Question

[Trần Thảo Anh]

tìm  x biết

a, \((\)\(6\frac{2}{7}\)x \(+\)\(\frac{3}{7}\)\()\)\(\times\)\(2\frac{1}{5}\)\(-\)\(\frac{3}{7}\)\(=\)-2

b, \(3\frac{1}{4}\)x \(+\)\((\)\(-\)\(\frac{7}{6}\)\()\)\(-\)\(1\frac{2}{3}\)\(=\)\(\frac{5}{12}\)

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

d, \((\)x\(+\)\(\frac{1}{5}\)\()\)² \(+\)\(\frac{17}{25}\)\(=\)\(\frac{26}{25}\)

e, \(-1\frac{5}{27}\)\(-\)\((\)\(3x\)\(-\)\(\frac{7}{9}\)\()\)³\(=\)\(-\frac{24}{27}\)

g,\(\frac{x}{1.2}\)\(+\)\(\frac{x}{2.3}\)\(+\)\(\frac{x}{3.4}\)\(+\).....\(+\)\(\frac{x}{99.100}\)\(=\)\(\frac{99}{100}\)

h, \(\frac{x}{3}\)\(+\)\(\frac{x}{6}\)\(+\)\(\frac{x}{10}\)\(+\)\(\frac{x}{15}\)\(=\)\(3x-1\)

Answer 1

[Huỳnh Quang Sang]

a) \(\left(6\frac{2}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)

=> \(\left(\frac{44}{7}x+\frac{3}{7}\right)\cdot\frac{11}{5}=-\frac{11}{7}\)

=> \(\frac{44}{7}x+\frac{3}{7}=-\frac{5}{7}\)

=> \(\frac{44}{7}x=-\frac{8}{7}\)

=> \(\frac{44x}{7}=-\frac{8}{7}\)

=> 44x = -8 => 11x = -2 => \(x=-\frac{2}{11}\)

b) \(3\frac{1}{4}x+\left(-\frac{7}{6}\right)-1\frac{2}{3}=\frac{5}{12}\)

=> \(\frac{13}{4}x-\frac{7}{6}-1\frac{2}{3}=\frac{5}{12}\)

=> \(\frac{13}{4}x-\frac{7}{6}=\frac{25}{12}\)

=> \(\frac{13}{4}x=\frac{13}{4}\)

=> x = 1

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

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

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

=> \(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

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

=> \(\left(3x-\frac{7}{9}\right)^3=-1\frac{5}{27}-\left(-\frac{24}{27}\right)=-\frac{32}{27}+\frac{24}{27}=-\frac{8}{27}\)

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

=> \(3x-\frac{7}{9}=-\frac{2}{3}\)

=> \(x=\frac{-\frac{2}{3}+\frac{7}{9}}{3}=\frac{1}{27}\)

g) \(\frac{x}{1\cdot2}+\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+...+\frac{x}{99\cdot100}=\frac{99}{100}\)

=> \(\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{99}-\frac{x}{100}=\frac{99}{100}\)

=> \(\frac{x}{1}-\frac{x}{100}=\frac{99}{100}\)

=> \(\frac{100x-x}{100}=\frac{99}{100}\)

=> \(\frac{99x}{100}=\frac{99}{100}\)

=> x = 1

h) \(\frac{x}{3}+\frac{x}{6}+\frac{x}{10}+\frac{x}{15}=3x-1\)

=> \(\frac{2x}{6}+\frac{2x}{12}+\frac{2x}{20}+\frac{2x}{30}=3x-1\)

=> \(\frac{2x}{2\cdot3}+\frac{2x}{3\cdot4}+\frac{2x}{4\cdot5}+\frac{2x}{5\cdot6}=3x-1\)

=> \(2\left(\frac{x}{2\cdot3}+\frac{x}{3\cdot4}+\frac{x}{4\cdot5}+\frac{x}{5\cdot6}\right)=3x-1\)

=> \(2\left(\frac{x}{2}-\frac{x}{6}\right)=3x-1\)

=> \(2\left(\frac{3x}{6}-\frac{x}{6}\right)=3x-1\)

=> \(2\cdot\frac{2x}{6}=3x-1\)

=> \(\frac{x}{3}=\frac{3x-1}{2}\)

=> 2x = 3(3x - 1)

=> 2x - 9x + 3  = 0

=> -7x = -3

=> x = 3/7