Question

Không tính \(\Delta \), hãy giải các phương trình:

a)   \({x^2} - 3x + 2 = 0\)

b)  \( - 3{x^2} + 5x + 8 = 0\)

c)   \(\frac{1}{3}{x^2} + \frac{1}{6}x - \frac{1}{2} = 0\)

Answer 1

a) 
\(x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b)
\(-3x^2+5x+8=0\\ \Leftrightarrow-\left(3x^2+3x\right)+\left(8x+8\right)=0\\ \Leftrightarrow-3x\left(x+1\right)+8\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(8-3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\8-3x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

c)
\(\dfrac{1}{3}x^2+\dfrac{1}{6}x-\dfrac{1}{2}=0\\ \Leftrightarrow\dfrac{2x^2}{6}+\dfrac{x}{6}-\dfrac{3}{6}=0\\ \Leftrightarrow\dfrac{2x^2+x-3}{6}=0\\ \Leftrightarrow2x^2+x-3=0\\ \Leftrightarrow\left(2x^2-2x\right)+\left(3x-3\right)=0\\ \Leftrightarrow2x\left(x-1\right)+3\left(x-1\right)=0\\ \Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-3\\x=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)

Vậy...
 

Answer 2

a) Ta có: \(a + b + c = 1 - 3 + 2 = 0\) nên phương trình có hai nghiệm phân biệt \({x_1} = 1;{x_2} = \frac{2}{1} = 2\).

b) Ta có: \(a - b + c =  - 3 - 5 + 8 = 0\) nên phương trình có hai nghiệm \({x_1} =  - 1;{x_2} =  - \frac{8}{{ - 3}} = \frac{8}{3}\).

c) Ta có: \(a + b + c = \frac{1}{3} + \frac{1}{6} - \frac{1}{2} = 0\) nên phương trình có hai nghiệm \({x_1} = 1;{x_2} = \frac{{ - \frac{1}{2}}}{{\frac{1}{3}}} =  - \frac{3}{2}\).