\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right).\)

\(\Rightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+4\left(x^2-5x-2012\right)^2=0\)

\(\Leftrightarrow\left[\left(2x^2+x-2013\right)-2\left(x^2-5x-2012\right)\right]^2=0\)(Hằng đẳng thức)

\(\Leftrightarrow2x^2+x-2013-2x^2+10x+4024=0\)

\(\Leftrightarrow11x=-2011\)

\(\Leftrightarrow x=\frac{-2011}{11}\)

quy đồng nha bạn

\(\left(x^5+x^3+x^2+1\right):\left(x^3+1\right)\)

\(=\left[x^2\left(x^3+1\right)+1\left(x^3+1\right)\right]:\left(x^3+1\right)\)

\(=\left(x^3+1\right)\left(x^2+1\right):\left(x^3+1\right)\)

\(=x^2+1\)

Gọi số cần tìm là \(\overline{abc}\)

Theo bài ra có:

\(\overline{9abc}+\overline{abc9}=10384\Rightarrow9000+\overline{abc}+10.\overline{abc}+9=10384\)

\(\Rightarrow11.\overline{abc}=1375\Rightarrow\overline{abc}=125\)

     \(x^3-7x^2-13x+91=0\)

\(\Rightarrow x^2\left(x-7\right)-13\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x^2-13\right)=0\)

\(\Rightarrow\left(x-7\right)\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

Tìm được \(x\in\left\{7;\sqrt{13};-\sqrt{13}\right\}\)

a) ĐKXĐ : x khác 2/5

\(\frac{2x+3}{2-5x}\le0\)

\(\Leftrightarrow2x+3\le2-5x\)

\(\Leftrightarrow7x\le-1\)

\(\Leftrightarrow x\le\frac{-1}{7}\left(\text{thỏa mãn}\right)\)

b) \(\left|5x+3\right|=\left|x+2\right|\)

\(\Rightarrow\orbr{\begin{cases}5x+3=x+2\\5x+3=-x-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x=-1\\6x=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{6}\end{cases}}\)

P.s: cái này chưa học có j sai góp ý hộ nha ^^

ĐKXĐ: \(x\ne\frac{2}{5}\)

\(\frac{2x+3}{x-5x}\le0\)

Xét 2 trường hợp

TH1: \(\hept{\begin{cases}2x+3\ge0\\2-5x< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge-\frac{3}{2}\\x>\frac{2}{5}\end{cases}}}\Leftrightarrow x>\frac{2}{5}\)

TH2: \(\hept{\begin{cases}2x+3\le0\\2-5x>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le-\frac{3}{2}\\x< \frac{2}{5}\end{cases}}}\Leftrightarrow x\le-\frac{3}{2}\)

Vậy \(\orbr{\begin{cases}x>\frac{2}{5}\\x\le-\frac{3}{2}\end{cases}}\)

P/S: chưa học => trình bày thiếu sót ( sai ) => sửa hộ~

\(ĐKXĐ:\hept{\begin{cases}x-3\ne0\\3x^2-6x-9\ne0\\3x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne3\\3\left(x^2-2x-3\right)\ne0\\3\left(x+1\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

\(M=\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right).\frac{x^2-2x-3}{x^2+x+2}\)

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

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

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

\(=\frac{3x^2+3x-6}{3\left(x+1\right)\left(x-3\right)}.\frac{x^2-2x-3}{x^2+x+2}\)

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

\(=\frac{x^2+x-2}{x^2+x+2}=1-\frac{4}{x^2+x+2}\)

b,\(\text{Với }x\ne-1\text{ và }x\ne3\text{ ta có:}\)

\(\text{Để }M=1-\frac{4}{x^2+x+2}< 1\)

\(\Leftrightarrow-\frac{4}{x^2+x+2}< 0\)

\(\Leftrightarrow\frac{4}{x^2+x+2}>0\)

\(\Leftrightarrow4>0\left(\text{hiển nhiên}\right)\)

Vậy ... đpcm