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KN
1
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
Lời giải:
a.
$x=\frac{7}{25}+\frac{-1}{5}=\frac{7}{25}+\frac{-5}{25}=\frac{7-5}{25}=\frac{2}{25}$
b.
$x=\frac{5}{11}+\frac{4}{-9}=\frac{5}{11}-\frac{4}{9}=\frac{45}{99}-\frac{44}{99}=\frac{1}{99}$
c.
$\frac{x}{-1}=\frac{-1}{3}-\frac{5}{9}=\frac{-3}{9}-\frac{5}{9}=\frac{-8}{9}$
$x=(-1).\frac{-8}{9}=\frac{8}{9}$
3 tháng 4 2021
a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)
hay \(x=\dfrac{6}{25}\)
Vậy: \(x=\dfrac{6}{25}\)
3 tháng 4 2021
b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)
hay \(x=\dfrac{5}{99}\)
Vậy: \(x=\dfrac{5}{99}\)
VV
24 tháng 7 2021
a, \(x\) : \(\dfrac{13}{3}\) = -2,5
\(x\) = -2,5 . \(\dfrac{13}{3}\)
\(x\) = \(\dfrac{65}{6}\)
b,\(\dfrac{3}{5}\)\(x\) = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)
\(\dfrac{3}{5}x\) = \(\dfrac{-3}{20}\)
\(x\) = \(\dfrac{-3}{20}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{-1}{4}\)
c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)
\(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)
\(\dfrac{12}{13}x=2\)
\(x=2:\dfrac{12}{13}\)
\(x=\dfrac{13}{6}\)
4 tháng 1 2022
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
I
1
1 tháng 6 2018
a) \(x=\frac{7}{25}+\frac{-1}{5}\)
\(\Rightarrow x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\)
b) \(x=\frac{5}{11}+\frac{4}{-9}\)
\(\Rightarrow x=\frac{1}{99}\)
Vậy \(x=\frac{1}{99}\)
c) \(\frac{5}{9}+x=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{3}-\frac{5}{9}\)
\(\Rightarrow x=\frac{-8}{9}\)
Vậy \(x=\frac{-8}{9}\)
d) \(\frac{3}{4}-x=-1\)
\(\Rightarrow x=\frac{3}{4}-\left(-1\right)\)
\(\Rightarrow x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
e) \(x+4=\frac{1}{5}\)
\(\Rightarrow x=\frac{1}{5}-4\)
\(\Rightarrow x=\frac{-19}{5}\)
Vậy \(x=\frac{-19}{5}\)
f) \(x-\frac{1}{5}=2\)
\(\Rightarrow x=2+\frac{1}{5}\)
\(\Rightarrow x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}\)
g) \(x+\frac{5}{3}=\frac{1}{81}\)
\(\Rightarrow x=\frac{1}{81}-\frac{5}{3}\)
\(\Rightarrow x=\frac{-134}{81}\)
Vậy \(x=\frac{-134}{81}\)
_Chúc bạn học tốt_
TT
12 tháng 5 2021
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
\(a,x=\frac{7}{25}-\frac{1}{5}\Rightarrow x=\frac{2}{25}\)
\(b,x=\frac{5}{11}-\frac{4}{9}\Rightarrow x=\frac{1}{99}\)
\(c,\frac{5}{9}-\frac{x}{1}=-\frac{1}{3}\)
\(\frac{x}{1}=\frac{8}{9}\Rightarrow\frac{9x}{9}=\frac{8}{9}\Rightarrow9x=8\Rightarrow x=\frac{8}{9}\)
a) \(x=\frac{7}{25}+\frac{-1}{5}\)
\(x=\frac{7}{25}+\frac{-5}{25}\)
\(x=\frac{2}{25}\)
b) \(x=\frac{5}{11}+\frac{4}{-9}\)
\(x=\frac{-45}{-99}+\frac{44}{-99}\)
\(x=\frac{-1}{-99}=\frac{1}{99}\)
c) \(\frac{5}{9}+\frac{x}{-1}=-\frac{1}{3}\)
\(\frac{x}{-1}=-\frac{1}{3}-\frac{5}{9}\)
\(\frac{x}{-1}=-\frac{3}{9}-\frac{5}{9}\)
\(\frac{x}{-1}=-\frac{8}{9}\)
\(x=\frac{8}{9}\)