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a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
Xin lỗi biết làm câu 1 thôi,thông cảm
Ta có A=:
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=\frac{2^2}{2^2}+\frac{3^2}{3^2}+...+\frac{100^2}{100^2}-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Mà \(\left(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}\right)< |\frac{100}{101}\)(tự tính)
\(\Rightarrow C>98\left(đpcm\right)\)
b,x - 3 = xy + 2y
=> -3 = xy + 2y - x
=> -5 = y(x + 2) - x - 2
=> -5 = y(x + 2) - (x + 2)
=> -5 = (y - 1)(x + 2)
=> x + 2 \(\in\)Ư(-5) = Ư(5) = {-1;1;-5;5}
Ta có bảng sau;
x + 2 | -1 | 1 | -5 | 5 |
x | -3 | -1 | -7 | 3 |
y - 1 | 5 | -5 | 1 | -1 |
y | 6 | -4 | 2 | 0 |
(x;y)\(\in\){(-3;6);(-1;-4);(-7;2);(3;0)}
Mà x,y thuộc N
=> (x;y) \(\in\){(3;0)}