Question

1​ Tìm x

a) ( 2x-3).(6-2x)=0

b) x:\(\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{-2}{3}\)

Answer 1

\(\left(2x-3\right)\left(6-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

Answer 2

a) \(\left(2x-3\right).\left(6-2x\right)=0\). bài này làm theo kiểu xét dấu nha. hiểu thì mình mới làm

b) \(x:\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{-2}{3}\)

\(x:\dfrac{3}{4}=\dfrac{-2}{3}-\dfrac{1}{4}\)

\(x:\dfrac{3}{4}\) = \(\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}.\dfrac{3}{4}\)

\(x=\dfrac{-5}{4}\)

tick nha

Answer 3

a, (2x - 3).(6 - 2x) = 0

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

Answer 4

b, \(x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)

\(\Rightarrow x:\dfrac{3}{4}=-\dfrac{5}{12}\)

\(\Rightarrow x=-\dfrac{5}{16}\)

Answer 5

a ) Nếu kết quả là 0 thì x = 0 vì 0 nhân với số nào cũng bằng 0

\(\Rightarrow x=0\)

b ) \(x:\dfrac{3}{4}+\dfrac{1}{4}=\dfrac{-2}{3}\)

\(x:\dfrac{3}{4}=\dfrac{-2}{3}-\dfrac{1}{4}\)

\(x:\dfrac{3}{4}=\dfrac{3}{12}+\dfrac{-8}{12}\)

\(x:\dfrac{3}{4}=\dfrac{-5}{12}\)

\(x=\dfrac{-5}{12}.\dfrac{3}{4}\)

\(x=\dfrac{-15}{48}\)

\(x=\dfrac{-5}{16}\)