1/4.12/13+1/4.1/13-3/25
1/4.(12/13+1/13)-3/25
1/4.1-3/25
1/4-3/25
1/8
\(\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-12\%=\frac{1}{4}\cdot\frac{12}{13}+\frac{1}{4}\cdot\frac{1}{13}-\frac{3}{25}=\frac{1}{4}\cdot\left(\frac{12}{13}+\frac{1}{13}\right)-\frac{3}{25}\)
\(=\frac{1}{4}\cdot1-\frac{3}{25}=\frac{1}{4}-\frac{3}{25}=\frac{13}{100}\)
Nhớ bài đây sửa đi sửa lại cũng vì do cái số " % " :(((
a) \(\left|\frac{2}{5}:x\right|=\frac{1}{4}\)
Trường hợp 1 : \(\frac{2}{5}\) : x = \(\frac{1}{4}\)
=> x = \(\frac{2}{5}:\frac{1}{4}=\frac{2}{5}\cdot4=\frac{8}{5}\)
Trường hợp 2 : \(\frac{2}{5}:x=-\frac{1}{4}\)
=> \(x=\frac{2}{5}:\left(-\frac{1}{4}\right)=\frac{2}{5}\cdot\left(-4\right)=-\frac{8}{5}\)
Vậy \(x=\pm\frac{8}{5}\)
b) \(\frac{x}{24}=-\frac{1}{3}-\frac{1}{8}=-\frac{11}{24}\)
=> x = -11
c) \(\frac{3}{x+3}=\frac{-7}{21}\)
=> \(\frac{3}{x+3}=\frac{-1}{3}\)
=> -1(x + 3) = 9
=> -x - 3 = 9
=> -x = 12
=> x = -12
\(\frac{1}{4}.\frac{12}{13}+\frac{1}{4}.\frac{1}{13}-12\%\)
=\(\left(\frac{1}{4}+\frac{1}{4}\right).\left(\frac{12}{13}+\frac{1}{13}\right)-\frac{3}{25}\)
=\(\frac{1}{2}.1-\frac{3}{25}\)
=\(\frac{1}{2}-\frac{3}{25}\)
=\(\frac{25}{50}-\frac{6}{50}\)
=\(\frac{19}{50}\)
Tìm x
a)\(\left|\frac{2}{5}:x\right|=\frac{1}{4}\)
=>\(\hept{\begin{cases}\frac{2}{5}:x=\frac{1}{4}\\\frac{2}{5}:x=\frac{-1}{4}\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{2}{5}:\frac{1}{4}\\x=\frac{2}{5}:\frac{-1}{4}\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{8}{5}\\x=\frac{-8}{5}\end{cases}}\)
+)Vậy x \(\varepsilon\left[\frac{8}{5};\frac{-8}{5}\right]\)
