f) \(4x^2-12x+9=0\)
<=> \(\left(2x-3\right)^2\) = 0
<=> \(2x-3=0\)
<=> \(2x=3\) <=> \(x=\dfrac{3}{2}\)
Vậy ...............
g) \(3x^2+7x+2=0\)
<=> \(\left(3x^2+6x\right)+\left(x+2\right)=0\)
<=> \(3x\left(x+2\right)+\left(x+2\right)=0\)
<=> \(\left(x+2\right)\left(3x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy ........................
h) \(x^2-4x+1=0\)
<=> \(\left(x^2-4x+4\right)-3=0\)
<=> \(\left(x-2\right)^2=3\)
<=> \(\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)
Vậy .........................
i) \(2x^2-6x+1=0\)
<=> \(2\left(x^2-3x+2,25\right)-3,5=0\)
<=> \(\left(x-1,5\right)^2=1,75\)
<=> \(\left[{}\begin{matrix}x-1,5=\sqrt{1,75}\\x-1,5=-\sqrt{1,75}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{1,75}+1,5\\x=-\sqrt{1,75}+1,5\end{matrix}\right.\)
Vậy ...................
j) \(3x^2+4x-4=0\)
<=> \(\left(3x^2+6x\right)-\left(2x+4\right)=0\)
<=> \(3x\left(x+2\right)-2\left(x+2\right)\) = 0
<=> \(\left(x+2\right)\left(3x-2\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....................................
f) \(4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy..
g) \(3x^2+7x+2=0\)
\(\Rightarrow3x^2+6x+x+2=0\)
\(\Rightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy..
h) \(x^2-4x+1=0\)
\(\Rightarrow x^2-4x+4-3=0\)
\(\Rightarrow\left(x-2\right)^2-3=0\)
\(\Rightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2-\sqrt{3}=0\\x-2+\sqrt{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
Vậy..
j) \(3x^2+4x-4=0\)
\(\Rightarrow3x^2+6x-2x-4=0\)
\(\Rightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..
f
cách 1
\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
cách 2
\(\Delta=\left(-12\right)^2-4.4.9=0\) nên phương trình có nghiệm kép
\(x_1=x_2=-\dfrac{-12}{2.4}=\dfrac{3}{2}\)
các câu khác làm tương tự
Sửa lại câu h
h) \(x^2-4x+1=0\)
<=> \(\left(x^2-4x+4\right)-3=0\)
<=> \(\left(x-2\right)^2=3\)
<=> \(\left[{}\begin{matrix}x-2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}+2\\x=2-\sqrt{3}\end{matrix}\right.\)
Vậy nghiệm của PT là S = \(\left\{2+\sqrt{3};2-\sqrt{3}\right\}\)
f, \(4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
g, \(3x^2+7x+2=0\)
\(\Rightarrow x=\dfrac{-7\pm\sqrt{7^2-4.3.2}}{2.3}=\dfrac{-7\pm49-24}{6}=\dfrac{-7\pm\sqrt{25}}{6}=\dfrac{-7\pm5}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-7+5}{6}\\x=\dfrac{-7-5}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-2\end{matrix}\right.\)
\(\Rightarrow x_1=\dfrac{-1}{3};x_2=-2\)
h, \(x^2-4x+1=0\)
\(\Rightarrow\dfrac{4\pm\sqrt{\left(-4\right)^2-4.1.1}}{2.1}=\dfrac{4\pm\sqrt{16-4}}{2}=\dfrac{4\pm\sqrt{12}}{2}=\dfrac{4\pm2\sqrt{3}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+2\sqrt{3}}{2}\\x=\dfrac{4-2\sqrt{3}}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow x_1=2+\sqrt{3};x_2=2-\sqrt{3}\)
i, \(2x^2-6x+1=0\)
\(\Rightarrow x=\dfrac{6\pm\sqrt{\left(-6\right)^2-4.2.1}}{2.2}=\dfrac{6\pm\sqrt{36-8}}{4}=\dfrac{6\pm\sqrt{28}}{4}=\dfrac{6\pm2\sqrt{7}}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6+2\sqrt{7}}{4}\\x=\dfrac{6-2\sqrt{7}}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{7}}{2}\\x=\dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\)
\(\Rightarrow x_1=\dfrac{3+\sqrt{7}}{2};x_2=\dfrac{3-\sqrt{7}}{2}\)
j, \(3x^2+4x-4=0\)
\(\Rightarrow x=\dfrac{-4\pm\sqrt{4^2-4.3.\left(-4\right)}}{2.3}=\dfrac{-4\pm\sqrt{16+68}}{6}=\dfrac{-4\pm\sqrt{64}}{6}=\dfrac{-4\pm8}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-4+8}{6}\\x=\dfrac{-4-8}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(\Rightarrow x_1=\dfrac{2}{3};x_2=-2\)