\(c,\left(x+2\right)^2+\left(x-1\right)\left(x+3\right)=2\left(x-4\right)\left(x+4\right)\\ \Leftrightarrow x^2+4x+4+x^2+2x-3=2\left(x^2-16\right)\\ \Leftrightarrow2x^2+6x+1=2x^2-32\\ \Leftrightarrow6x=-33\\ \Leftrightarrow x=-\dfrac{11}{2}\)
c, ĐKXĐ:\(x\ne-1\)
\(\dfrac{1}{x+1}+\dfrac{2x^2+1}{x^3+1}+\dfrac{2x^3-2x^2}{x^2-x+1}=2x\\ \Leftrightarrow\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x^2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-x+1+2x^2+1+2x^2\left(x^2-1\right)-2x\left(x^3+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=0\)
\(\Rightarrow x^2-x+1+2x^2+1+2x^4-2x^2-2x^4-2x=0\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
a) \(\left(x+2\right)^2+\left(x-1\right)\left(x+3\right)=2.\left(x+4\right)\left(x-4\right)-3\)
\(\Rightarrow x^2+4x+4+x^3+2x-3=2x^2-32-3\)
\(\Rightarrow6x=-36\)
\(\Rightarrow x=-6\)
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