Ta có : \(\left(2x+3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-3\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
Tìm x,biết :
\(a,\left(x+\frac{1}{5}\right)-4=-2\)
\(\left(x+\frac{1}{5}\right)=2\)
\(x+\frac{1}{5}=2\)
\(x=\frac{9}{5}\)
b,\(\left(2x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
\(c,\frac{31}{9}x-\frac{5}{2}=\frac{8}{3}\)
\(\frac{31}{9}x=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}x=\frac{31}{6}\)
\(x=\frac{3}{2}\)
\(a,\left(x+\frac{1}{5}\right)-4=-2\)
\(\left(x+\frac{1}{5}\right)=2\)
\(x+\frac{1}{5}=2\)
\(x=\frac{9}{5}\)
b,\(\left(2x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
\(c,\frac{31}{9}x-\frac{5}{2}=\frac{8}{3}\)
\(\frac{31}{9}x=\frac{8}{3}+\frac{5}{2}\)
\(\frac{31}{9}x=\frac{31}{6}\)
\(x=\frac{3}{2}\)
(2x+3)*(x-7)=0
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
vậy x nhận 2 giá trị là -3/2 và 7
a) (x+1/5)-4=-2
x+1/5=-2+4
x+1/5=2
x=2-1/5
x=1,8
b) (2x+3)*(x-7)=0
2x+3=0 x-7=0
2x=3 x=0+7
x=3/2 x=7
c) 31/9(x)-5/2=8/3
31/9(x)=8/3+5/2
31/9(x)=31/6
x=3/2
