3*4+1=?
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1 - \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) = \(\dfrac{4}{4}\) - \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) = \(\dfrac{2}{4}\) = \(\dfrac{1}{2}\)
Vậy a, S còn b, Đ
Thực hiện phép trừ rồi điền kết quả vào chỗ trống.
4 - 1 = 3 4 - 2 = 2 3 + 1 = 4 1 + 2 = 3
3 - 1 = 2 3 - 2 = 1 4 - 3 = 1 3 - 1 = 2
2 - 1 = 1 4 - 3 = 1 4 - 1 = 3 3 - 2 = 1
Lời giải chi tiết:
4 + 1 > 4 | 5 – 1 < 5 | 3 + 0 = 3 |
4 + 1 = 5 | 5 – 0 = 5 | 3 + 1 = 4 |
4 – 1 < 4 | 3 + 1 > 3 | 3 + 1 < 5 |
Lời giải chi tiết:
2 < 4 – 1 | 3 – 2 < 3 – 1 |
3 = 4 – 1 | 4 – 1 > 4 – 2 |
4 > 4 – 1 | 4 – 1 = 3 + 0 |
a)\(\dfrac{2}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right).\dfrac{4}{5}=1.\dfrac{4}{5}=\dfrac{4}{5}\)
b)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{4}{5}\times1=\dfrac{4}{5}\)
b) \(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{3}=\dfrac{4}{15}\)
c) \(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}\)
d) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)
a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)
d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)
Lời giải chi tiết:
4 – 1 > 2 | 4 – 3 < 4 – 2 |
4 – 2 = 2 | 4 – 1 < 3 + 1 |
3 – 1 = 2 | 3 – 1 > 3 – 2 |
Lời giải chi tiết:
4 – 1 > 2 | 4 – 3 < 4 – 2 |
4 – 2 = 2 | 4 – 1 < 3 + 1 |
3 – 1 = 2 | 3 – 1 > 3 – 2 |
a)Ta có :
\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)
\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)
\(3A=1-\dfrac{1}{4^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)
~ Chúc bn học tốt ~
3*4+1=12+1
=13
3 x 4 + 1
= 12 + 1
= 13