Question

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

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

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

Answer 3

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\\dfrac{3}{4}x=-1\end{matrix}\right.\)

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

Vậy \(x=-\dfrac{4}{3}\) hoặc \(x=\dfrac{3}{2}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{6}\) hoặc \(x=\dfrac{1}{5}\)

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

\(\Leftrightarrow\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\left(đk:x\ne0\right)\)

\(\Leftrightarrow\dfrac{3}{7}+\dfrac{1}{7}\cdot\dfrac{1}{x}=\dfrac{3}{14}\)

\(\Leftrightarrow\dfrac{3}{7}+\dfrac{1}{7x}=\dfrac{3}{14}\)

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

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

\(\Leftrightarrow14=-21x\)

\(\Leftrightarrow-21x=14\)

\(\Leftrightarrow x=-\dfrac{2}{3}\left(đk:x\ne0\right)\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy \(x=-\dfrac{2}{3}\)

Answer 4

a) Ta có

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

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

TH1: 2x - 3 = 0

2x = 0 + 3

2x = 3

x = 3:2

x = \(\dfrac{3}{2}\)

TH2: \(\dfrac{3}{4}x+1=0\)

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

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

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

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

Vậy \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{-4}{3}\end{matrix}\right.\)là giá trị cần tìm

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

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

Th1: 5x - 1 = 0

5x = 0+1

5x = 1

x = 1 : 5

x = \(\dfrac{1}{5}\)

Th2: \(2x-\dfrac{1}{3}\) = 0

2x = 0 + \(\dfrac{1}{3}\)

2x = \(\dfrac{1}{3}\)

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

x = \(\dfrac{1}{6}\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\) là giá trị cần tìm

c) Ta có

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

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

\(\dfrac{1}{7}+x\) = \(\dfrac{1}{2}\)

x = \(\dfrac{1}{2}-\dfrac{1}{7}\)

x = \(\dfrac{5}{14}\)

Vậy x = \(\dfrac{5}{14}\) là giá trị cần tìm

Answer 5

lộn

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

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

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

x = \(\dfrac{-2}{3}\)

Vậy ...

Answer 6

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

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

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

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}5x=1\Rightarrow x=\dfrac{1}{5}\\2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\end{matrix}\right.\)

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

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

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

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

\(x=\dfrac{-2}{3}\)

Answer 7

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

\(\Leftrightarrow2x-3=0\Rightarrow2x=3\Rightarrow x=\dfrac{3}{2}\)

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

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

\(\Leftrightarrow5x-1=0\Rightarrow5x=1\Rightarrow x=\dfrac{1}{5}\)

\(2x-\dfrac{1}{3}=0\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)

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

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

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

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