) đề như vầy hả :\(\left\{{}\begin{matrix}\left(x^2-1\right)y+\left(y^2-1\right)x=2\left(xy-1\right)\left(1\right)\\4x^2+y^2+2x-y-6=0\left(2\right)\end{matrix}\right.\)

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

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

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

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

*xét \(xy=1\Leftrightarrow x=\dfrac{1}{y}\)thế vào Pt (2):\(\dfrac{4}{y^2}+y^2+\dfrac{2}{y}-y-6=0\)

\(\Leftrightarrow\dfrac{4+2y}{y^2}+\left(y+2\right)\left(y-3\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(\dfrac{2}{y^2}+y-3\right)=0\)

\(\Leftrightarrow\left(y+2\right)\left(y^3-3y^2+2\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(y-1\right)\left(y^2-2y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=1\\y=1-\sqrt{3}\\y=1+\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\\x=-\dfrac{1+\sqrt{3}}{2}\\x=\dfrac{-1+\sqrt{3}}{2}\end{matrix}\right.\)

* xét x+y=2(tương tự thay x=2-y vào Pt (2))

câu 2:

ta đưa về PT ẩn x:\(x^2-x\left(y+1\right)+y^2-y-2=0\)

Pt phải có nghiệm ,xét \(\Delta=\left(y+1\right)^2-4\left(y^2-y-2\right)\ge0\)

\(\Leftrightarrow y^2-2y-3\le0\Leftrightarrow\left(y+1\right)\left(y-3\right)\le0\)

\(\Leftrightarrow-1\le y\le3\).

vì x,y thuộc Z ,lần luợt thay các giá trị của y vừa tìm được vào PT ban đầu ta được các cặp (x,y) t/m là (0;-1);(-1;0);(2;0);(0;2);(3;2);(2;3)

bài 3:

DKXĐ:\(\left\{{}\begin{matrix}2x^2-x\ge0\\2x-x^2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le0\end{matrix}\right.\\0\le x\le2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}\le x\le2\end{matrix}\right.\)

bình phương , self study

Ptrình này vô nghiệm bn ạ