\(a,\text{ }\frac{2}{5}\text{ x }\left(2x+4\right)=0\)
Vì \(\frac{2}{5}\ne0\) \(\Rightarrow\text{ }2x+4=0\)
\(2x=0-4\)
\(2x=-4\)
\(x=-4\text{ : }2\)
\(x=-2\)
\(a,\text{ }\frac{2}{5}\text{ x }\left(2x+4\right)=0\)
\(\Rightarrow\text{ }2x+4=0\) ( Vì \(\frac{2}{5}\ne0\) )
\(2x=0-4\)
\(2x=-4\)
\(x=-4\text{ : }2\)
\(x=-2\)
\(b,\text{ }\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0+2\\x=0-3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }2\right\}\)
\(b,\text{ }\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0+2\\x=0-3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-3\text{ ; }2\right\}\)
a) \(\frac{2}{5}.\left(2x+4\right)=0\)
\(\Rightarrow2x+4=0\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
b) \(\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-6,25=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2-6,25=0\)
\(\Rightarrow x-\frac{1}{2}=\pm6,25\)
\(\Rightarrow x=\frac{27}{4};x=-\frac{23}{4}\)
c) \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x^2+4x+4-x^2+4=0\)
\(\Rightarrow4x+8=0\)
\(\Rightarrow4x=-8\)
\(\Rightarrow x=-2\)
Study well
Sorry !
Bạn ơi chỉnh 25 thành \(\frac{2}{5}\) nha !
Mình nhầm tí !
c) (x + 2)^2 - (x - 2)(x + 2) = 0
<=>(x + 2)(x + 2 - x +2) = 0
<=>4(x + 2) = 0
<=>4x + 8 = 0
<=> 4x = -8
<=> x = -2
\(S\in\left\{-2\right\}\)
\(c,\text{ }\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)
\(\left(x+2\right)\left[x+2-x+2\right]=0\)
\(\left(x+2\right)\left[\left(x-x\right)+\left(2+2\right)\right]=0\)
\(\left(x+2\right)\cdot4=0\)
\(\Rightarrow\text{ }\left(x+2\right)=0\text{ ( Vì }4\ne0\text{ ) }\)
\(x=0-2\)
\(x=-2\)
\(c,\text{ }\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\)
\(\left(x+2\right)\left[x+2-x+2\right]=0\)
\(\left(x+2\right)\left[\left(x-x\right)+\left(2+2\right)\right]=0\)
\(\left(x+2\right)\cdot4=0\)
\(\Rightarrow\text{ }\left(x+2\right)=0\text{ ( Vì }4\ne0\text{ ) }\)
\(x=0-2\)
\(x=-2\)
\(a.\frac{2}{5}.\left(2x+4\right)=0\)
\(=>2x+4=0\)
\(2x=-4\)
\(x=-2\)
\(b.\left(x-2\right)\left(x+3\right)=0\)
TH1: \(x-2=0\)
\(=>x=2\)( thỏa mãn )
TH2: \(x+3=0\)
\(=>x=-3\)
\(c.\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
Vậy \(\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)\)
\(\left(x-2\right)\left(x+2\right)=x.\left(x+2\right)-2\left(x+2\right)\)
\(=x^2+x.2-2.x+4\)
\(=x^2+4\)
=> \(\left(x+2\right)^2=x^2+4\)
Xét vế 1: \(\left(x+2\right)^2=0\)
\(=>x=-2\)
Xét vế 2: \(\left(x-2\right)\left(x+2\right)=0\)
Nếu x = -2 thì \(\left(-2-2\right)\left(-2+2\right)=0\)
Vậy x = -2
P/S: Mk ko chắc có đúng ko nha !
a) \(\frac{2}{5}\left(2x+4\right)=0\)
\(\frac{2\left(2x+4\right)}{5}=0\)
\(\frac{2.2\left(x+2\right)}{5}=0\)
\(\frac{4\left(x+2\right)}{5}=0\)
\(4\left(x+2\right)=0.5\)
\(4\left(x+2\right)=0\)
\(x+2=0:4\)
\(x+2=0\)
\(x=-2\)
b) \(\left(x-2\right)\left(x+3\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
c) \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left[x+2-\left(x-2\right)\right]=0\)
\(x+2=0\)
\(x=-2\)