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

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

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

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

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

\(\Leftrightarrow\)\(0=0,5\) ( 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

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

\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

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

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)

a, \(-\frac{22}{15}x+\frac{1}{3}=\left|-\frac{2}{3}+\frac{1}{5}\right|=\left|-\frac{7}{15}\right|=\frac{7}{15}\)

\(\Rightarrow\frac{-22}{15}x=\frac{7}{15}-\frac{1}{3}=\frac{2}{15}\)

\(\Rightarrow x=\frac{2}{15}:\frac{-22}{15}=\frac{2}{15}.\frac{15}{-22}=-\frac{1}{11}\)

b,\(x:15=8:24\)

Vậy x=5

\(\Rightarrow x:15=\frac{8}{24}\Rightarrow x=\frac{8}{24}.15=\frac{120}{24}=5\)