`(x-3)^2-2(x-3)=4x-14`
`<=>x^2-6x+9-2x+6=4x-14`
`<=>x^2-8x+15=4x-14`
`<=>x^2-12x+29=0`
`∆'=6^2-1.29=7>0`
`=>` Phương trình có hai nghiệm phân biệt: $\left[\begin{array}{} x=\frac{6+\sqrt7}{1}=6+\sqrt7\\x=\frac{6-\sqrt7}{1}=6-\sqrt7 \end{array} \right.$
\(\left(x-3\right)^2-2\left(x-3\right)=4x-14\)
\(\Leftrightarrow x^2-6x+9-2x+6=4x-14\)
\(\Leftrightarrow x^2-12x+29=0\)
\(\Delta'=\left(-6\right)^2-1\cdot29=36-29=7>0\)
\(\rightarrow\) PT có 2 nghiệm pb
\(x_1=\dfrac{6+\sqrt{7}}{1}=6+\sqrt{7}\)
\(x_2=\dfrac{6-\sqrt{7}}{1}=6-\sqrt{7}\)
Vậy S = \(\left\{6+\sqrt{7};6-\sqrt{7}\right\}\)
\(\left(x-3\right)^2-2\left(x-3\right)=4x-14\)
\(\Leftrightarrow x^2-6x+9-2x+6=4x-14\)
\(\Leftrightarrow x^2-8x+15=4x-14\)
\(\Leftrightarrow x^2-8x+15-4x+14=0\)
\(\Rightarrow x^2-12x+29=0\)
\(\left(a=1;b=-12;c=29\right)\)
\(\Delta=b^2-4ac\)
\(=\left(-12\right)^2-4.1.29\)
\(=28>0\)
Vậy pt có 2 n0 phân biệt
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{12+\sqrt{28}}{2.1}=6+\sqrt{7}\)
\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{12-\sqrt{28}}{2.1}=6-\sqrt{7}\)
Vậy...
