Question

giải pt

\(a,x^2-2005x-2006=0\)

\(b,\left|x-2\right|+\left|x-3\right|+\left|2x-8\right|=9\)

 

Answer 1

\(a,x^2-2005x-2006=0\)

\(\Leftrightarrow x^2+x-2006x-2006=0\)

\(\Leftrightarrow x\cdot\left(x+1\right)-2006\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2006\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2006=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2006\end{cases}}}\)

Answer 2

a) \(x^2-2005x-2006=0\)

Ta có: \(2005^2+4.2006=4028049\)

pt có 2 nghiệm:

\(x_1=\frac{2005+\sqrt{4028049}}{2}\);\(x_2=\frac{2005-\sqrt{4028049}}{2}\)

Vậy tập nghiệm của pt là \(S=\left\{\frac{2005+\sqrt{4028049}}{2};\frac{2005-\sqrt{4028049}}{2}\right\}\)

Answer 3

a) \(x^2-2005x-2006=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2006\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-2006=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2006\end{cases}}\)

Answer 4

a, \(x^2-2005x-2006=0\)

Ta có : \(\Delta=b^2-4ac=\left(-2005\right)^2-4.1.\left(-2006\right)=4028049\)

Vì \(\Delta>0\)nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-\left(-2005\right)-\sqrt{4028049}}{2.1}=\frac{2005-2007}{2}=-\frac{2}{2}=-1\)

\(x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-\left(-2005\right)+\sqrt{4028049}}{2.1}=\frac{2005+2007}{2}=2006\)