$1) x^5+2x^4+3x^3+3x^2+2x+1=0$
\Leftrightarrow$x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0$
\Leftrightarrow$x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0$
\Leftrightarrow$(x+1)(x^4+x^3+2x^2+x+1)=0$
\Leftrightarrow$(x+1)[(x^4+x^3+x^2)+(x^2+x+1)]=0$
\Leftrightarrow$(x+1)(x^2+x+1)(x^2+1)=0$
\Leftrightarrow $x+1=0$\Leftrightarrow$x=1$
$2)$ Chia cả 2 vế cho $x^2$ không âm, ta được:
$x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0$
\Leftrightarrow $(x^2+2+\dfrac{1}{x^2})-3(x+\dfrac{1}{x})+2=0$
\Leftrightarrow $((x+\dfrac{1}{x})^2-3(x+\dfrac{1}{x})+2=0$
Đặt $y=x+\dfrac{1}{x}$ ta được:
$y^2-3y+2=0$ \Leftrightarrow $(y^2-y)-(2y-2)=0$
\Leftrightarrow $y(y-1)-2(y-1)=0$ \Leftrightarrow $(y-2)(y-1)=0$
\Rightarrow $y=2 hoặc y=1$
Với $y=2$\Rightarrow $x+\dfrac{1}{x}=2$
\Leftrightarrow $x^2+1=2x$\Leftrightarrow $x^2-2x+1=0$
\Leftrightarrow $(x-1)^2=0$\Leftrightarrow$x=1$
Với $y=1$\Rightarrow $x+\dfrac{1}{x}=1$
\Leftrightarrow $x^2+1=x$\Leftrightarrow $x^2-x+1=0$ (vô nghiệm)
Vậy, phương trình có nghiệm x=1
3) Chia cả 2 vế cho số $x^2$ không âm, ta có:
$6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}=0$
\Leftrightarrow$(6x+12+\dfrac{6}{x^2})+(5x+\dfrac{5}{x})-50=0$
\Leftrightarrow$6(x+\dfrac{1}{x})^2)+5(x+\dfrac{1}{x})-50=0$
Đặt $y=x+\dfrac{1}{x}$ ta được:
$6y^2+5y-50=0$
\Leftrightarrow$y^2+\dfrac{5y}{6}-\dfrac{25}{3}=0$
\Leftrightarrow$(y^2+2.y.\dfrac{5}{12}+\dfrac{25}{144})-\dfrac{1225}{144}=0$
\Leftrightarrow$(y+\dfrac{5}{12})^2-(\dfrac{35}{12})^2=0$
\Leftrightarrow$(y+\dfrac{5}{12}+\dfrac{35}{12})(y+\dfrac{5}{12}-\dfrac{35}{12}=0$
\Leftrightarrow$(y+\dfrac{10}{3})(y-\dfrac{5}{2})=0$
\Leftrightarrow $y=\dfrac{-10}{3} hoặc y=\dfrac{5}{2}$
Với $y=\dfrac{-10}{3}$\Rightarrow $x+\dfrac{1}{x}=\dfrac{-10}{3}$
\Leftrightarrow $x^2+1=\dfrac{-10}{3}x$
\Leftrightarrow $x^2+\dfrac{10}{3}x+1=0$
\Leftrightarrow $x^2+2.x.\dfrac{5}{3}+\dfrac{25}{9})-\dfrac{16}{9}=0$
\Leftrightarrow $(x+\dfrac{5}{3})^2 - (\dfrac{4}{3})^2=0$
\Leftrightarrow $(x+\dfrac{5}{3}+\dfrac{4}{3})(x+\dfrac{5}{3}-\dfrac{4}{3}=0$
\Leftrightarrow$(x+3)(x-\dfrac{1}{3})=0$
\Leftrightarrow $x=-3 hoặc x=\dfrac{-1}{3}$
Với $y=\dfrac{5}{2}$\Rightarrow $x+\dfrac{1}{x}=\dfrac{5}{2}$
\Leftrightarrow $x^2+1=\dfrac{5}{2}x$
\Leftrightarrow $x^2-\dfrac{5}{2}x+1=0$
\Leftrightarrow $x^2-2.x.\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{9}{16}=0$
\Leftrightarrow $(x-\dfrac{5}{4})^2-(\dfrac{3}{4})^2=0$
\Leftrightarrow$(x-\dfrac{5}{4}-\dfrac{3}{4})(x-\dfrac{5}{4}+\dfrac{3}{4}=0$
\Leftrightarrow $(x-2)(x-\dfrac{1}{2})=0$
\Leftrightarrow $x=2 hoặc x=\dfrac{1}{2}$
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