Hình bạn đánh lộn số \(3\) thành \(2\) hả?
\(a,\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) \(Đkxđ:x\ne0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+x1\right)}{\left(x^2+x+1\right)\left(x^2+1-x\right)}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{x^3+1-\left(x^3-1\right)}{\left(x^2+1\right)^2-x^2}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{2}{x^4+2x^2+1-x^2}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{2}{x^4+x^2+1}-\frac{3}{x\left(x^4+x^2+1\right)}=0\)
\(\Leftrightarrow\frac{2x-3}{x\left(x^4+x^2+1\right)}=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\left(tmđk\right)\)
\(b,x^3+2x^2+3x+2=y^3\)
Ta có: \(2x^2+3x+2=2\left(x+\frac{3}{4}\right)^2+\frac{7}{8}>0\forall x\Rightarrow y>x\)
Và: \(4x^2+9x+6=\left(2+\frac{9}{4}\right)^2+\frac{15}{16}>0\)
Nên: \(y^3=\left(x+2\right)^3-\left(4x^2+9x+6\right)< \left(x+2\right)^2\)
\(\Leftrightarrow y< x+2\)
\(\Rightarrow x< y< x+2\)
\(\Rightarrow y=x+1\)
\(x^3+2x^2+3x+2=\left(x+1\right)^3\)
\(\Rightarrow x^2-1=0\)
\(\Leftrightarrow x=\pm1\)
Vậy \(\left(x,y\right)=\left\{\left(1;2\right);\left(-1;0\right)\right\}\)