2) \(\dfrac{x}{2}\)-\(\dfrac{x}{10}\)<\(\dfrac{1}{2}-\dfrac{1}{3}\)

<=>\(\dfrac{x}{2}\)-\(\dfrac{x}{10}\)<\(\dfrac{1}{6}\)

=>15x-3x<5

<=>12x<5

<=>x<\(\dfrac{5}{12}\)

=> S={x|x<\(\dfrac{5}{12}\)}

\(\Leftrightarrow\left\{{}\begin{matrix}x^3+8y^3-4xy^2=1\\2x+y=2x^4+8y^4\end{matrix}\right.\)

Nhân vế với vế:

\(\left(2x+y\right)\left(x^3+8y^3-4xy^2\right)=2x^4+8y^4\)

\(\Leftrightarrow12xy^3-8x^2y^2+x^3y=0\)

\(\Leftrightarrow xy\left(12y^2-2xy+x^2\right)=0\)

\(\Leftrightarrow xy=0\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=...\\y=0\Rightarrow x=...\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\y\le\frac{16}{3}\end{matrix}\right.\)

\(2x^2-\left(3y-6\right)x+y^2-8y-20=0\)

\(\Delta=\left(3y-6\right)^2-8\left(y^2-8y-20\right)=y^2+28y+196=\left(y+14\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{3y-6+y+14}{4}=y+2\\x=\frac{3y-6-y-14}{4}=\frac{y-10}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=x-2\\y=2x+10\end{matrix}\right.\)

- Với \(y=2x+10\ge-2.2+10=6>\frac{16}{3}\) ko phù hợp ĐKXĐ (loại)

- Với \(y=x-2\)

\(4\sqrt{x+2}+\sqrt{22-3x}=x^2+8\)

\(\Leftrightarrow x^2+8-4\sqrt{x+2}-\sqrt{22-3x}=0\)

\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(x+4-3\sqrt{x+2}\right)+\frac{1}{3}\left(14-x-3\sqrt{22-3x}\right)=0\)

\(\Leftrightarrow x^2-x-2+\frac{4}{3}\left(\frac{x^2-x-2}{x+4+3\sqrt{x+2}}\right)+\frac{1}{3}\left(\frac{x^2-x-2}{14-x+3\sqrt{22-3x}}\right)=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(....\right)=0\) (ngoặc phía sau luôn dương)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\Rightarrow y=-3\\x=2\Rightarrow y=0\end{matrix}\right.\)

a: x^2-2x+y^2-8y+17=0

=>x^2-2x+1+y^2-8y+16=0

=>(x-1)^2+(y-4)^2=0

=>x=1 và y=4

b: Sửa đề: 4x^2-4xy+y^2+y^2+4y+4=0

=>(2x-y)^2+(y+2)^2=0

=>y=-2 và x=-1

\(x^2-y^2+2x-4y-10=0\)

\(\Leftrightarrow x^2+2x+1-y^2-4y-4-7=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Leftrightarrow\left(x-y+1-2\right)\left(x+y+1+2\right)=7\)

\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)

Xét bảng tìm x; y là xong

a) \(2x^2+20x+52=0\Rightarrow x^2+10x+26=0\Rightarrow\left(x+5\right)^2+1=0\)

\(\Rightarrow\) vô nghiệm

b) ĐK: \(x\ne1;-1\)

\(\dfrac{2x-19}{5x^2-5}-\dfrac{17}{x-1}=\dfrac{8}{1-x}\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{17}{x-1}+\dfrac{8}{x-1}=0\)

\(\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{9}{x-1}=0\Rightarrow\dfrac{2x-19-45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}=0\)

\(\Rightarrow-43x-64=0\Rightarrow x=-\dfrac{64}{43}\)

a)  Ta có: \(\Delta'=100-104=-4< 0\)

Vậy phương trình vô nghiệm.

b) ĐKXĐ: \(x\ne1;x\ne-1\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x^2-1\right)}=\dfrac{17}{x-1}-\dfrac{8}{x-1}\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{9}{x-1}\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow2x-19=45x+45\)

\(\Leftrightarrow43x=-64\)

\(\Leftrightarrow x=-\dfrac{64}{43}\)(TM)

Vậy phương trình có nghiệm là: \(x=-\dfrac{64}{43}\)