So sánh:
a/ 23 với 32
b/ ( 23) 2019 với (32) 2019
Giúp mình với ạ!! Mình tick cho!!
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Khách
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NT
1
Những câu hỏi liên quan
5 tháng 6 2023
\(\dfrac{7}{19}x\dfrac{8}{23}+\dfrac{7}{19}x\dfrac{15}{23}+1\dfrac{7}{19}\)
= \(\dfrac{7}{19}x\left(\dfrac{8}{23}+\dfrac{15}{23}\right)+1+\dfrac{7}{19}\)
=\(\dfrac{7}{19}x1+1+\dfrac{7}{19}\)
= \(\dfrac{7}{19}+1+\dfrac{7}{19}=1\dfrac{14}{19}\) = \(\dfrac{33}{19}\)
\(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{3}{21}-\dfrac{17}{32}\)
= \(\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{17}{32}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{49}{32}-\dfrac{17}{32}\right)\)
= 1 + 1 + 1 = 3
\(\dfrac{8}{9}x\dfrac{15}{16}x\dfrac{24}{25}x\dfrac{35}{36}x\dfrac{48}{49}x\dfrac{63}{64}\)
= \(\dfrac{3}{4}\) *Câu này bạn tự sử dụng gạch nhé!
5 tháng 6 2023
`1,`
`a,`
`7/19 \times 8/23 + 7/19 \times 15/23 + 1 7/19`
`= 7/19 \times 8/23 + 7/19 \times 15/23 + 1 + 7/19`
`= 7/19 \times (8/23 + 15/23 + 1) + 1`
`= 7/19 \times 2 + 1`
`=14/19 + 1`
`= 33/19`
`b,`
`75/100 + 18/21 + 49/32 + 1/4 + 3/21 - 17/32`
`= 75/100 + (18/21 + 3/21) + (49/32 - 17/32) + 1/4`
`= 0,75 + 1 + 1 + 0,25`
`= (0,75 + 0,25) + 1 + 1`
`= 1+1+1=3`
`c,`
`8/9 \times 15/16 \times 24/25 \times 35/36 \times 48/49 \times 63/64`
`=` \(\dfrac{2\times3}{3\times3}\times\dfrac{3\times5}{4\times4}\times\dfrac{3\times4\times2}{5\times5}\times\dfrac{5\times7}{6\times6}\times\dfrac{6\times8}{7\times7}\times\dfrac{7\times9}{8\times8}\)
`= 3/4` (bạn sử dụng gạch, rút gọn các số là được nhé).
DV
2
14 tháng 10 2023
Đặt \(A=1+5^2+5^4+...+5^{40}\)
\(\Rightarrow25A=5^2+5^4+5^6+...+5^{42}\)
Lấy \(25A-A=\left(5^2+5^4+5^6+...+5^{42}\right)-\left(1+5^2+5^4+...+5^{40}\right)\)
\(\Rightarrow24A=5^{42}-1\)
\(\Rightarrow A=\dfrac{5^{42}-1}{24}\)
ND
Nguyễn Đăng Nhân
CTVHS
15 tháng 9 2023
a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)
b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)
c) \(27^{40}=3^{3^{40}}=3^{120}\)
\(64^{60}=8^{2^{60}}=8^{120}\)
Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)
PN
8 tháng 8 2020
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
T
2 tháng 1
\(A=2+2^2+2^3+\dots+2^{60}\\2A=2^2+2^3+2^4+\dots+2^{61}\\2A-A=(2^2+2^3+2^3+\dots+2^{61})-(2+2^2+2^3+\dots+2^{60})\\A=2^{61}-2\)
Ta thấy: \(2^{61}-2< 2^{61}\)
\(\Rightarrow A< B\)
2 tháng 1
A=2+22+23+...+260
\(\Rightarrow\)2A=22+23+24+...+261
\(\Rightarrow\)2A-A=(22+23+24+...+261)-(2+22+2324+...+260)
\(\Rightarrow\)A=261-2
Mà 261-2<261 nên A<B
Vậy A<B
AH
Akai Haruma
Giáo viên
28 tháng 12 2023
Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$
$\frac{-5}{9}=\frac{-35}{63}$
Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$
$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$
---------
b.
$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$
$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$
$\Rightarrow -0,625> \frac{-19}{50}$
c.
$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$
22 tháng 7 2018
\(\text{a.61 x 23 + 39 x 22}\)
\(=61.22+1+39.22\)
\(=22.\left(61+39\right)+1\)
\(=22.100+1\)
\(=2200+1\)
\(=2201\)
mình chỉ làm được 1 câu thôi
hok tốt
LT
Lương Thị Vân Anh
CTVHS
25 tháng 7 2023
Ta có \(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\)
\(2A-A=\left(1+\dfrac{2}{2}+\dfrac{3}{2^2}+...+\dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\)\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\)
Đặt B = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
2B = \(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
2B - B = \(\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)B = 2 - \(\dfrac{1}{2^{2022}}\)
Suy ra A = 2 - \(\dfrac{1}{2^{2022}}\) - \(\dfrac{2023}{2^{2023}}\) < 2
Vậy A < 2
25 tháng 7 2023
\(A=\dfrac{1}{2}+\dfrac{2}{2^{2}}+\dfrac{3}{2^{3}}+...+\dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\)
\(2A=1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\\2A-A=\left(1+\dfrac22+\dfrac3{2^2}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2021}}+\dfrac{2023}{2^{2022}}\right)-\left(\dfrac12+\dfrac2{2^2}+\dfrac3{2^3}\ +\,.\!.\!.+\ \dfrac{2022}{2^{2022}}+\dfrac{2023}{2^{2023}}\right)\\A=1+\dfrac12+\dfrac1{2^3}\ +\,.\!.\!.+\ \dfrac1{2^{2021}}+\dfrac1{2^{2022}}-\dfrac{2023}{2^{2023}}\\2\left(A+\dfrac{2023}{2^{2023}}\right)=2+1+\dfrac12+\dfrac1{2^2}\ +\,.\!.\!.+\ \dfrac1{2^{2020}}+\dfrac1{2^{2021}}\\A+\dfrac{2023}{2^{2023}}=2-\dfrac1{2^{2022}}\\A=2-\dfrac1{2^{2022}}+\dfrac{2023}{2^{2023}}<2\)
VE
2
4 tháng 7 2021
\(8^2=64=32+2\sqrt{16^2}\)
\(\left(\sqrt{15}+\sqrt{17}\right)^2=32+2\sqrt{15.17}=32+2\sqrt{\left(16-1\right)\left(16+1\right)}\)
\(=32+2\sqrt{16^2-1}\)
\(< =>8^2>\left(\sqrt{15}+\sqrt{17}\right)^2\)
\(8>\sqrt{15}+\sqrt{17}\)
\(\left(\sqrt{2019}+\sqrt{2021}\right)^2=4040+2\sqrt{2019.2021}\)
\(=4040+2\sqrt{\left(2020-1\right)\left(2020+1\right)}=4040+2\sqrt{2020^2-1}\)
\(\left(2\sqrt{2020}\right)^2=8080=4040+2\sqrt{2020^2}\)
\(< =>\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)
4 tháng 7 2021
mik chọn điền
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mik lười chép ại đề bài
a, Ta có : \(2^3=8\)
\(3^2=9\)
Mà \(8< 9\) \(\Rightarrow\) \(2^3< 3^2\)
b, Ta có : \(\left(2^3\right)^{2019}=8^{2019}\)
\(\left(3^2\right)^{2019}=9^{2019}\)
Mà \(8^{2019}< 9^{2019}\) \(\Rightarrow\) \(\left(2^3\right)^{2019}< \left(3^2\right)^{2019}\)