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Bài 1:
a) x + 1919 - 3535 = 3636 b) 3434 - x + 611611 = 5656
x + 1919 = 3636 + 3535 3434 - x = 5656 - 611611
x + 1919 = 11101110 3434 - x = 19661966
x = 11101110 - 1919 x = 19661966 + 3434
x = 89908990 x = 137132137132
Bài 2
a) x : 13/16 = 5/8 b)x - 14/28 = 6/9 + 8/25
x = 5/8 * 13/16 x - 14/28 = 74/75
x = 65/128 x = 74/75 + 14/28
x = 223/150
Bài 3
a)62/7 * x = 29/9 : 3/56 b)1/5 : x = 1/5 + 1/7
62/7 * x = 1624/27 1/5 : x = 12/35
x = 1624/27 : 62/7 x = 12/35 * 1/5
x = 5684/637 x = 12/175
Bài 1:
a) x + \(\frac{1}{9}\) - \(\frac{3}{5}\) = \(\frac{3}{6}\) b) \(\frac{3}{4}\) - x + \(\frac{6}{11}\) = \(\frac{5}{6}\)
x + \(\frac{1}{9}\) = \(\frac{3}{6}\) + \(\frac{3}{5}\) \(\frac{3}{4}\) - x = \(\frac{5}{6}\) - \(\frac{6}{11}\)
x + \(\frac{1}{9}\) = \(\frac{11}{10}\) \(\frac{3}{4}\) - x = \(\frac{19}{66}\)
x = \(\frac{11}{10}\) - \(\frac{1}{9}\) x = \(\frac{19}{66}\) + \(\frac{3}{4}\)
x = \(\frac{89}{90}\) x = \(\frac{137}{132}\)
Bài 2
a) x : 13/16 = 5/8 b)x - 14/28 = 6/9 + 8/25
x = 5/8 * 13/16 x - 14/28 = 74/75
x = 65/128 x = 74/75 + 14/28
x = 223/150
Bài 3
a)62/7 * x = 29/9 : 3/56 b)1/5 : x = 1/5 + 1/7
62/7 * x = 1624/27 1/5 : x = 12/35
x = 1624/27 : 62/7 x = 12/35 * 1/5
x = 5684/637 x = 12/175
Bài 1:
a ) \(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
\(x+\frac{1}{9}=\frac{3}{6}+\frac{3}{5}\)
\(x+\frac{1}{9}=\frac{11}{10}\)
\(x=\frac{11}{10}-\frac{1}{9}=\frac{89}{90}\)
Vậy \(x=\frac{89}{90}\)
b) \(\frac{3}{4}-x+\frac{6}{11}=\frac{5}{6}\)
\(\frac{3}{4}-x=\frac{5}{6}-\frac{6}{11}\)
\(\frac{3}{4}-x=\frac{19}{66}\)
\(x=\frac{3}{4}-\frac{19}{66}=\frac{61}{132}\)
Vậy \(x=\frac{61}{132}\)
Bài 2 :
a) \(x:\frac{13}{16}=\frac{5}{-8}\)
\(x=\frac{5}{-8}.\frac{13}{16}=-\frac{65}{128}\)
Vậy \(x=-\frac{65}{128}\)
b) \(x.\frac{-14}{28}=\frac{6}{-9}-\frac{2}{15}\)
\(x.\frac{-14}{28}=-\frac{4}{5}\)
\(x=-\frac{4}{5}:\frac{-14}{28}=\frac{8}{5}\)
Vậy \(x=\frac{8}{5}\)
2, tìm x thuộc Z biết :
a, x^2 -(-3 )^2 =16
x^2-9 = 16
x^2 = 25
=> x = 5
b, x^2 + (-4) ^2 =0
x^2 + 16 = 0
x^2 = -16
=> x= -4
a. (x2 - 4).(x+3/5) = 0
TH1: x2 - 4 = 0
x2 = 4
x2 = 22
-22
=> x = 2
-2
Vậy x \(\in\){-2;2}