a) - 0,52 : x = - 9,36 : 16,38

- 0,52 : x = \(-\frac{4}{7}\)

\(x=\left(-0,52\right):\left(-\frac{4}{7}\right)=\frac{91}{100}=0,91\)

b) \(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)

\(\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)

\(\frac{x}{1,61}=\frac{34}{23}\)

\(x=\frac{34}{23}.1,61=2,38\)

Ta có :

\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)

\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}\right)=0\)

\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)

\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)

Mà : \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\)

\(\Leftrightarrow424-x=0\)

\(\Leftrightarrow x=424\)

Vậy x = 424

\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)

\(\Leftrightarrow\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)

\(\Leftrightarrow\frac{34}{23}=\frac{x}{1,61}\)

\(\Leftrightarrow x=\frac{34}{23}.1,61\)

\(\Leftrightarrow x=\frac{119}{50}\)

Có \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)+x=\frac{23}{45}\)

Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)

Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)

\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)

Thay vào phép tính trên ta được:

\(\frac{11}{45}\cdot x=\frac{23}{45}\\ x=\frac{23}{45}:\frac{11}{45}\\ x=\frac{23}{11}\)

Vậy \(x=\frac{23}{11}\)

\(=\left(\frac{x-1}{12}-1\right)+\left(\frac{2x-3}{23}-1\right)+\left(\frac{3x-5}{34}-1\right)+\left(\frac{4x-7}{45}-1\right)\)

\(=\frac{x-13}{12}+2.\frac{x-13}{23}+3.\frac{x-13}{34}+4.\frac{x-13}{45}\)

\(=\left(x-13\right)\left(\frac{1}{12}+\frac{2}{23}+\frac{3}{34}+\frac{4}{45}\right)\)

Đề thiếu phải không??? :))

ko, đúng mak bn

Nguyễn Lam Giang

Tìm x mà không có vế phải thì tìm bằng niềm tin ak