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

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

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

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

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

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

Vậy không có x thoả mãn đề bài 

x = 2 nhé .

Em mới học lớp 5 thôi .

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

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

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

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

\(\Rightarrow x^2-x=-1\)

\(\Rightarrow x\left(x-1\right)=-1\)

:vvv

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: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))

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

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

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

\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)

\(\Leftrightarrow-\frac{1}{4}x=\frac{4}{9}\)

\(\Leftrightarrow x=-\frac{16}{9}\)

\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)

\(< =>\frac{1}{2}x-\frac{3}{4}x=\left(-\frac{2}{3}\right)^2\)

\(< =>-\frac{1}{4}x=\frac{4}{9}\)

\(=>x=\frac{-16}{9}\)

\(0,5x-75\%x=\left(\frac{1}{3}-1\right)^2\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}x=\left(\frac{-2}{3}\right)^2\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{3}{4}\right)x=\frac{4}{9}\)

\(\Leftrightarrow\frac{-1}{4}x=\frac{4}{9}\)

\(\Leftrightarrow x=\frac{4}{9}:\frac{-1}{4}=\frac{-16}{9}\)

~ Hok tốt ~