\(a)\) \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Leftrightarrow\)\(\left(2x+1\right)\left(0,5x+2\right)=\left(x+1\right)\left(x+3\right)\)

\(\Leftrightarrow\)\(2x\left(0,5x+2\right)+0,5x+2=x\left(x+1\right)+3\left(x+1\right)\)

\(\Leftrightarrow\)\(x^2+4x+0,5+2=x^2+x+3x+3\)

\(\Leftrightarrow\)\(x^2+4x-x^2-4x=3-0,5-2\)

\(\Leftrightarrow\)\(0=0,5\) ( vô lí :'< )

Vậy không có x thoả mãn đề bài 

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x+1}{2x+1}=\frac{0,5+2}{x+3}=\frac{\left(x+1\right)-2.\left(0,5x+2\right)}{\left(2x+1\right)-2.\left(x+3\right)}=\frac{x+1-x-4}{2x+1-2x-6}=\frac{-3}{-5}=\frac{3}{5}\)

suy ra:

\(\frac{x+1}{2x+1}=\frac{3}{5}\Rightarrow5.\left(x+1\right)=3.\left(2x+1\right)\)

=>5x+5=6x+3

5x-6x=3-5

-x=-2

x=2

mình viết nhầm, mình sửa lại bài nhé

\(\frac{x+1}{2x+1}-\frac{0,5x+2}{x+3}\)

\(\Rightarrow\) (x+1)(x+3) = (0,5x+2)(2x+1)

\(\Rightarrow\) x² + 3x + x + 3 = x² + 0,5x + 4x + 2

\(\Rightarrow\) x² + 4x + 3 = x² + 4,5x + 2

\(\Rightarrow\) x² - x² + 4x - 4,5x = 2 - 3

\(\Rightarrow\) -0,5x = -1

\(\Rightarrow\) x = \(\frac{-1}{-0,5}\)

\(\Rightarrow\) x = 2

Vậy x = 2

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\Rightarrow\) (x + 1)(x + 3) = (2x + 1)(0,5x + 2)

\(\Rightarrow\) x² + 3x + x + 3 = x² + 4x + 0,5x + 2

\(\Rightarrow\) x² + 3x + x + 3 - x² - 4x - 0,5x - 2 = 0

\(\Rightarrow\) 0,5x + 1 = 0

\(\Rightarrow\) 0,5x = 0 - 1

\(\Rightarrow\) 0,5x = -1

\(\Rightarrow\) x = -1 : 0,5

\(\Rightarrow\) x = -2

Vậy x = -2

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+5}\)

\(\Rightarrow\) (x+1)(x+3) = (0,5x+2)(2x+1)

\(\Rightarrow\) x² + 3x + x + 3 = x² + 0,5x + 4x + 2

\(\Rightarrow\) x² + 4x + 3 = x² + 4,5x + 2

\(\Rightarrow\) x² - x² + 4x - 4,5x = 2 - 3

\(\Rightarrow\) -0,5x = -1

\(\Rightarrow\) x = \(\frac{-1}{-0.5}\)

\(\Rightarrow\) x =2

Vậy x = 2

Áp dụng tính chất của tỉ lệ thức

a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)

=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)

=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)

b) \(2x-\left|x+1\right|=\frac{1}{2}\)

=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))

=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)

=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)

a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)

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

\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)

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

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

\(\Leftrightarrow-3x=-9\)

\(\Leftrightarrow x=3\)

Vậy x = 3

b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)

                            \(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)

                             \(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)

                              \(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)

                             \(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)

Vậy x = 2

bài này sử dụng tích chéo nha bạn

z cau lamđi

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

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

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

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

e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)

\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)

\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)

Vậy ....