x2- 2x +1=4
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\(2x^2-32=0\)
\(\Leftrightarrow x^2-16=0\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
a) -7x-2x2-2=0
x (-7-2x)-2=0
TH1: x=0
TH2: (-7-2x)-2=0
-7-2x =2
-2x =2+7
-2x =9
x =9:(-2)
x =\(-\frac{9}{2}\)
Vậy x=0 hoặc x=\(-\frac{9}{2}\)
b) 2x2-2x-1=0
x(2x-2)-1 =0
TH1: x=0
TH2: (2x-2)-1=0
2x-2 =1
2x =1+2
2x =3
x =3:2
x =\(\frac{3}{2}\)
Vậy x=0 hoặc x= \(\frac{3}{2}\)
nếu là bài toán lớp 7 thì trình bày như thế còn là bài toán lớp 8 thì nhớ làm giống phương trình nha :)
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(6x^2-15x\)
b) \(x^2+5x+4\)
c) \(49-x^2\)
d) \(x^2+4x+4\)
e) \(9-12x+4x^2\)
f) \(x^3-8\)
\(a,=6x^2-15x\\ b,=x^2+5x+4\\ c,=49-x^2\\ d,=x^2+4x+4\\ e,=9-12x+4x^2\\ f,=x^3-8\)
\(x^2-2x+1=4\)
\(\left(x-1\right)^2-2^2=0\)
\(\left(x-1-2\right)\left(x-1+2\right)=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(x^2-2x+1=4\) \(\left(x-1\right)^2=2^2\) \(\left(x-1\right)^2-2^{^{ }2}=0\) \(\left(x-3\right)\left(x-1\right)=0\) \(\Rightarrow x=3;x=1\)