2mu17+15mu4).(3mu19-2mu17).(2mu4-4mu2)
5mu6 :5mu3+2mu2 .2mu3
19 mu7 : 19mu5+4.4mu2
2mu5. 7+3mu19 :3 mứ-2018
17.3mu2 -5mu15 :5mu13+39mu0
`@` `\text {Ans}`
`\downarrow`
`5^6 \div 5^3 + 2^2*2^3`
`= 5^3 + 2^5`
`= 125+32`
`=``157`
`19^7 \div 19^5 + 4*4^2`
`= 19^(7-5) + 4^(1+2)`
`= 19^2 + 4^3`
`= 361 + 64`
`=``425`
\(2^5\cdot7+3^{2019}\div3^{2018}?\)
`= 224 + 3`
`= 227`
`17*3^2 - 5^15 \div 5^13 + 39^0`
`= 153 - 5^2 + 1`
`= 153 - 125 + 1`
`= 153 - 124`
`=29`
Cho A=1/2mu2+1/3mu2+1/4mu2+...+1/100mu2
B=1/4mu2+1/6mu2+1/8mu2+...+1/200mu2
Tính A/B
4mu2:4.3mu3-2mu2+7
42:4.33-22+7
⇒ 4.27-4+7
⇒ 108-4+7
⇒ 111
4\(^2\) : 4.3\(^3\) - 2\(^2\) + 7
= 16 : 4 . 27 - 4 + 7
= 4 . 27 - 4 + 7
= 108 - 4 + 7
= 104 + 7
= 111
-Tick cho mình nha mình cảm ơn ạ. <3
(x-140):35=3mu2+4mu2
(x-140):35=3^2+4^2
(x-140):35=25
x-140=25x35=875
x=875+140=1015
A=4+2mu2+2mu3+2mu4+...+2mu20
A= 4+2.2+2.2.2+2.2.2.2+.......+{2.2.2.2.2.....} có 20 thừa số 2
Có số số hạng ở trong khoảng số 2 là:
(20-2)+1=19(số)
Có 20 thừa số 2 suy ra:20.2=40
Tổng là:
(40+2)*19:2=399
A=4+399
A=403
**** nhé Hương Linh xinh xắn
[3x-2mu4]*7mu5=27mu6*1/2020mu0
A=2mu2+4mu2+6mu2+...+(2K)mu2
3mu4 .5mu2 -128.2mu3 +1 mu 17
24-4mu2:4.2+ 3
`@` `\text {Ans}`
`\downarrow`
`3^4*5^2 - 128*2^3 + 1^17`
`= 9^2*5^2 - 2^7*2^3 + 1`
`= (9*5)^2 - 2^10+1`
`= 45^5-2^10 + 1`
`= 2025 - 1024 + 1`
`= 2025 - 1023`
`= 1002`
\(24-4^2\div4\cdot2+3\)
`= 24 - 4*2 + 3`
`= 24 - 8 + 3`
`= 24 - 5`
`= 19`
`@` `\text {Kaizuu lv u}.`
1440/[41-(x-5)=2mu4 *3
\(\frac{1440}{41-\left(x-5\right)}=2^4.3\)
\(\Rightarrow\left[41-\left(x-5\right)\right]2^4.3=1440\)
\(\Rightarrow\left[41-\left(x-5\right)\right].48=1440\)
\(\Rightarrow41-\left(x-5\right)=30\)
\(\Rightarrow x-5=11\)
\(\Rightarrow x=16\)
Vậy \(x=16\)