Question

GIAI CÁC PHƯƠNG TRINH SAU

A) \(X^4+X^3+2X^2+X+1=0\)

B) \(\frac{X+106}{3}+\frac{X+116}{4}+\frac{X+130}{5}+\frac{X+148}{6}=20\)

Answer 1

a) \(\Leftrightarrow\left(x^2+1\right)\left(x^2+x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2+x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\end{cases}}}\Rightarrow\)Vô lí

b)\(\Leftrightarrow\left(\frac{x+106}{3}-2\right)+\left(\frac{x+116}{4}-4\right)+\left(\frac{x+130}{5}-6\right)+\left(\frac{x-148}{6}-8\right)=0\Leftrightarrow\frac{x+100}{3}+\frac{x+100}{4}+\frac{x+100}{5}+\frac{x+100}{6}=0\Leftrightarrow\left(x+100\right)\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)=0\Leftrightarrow x+100=0\Leftrightarrow x=-100\)