1 3/4 = ......../............
đúng ghi Đ sai ghi S
a)1-3/4+1/4=1-3+1/4 b)1-3/4+1/4=4/4-3/4+1/4
=1-1=0[ ] 1/4+1/4=1/2[ ]
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
Tính:
4 - 1 = 4 - 2 = 3 + 1 = 1 + 2 =
3 - 1 = 3 - 2 = 4 - 3 = 3 - 1 =
2 - 1 = 4 - 3 = 4 - 1 = 3 - 2 =
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
>, <, = ?
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:
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 |
Dễ nhỉ !
Ai đồng tình thì cho mình nha !
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 |
>, <, =?
2 … 4 – 1 | 3 – 2 … 3 – 1 |
3 … 4 – 1 | 4 – 1 …4 – 2 |
4 … 4 – 1 | 4 – 1 … 3 + 0 |
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 |
< <
= >
câu trả lời đây
Tính (13+1/4).(33+1/4).(53+1/4).....(20153+1/4) / (23+1/4).(43+1/4).....(20163+1/4)
1.tính thuận tiện
a. 2/3 x 4/5 + 1/3 x 4/5
b. 2/3 x 4/5 - 1/3 x 4/5
c. 1/2 : 3/4 + 1/6 : 3/4
d. 1/2 : 3/4 - 1/6 : 3/4
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) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)
c)\(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}+\dfrac{1}{6}.\dfrac{4}{3}=\left(\dfrac{1}{2}+\dfrac{1}{6}\right).\dfrac{4}{3}=\dfrac{2}{3}.\dfrac{4}{3}=\dfrac{8}{9}\)
d)\(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}-\dfrac{1}{6}.\dfrac{4}{3}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right).\dfrac{4}{3}=\dfrac{1}{3}.\dfrac{4}{3}=\dfrac{4}{9}\)
a,7/4+-3/5
b,2021-(1/3)^2 x 3^2
c,7,5 x(-3/5)
d,(-1/4)^2 x 4/11+7/11 x(-1/4)^2
e, A= 1/4 +1/4^2+1/4^3+...+1/4^2020<1/3
a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)
d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)
>, <, = ?
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 |
4-1 > 2
4-2 = 2
4-3 < 4-2
4-1 < 3+1
4 – 1 > 2 | 4 – 3 < 4 – 2 |
4 – 2 = 2 | 4 – 1 < 3 + 1 |
3 – 1 = 2 | 3 – 1 > 3 – 2 |
>, <, = ?
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 ) A = 1/4 + 1/4^2 +1/4^3 +.........+ 1/4^100 + 1/3.4^100
b) B = 1/3 - 1/3^2 + 1/3^3 - 1/3^4 +.........+ 1/ 3^99
a)Ta có :
\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)
\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)
\(3A=1-\dfrac{1}{4^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)
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