a/ \(-x-\dfrac{2}{3}=-\dfrac{6}{7}\)

\(\Leftrightarrow-x=-\dfrac{6}{7}+\dfrac{2}{3}\)

\(\Leftrightarrow-x=-\dfrac{18}{21}+\dfrac{14}{21}\)

\(\Leftrightarrow-x=-\dfrac{4}{21}\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy ....

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 ..

a: \(\dfrac{31-2x}{x+23}=\dfrac{9}{4}\)

=>121-8x=9x+207

=>-17x=86

hay x=-86/17

b: \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)

=>|2x-1|=9/5

=>2x-1=9/5 hoặc 2x-1=-9/5

=>2x=14/5 hoặc 2x=-4/5

=>x=7/5 hoặc x=-2/5

a) Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}=\dfrac{3x+2-3x+1}{5x+7-5x-1}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5x+7}{2}\\3x-1=\dfrac{5x+1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+4=5x+7\\6x-2=5x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)(Nhận)

Vậy x = 3

b) Giống vậy :)

a, \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)

\(\Leftrightarrow15x^2+13x=15x^2+16x-9\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

Vậy x = 3

b, Tương tự

a,

\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)

b,

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

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

c,

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)

\(\Rightarrow5\left|x+1\right|^2=180\)

\(\Rightarrow\left|x+1\right|^2=36\)

Mà \(\left|x+1\right|\ge0\)

=> x + 1 = 6 <=> x = 7

Giải:

a) \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x-3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, có:

\(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x-3}=\dfrac{3x+2-3x+1}{5x+7-5x+3}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{3x+2}{5x+7}=\dfrac{3}{10}\)

\(\Leftrightarrow30x+20=15x+21\)

\(\Leftrightarrow15x=1\)

\(\Leftrightarrow x=\dfrac{1}{15}\)

Vậy ...

b) \(\dfrac{\left|2x-1\right|}{\dfrac{1}{2}}=\dfrac{18}{5}\)

\(\Leftrightarrow5\left|2x-1\right|=9\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{9}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{9}{5}\\2x-1=-\dfrac{9}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Vậy ...

1: \(\Leftrightarrow\left(x+1\right)^2=4\)

=>x+1=2 hoặc x+1=-2

=>x=1 hoặc x=-3

2: \(\Leftrightarrow7x-21=5x+25\)

=>2x=46

=>x=23

3: \(\Leftrightarrow x^2+4x+3=x^2+0.5x+4x+2\)

=>4,5x+2=4x+3

=>x=1

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

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy ...

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

\(\Rightarrow\) TH1 : 5x-1=0 \(\Rightarrow\) 5x=1 \(\Rightarrow\) x=\(\dfrac{1}{5}\)

\(\Rightarrow\) TH2 : 2x-\(\dfrac{1}{3}\) = 0 \(\Rightarrow\) 2x=\(\dfrac{1}{3}\) \(\Rightarrow\) x=\(\dfrac{1}{6}\)

Vậy x \(\in\) \(\left\{\dfrac{1}{5};\dfrac{1}{6}\right\}\)

a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)

=>(x+1)(2x+6)=(2x+1)(x+4)

\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)

=>9x+4=8x+6

=>x=2

b: \(x^2+5x=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5

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

\(\dfrac{2}{3}-\dfrac{1}{3}x-\dfrac{1}{2}-x+\dfrac{1}{2}=5\)

\(\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}x-x=5\)

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

\(\dfrac{2}{3}-\dfrac{4}{3}x=5\)

\(\dfrac{4}{3}x=\dfrac{2}{3}-5\)

\(\dfrac{4}{3}x=-\dfrac{13}{3}\)

\(x=-\dfrac{13}{3}:\dfrac{4}{3}\)

\(x=-\dfrac{13}{4}\)

Vậy...............

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

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

Vậy................

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

\(\Rightarrow2x-1=0\)

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

Vậy.............