a) \(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x-3=0\\x+1=0\end{array}\right.\)\(\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x=3\\x=-1\end{array}\right.\)
a.
\(x^2-2x-3=0\)
\(x^2-2\times x+1^2-1^2-3=0\)
\(\left(x-1\right)^2-4=0\)
\(\left(x-1\right)^2=4\)
\(\left(x-1\right)^2=\left(\pm2\right)^2\)
\(x-1=\pm2\)
TH1:
x - 1 = 2
x = 2 + 1
x = 3
TH2:
x - 1 = -2
x = -2 + 1
x = -1
Vậy x = 3 hoặc x = -1
b.
\(2x^2+5x-3=0\)
\(2\times\left(x^2+2\times x\times\frac{5}{4}+\left(\frac{5}{4}\right)^2-\left(\frac{5}{4}\right)^2-\frac{3}{2}\right)=0\)
\(\left(x+\frac{5}{4}\right)^2-\frac{49}{16}=0\)
\(\left(x+\frac{5}{4}\right)^2=\frac{49}{16}\)
\(\left(x+\frac{5}{4}\right)^2=\left(\pm\frac{7}{4}\right)^2\)
\(x+\frac{5}{4}=\pm\frac{7}{4}\)
TH1:
x + 5/4 = 7/4
x = 7/4 - 5/4
x = 2/4
x = 1/2
TH2:
x + 5/4 = -7/4
x = -7/4 - 5/4
x = -12/4
x = -3
Vậy x = -3 hoặc x = 1/2
Chúc bạn học tốt ^^
a) \(x^2-2x-3=\left(x+1\right)\left(x-3\right)=0\)<=> \(\left[\begin{array}{nghiempt}x+1=0\\x-3=0\end{array}\right.\)
<=> x=-1 hoặc x=3
b) \(2x^2+5x-3=\left(x+3\right)\left(x-\frac{1}{2}\right)=0\)
<=> \(\left[\begin{array}{nghiempt}x-\frac{1}{2}=0\\x+3=0\end{array}\right.\)<=> x=1/2 hoặc x=-3
a. \(x^2-2x-3=0\)
\(x^2-2x+1-4=0\)
\(\left(x-1\right)^2-4=0\)
\(\left(x-1\right)^2=4\)
\(\left(x-1\right)=\pm2\)
\(\left[\begin{array}{nghiempt}x-1=2\\x-1=-2\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=-1\end{array}\right.\)
b, \(2x^2+5x-3=0\)
\(\Leftrightarrow x^2+\frac{5}{2}x-\frac{3}{2}=0\)
\(\Leftrightarrow x^2-\frac{1}{2}x+3x-\frac{3}{2}=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)+3\left(x-\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-\frac{1}{2}=0\\x+3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-3\end{array}\right.\)
b) \(\Leftrightarrow2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x+3=0\\2x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=\frac{1}{2}\end{array}\right.\)
