a)
\(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2-x+3x-1=0\)
\(\Leftrightarrow x\left(3x-1\right)+\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
b)
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
a, \(3x^2+2x-1=0\)
\(\Rightarrow3x^2-x+3x-1=0\)
\(\Rightarrow\left(3x^2-x\right)+\left(3x-1\right)=0\)
\(\Rightarrow x.\left(3x-1\right)+\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right).\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=1\\x=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
Vậy......
b, \(x^2-5x+6=0\)
\(\Rightarrow x^2-3x-2x+6=0\)
\(\Rightarrow\left(x^2-3x\right)-\left(2x-6\right)=0\)
\(\Rightarrow x.\left(x-3\right)-2.\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right).\left(x-2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy......
Chúc bạn học tốt!!!
\(a,3x^2+2x-1=0\Leftrightarrow x^2-1+2x^2-2x\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\\dfrac{-1}{3}\end{matrix}\right.\)Vậy phương trình có tập nghiệm S=\(\left\{1;\dfrac{-1}{3}\right\}\)
\(b,x^2-5x+6=0\Leftrightarrow x^2-2x-3x+6=0\)\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S=\(\left\{2;3\right\}\)
\(c,x^2-3x+2=0\Leftrightarrow x^2-x-2x+2=0\)\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{1;2\right\}\)
c, \(x^2-3x+2=0\)
\(\Rightarrow x^2-2x-x+2=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(x-2\right)=0\)
\(\Rightarrow x.\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right).\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy......
d, \(2x^2-6x+1=0\)
Bạn xem lại đề câu d nha!!
a, \(3x^2+2x-1=0\)
\(\Rightarrow x=\dfrac{-2\pm\sqrt{2^2-4.3.\left(-1\right)}}{2.3}=\dfrac{-2\pm\sqrt{4+12}}{6}=\dfrac{-2\pm\sqrt{16}}{6}\)
\(\Rightarrow x=\dfrac{-2\pm4}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2+4}{6}\\x=\dfrac{-2-4}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)
b, \(x^2-5x+6=0\)
\(\Rightarrow x=\dfrac{5\pm\sqrt{\left(-5\right)^2-4.1.6}}{2.1}=\dfrac{5\pm\sqrt{25-24}}{2}\)
\(\Rightarrow x=\dfrac{5\pm\sqrt{1}}{2}=\dfrac{5\pm1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5+1}{2}\\x=\dfrac{5-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
c, \(x^2-3x+2=0\)
\(\Rightarrow x=\dfrac{3\pm\sqrt{\left(-3\right)^2-4.1.2}}{2.1}=\dfrac{3\pm\sqrt{\left(9-8\right)}}{2}=\dfrac{3\pm\sqrt{1}}{2}=\dfrac{3\pm1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+1}{2}\\x=\dfrac{3-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
d, \(2x^2-6x+1=0\)
\(\Rightarrow x=\dfrac{6\pm\sqrt{\left(-6\right)^2-4.2.1}}{2.2}=\dfrac{6\pm\sqrt{36-8}}{4}=\dfrac{6\pm\sqrt{28}}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6+2\sqrt{7}}{4}\\x=\dfrac{6-2\sqrt{7}}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{7}}{4}\\x=\dfrac{3-\sqrt{7}}{4}\end{matrix}\right.\)
