Bài 4:
a, (x-3).(7-x)=0
=> x-3=0 hoặc 7-x=0
=> x-3=0 => x=3 hoặc
7-x=0 => x=7
Vậy x=3 hoặc x=7
4b, Ta có (x+1).(x2-4)=0
=> x+1 =0 hoặc x-4=0
Th1: x+1=0
=> x=-1
Th2: x2-4=0
=> x2=4=22
=> x=2
Vậy x=1 hoặc x=2 tại (x+1).(x2-4)=0
3.
c) \(25-\left(30+x\right)=1-\left(27-8\right)\)
\(25-30-x=1-19\)
\(-5-x=-18\)
\(x=13\)
4.
a) \(\left(x-3\right)\left(7-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
Vậy x = 3 hoặc x = 7
b) \(\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=4=\left(\pm2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Vậy..
c) \(\left(x-3\right)\left(7-x\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\7-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}7-x< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x< 7\end{matrix}\right.\\\left\{{}\begin{matrix}x>7\\x< 3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3< x< 7\)
Vậy các số t/m điều kiện trên đều được.
tik mik nha !!!
3c,
Ta có 25-(30+x)=1-(27-8)
=> 25-30-x=1-27+8
=> -5-x=-26+8
=> -5-x=-18
=> x=-5+18
=> x= 13
3) dễ tự làm :v
\(\left(x-3\right)\left(7-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\7-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
\(\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x^2-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x^2=4\Rightarrow x=\pm2\end{matrix}\right.\)
\(\left(x-3\right)\left(7-x\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\Rightarrow x>3\\7-x>0\Rightarrow x< 7\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\Rightarrow x< 3\\7-x< 0\Rightarrow x>7\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3< x< 7\Rightarrow x\in\left\{4;5;6\right\}\)