Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
k) \(2x-49=5.3^2\)
\(2x-49=45\)
\(2x=49+45\)
\(2x=94\)
\(x=47\)
l) \(3^2.\left(x+14\right)-5^2=5.2^2\)
\(9.\left(x+14\right)-25=20\)
\(9.\left(x+14\right)=45\)
\(x+14=5\)
\(x=-9\)
m) \(6x+x=5^{11}:5^9+3^1\)
\(7x=5^{11-9}+3\)
\(7x=5^2+3\)
\(7x=28\)
\(x=4\)
n) \(7x-x=5^{21}:5^{19}+3.2^2-\left(-7^{-0}\right)\)
\(6x=5^{21-19}+12-1\)
\(6x=5^2+11\)
\(6x=36\)
\(x=6\)
o) \(7x-2x=6^{17}:6^{15}+44:11\)
\(5x=6^{17-15}+4\)
\(5x=6^2+4\)
\(5x=40\)
\(x=8\)
o)7x-x=521:519+3.22-70
=> 6x = 5^2 + 12 -1
=> 6x = 36
=> x = 36/6 = 6
Kết quả 6
Học tốt
a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
a) 6x + x = 511 : 59 + 31
6x + x = 52 + 31
7x = 25 + 3
7x = 28
x = 28 : 7
x = 4
b) 7x - x = 521 : 59 + 3 . 22 - 70
6x = 512 + 3 . 4 - 1
6x = 244140625 + 12 - 1
6x = 244140636
x = 244140636 : 6
x = 40690106
a)6x +x = 511 :59 +31
(6+1)x=52 +3
7x =25 +3=28
x = 28 :7
x=4
Vậy x=4
b) 7x-x = 521 : 59 +3 . 22 - 7
(7-1)x = 512 +3.4-1
6x= 244140625 + 12 - 1
6x = 244140637-1=244140636
x=244140636:6
x=40690106
Vậy x= 40690106
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
Ta có :abcdeg=ab.10000+cd.100+eg
=9999.ab+99.cd+ab+cd+eg
=﴾9999ab+99cd﴿+﴾ab+cd+eg﴿
Vì 9999ab+99cd chia hết cho 11 và ab+cd+eg chia hết cho 11
=>abcdeg chia hết cho 11
Vậy nếu có ab+cd+egchia hết cho 11 thì abcdeg chia hết cho 11
\(\left(10^2-5^2\right)+\left(9^2-4^2\right)+..+\left(6^2-1^2\right)\)
\(\left(10-5\right)\left(10+5\right)+...+\left(6-1\right)\left(6+1\right)\)
\(5.15+5.13+5.11+5.9+5.7\)
\(5\left(15+13+11+9+7\right)\)
\(5.\frac{\left(15+7\right).5}{2}=\frac{5.22.5}{2}=11.25=25+250=275\)
\(3^2\cdot5\left[7:1^{2022}-\left(11-9\right)^2\right]+2022^0\)
\(=9\cdot5\cdot\left(7:1-2^2\right)+1\)
\(=9\cdot5\cdot\left(7-4\right)+1\)
\(=45\cdot3+1\)
\(=135+1\)
\(=136\)