Question

[Phạm Nhật Minh]

Bài 1 ; Tìm x

a. ( 2x - 3 ) \(\left(\dfrac{3}{4}x+1\right)\) = 0

b. ( 5x - 1 ) \(\left(2x-\dfrac{1}{3}\right)\) = 0

c. \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)

Answer 1

[Đức Hiếu]

a, \(\left(2x-3\right)\left(\dfrac{3}{4}x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-4}{3}\end{matrix}\right.\)

Vậy......

b, \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy......

c, \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)

\(\Rightarrow\dfrac{1}{7}:x=\dfrac{-3}{14}\)

\(\Rightarrow x=\dfrac{1}{7}:\dfrac{-3}{14}=\dfrac{-2}{3}\)

Vậy.....

Chúc bạn học tốt!!!

Answer 2

[Đỗ Ngọc Khánh Ly]

a,(2x-3)(\(\dfrac{3}{4}\)x+1)=0

*2x-3=0\(\rightarrow\)2x=3\(\rightarrow\)x=\(\dfrac{3}{2}\)

*\(\dfrac{3}{4}\)x+1=0\(\rightarrow\)\(\dfrac{3}{4}\)x=\(-1\)\(\rightarrow\)x=\(\dfrac{-3}{4}\)

Vậy x\(\in\){\(\dfrac{-3}{4};\dfrac{3}{2}\)}

b,(5x-1)(2x-1/3)=0

*5x-1=0\(\rightarrow\)5x=1\(\rightarrow\)x=1/5

*2x-1/3=0\(\rightarrow\)2x=1/3\(\rightarrow\)x=1/6

Vậy x\(\in\){1/5;1/6}

c,3/7+1/7:x=3/14

1/7:x=-3/14

x=-14/21 Vậy x=-14/21