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\(\left(x-1\right)^{10}=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
1/ tính :
a/ A = 341 . 67 + 341 . 16 + 659 . 83
A = 341 . ( 67 + 16 ) + 659 . 83
A = 341 . 83 + 659 . 83
A = 83 . ( 341 + 659 )
A = 83 . 1000
A = 83 000
b/ B = 42 . 53 + 47 . 156 - 47 . 114
B = 42 . 53 + 47 . ( 156 - 114 )
B = 42 . 53 + 47 . 42
B = 42 . ( 53 + 47 )
B = 42 . 100
B = 4 200
2/ thu gọn tổng :
A = 3 + 32 + 33 + ... + 3100
3A = 3^2 + 3^3 + 3^4 + ...+ 3^101
3A - A = ( 3^2 + 3^3 + 3^4 + ...+ 3^101 ) - ( 3 + 32 + 33 + ... + 3100 )
2A = 3^101 - 3
A = 3^101 - 3 / 2
Bài 1:
\(A=341.67+341.16+659.83.\)
\(=341.\left(67+16\right)+659.83\)
\(=341.83+659.83\)
\(=83.\left(341+659\right)\)
\(=83.1000=83000\)
\(B=42.53+47.156-47.114\)
\(=42.53+47.\left(156-114\right)\)
\(=42.53+47.42\)
\(=42.\left(47+53\right)\)
\(=42.100=4200\)
Bài 2:
\(A=3+3^2+3^3+3^4+....+3^{100}\)
\(\Rightarrow3A=3^2+3^3+3^4+3^5+...+3^{101}\)
\(2A=3A-A=\left(3^2+3^3+3^4+3^5+....+3^{101}\right)-\left(3+3^2+3^3+3^4+....+3^{100}\right)\)
\(\Rightarrow2A=3^{101}-3\)
\(\Rightarrow A=\frac{3^{101}-3}{2}\)
Bài 3:
\(S=1+2+3+4+...+2018\)
\(=\frac{\left[1+2018\right].\left[\left(2018-1\right)+1\right]}{2}\)
\(=\frac{2019.2018}{2}=2037171\)
\(P=1+3+5+7+...+2017\)
\(=\frac{\left[2017+1\right].\left[\left(2017-1\right):2+1\right]}{2}\)
\(=\frac{2018.1009}{2}\)
\(=1018081\)
Bài 4:
\(\text{Vì ab = 0 }\)\(\Rightarrow\)\(a=0\)\(\text{hoặc}\)\(b=0\)
\(\text{Th1 : ( a = 0)}\)
\(a+4b=16\)
\(0+4b=16\)
\(4b=16\Leftrightarrow b=4\)
\(\text{Th2: ( b = 0)}\)
\(a+4b=16\)
\(a+4.0=16\)
\(a+0=16\Leftrightarrow a=16\)
\(\text{Vậy :}\)\(a;b\in\left\{0;4\right\};\left\{16;0\right\}\)
Bài 5:
\(A=\frac{10^2+11^2+12^2}{13^2+14^2}=\frac{365}{365}=1\)
\(B=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(12.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{12^2.2^{32}}{11.2^{13}.4^{11}-16^9}=....=2\)
a)2019-5{70+9[37-(32+1)]}=2019-5{70+9[37-33]}=2019-5{70+9.4}=2019-5.106=2019-530=1489
b)3[25-(9+8)]-25(8+1)=3[25-17]-25.9=3.8-225=-201
1 Ta có: 201810 + 20189 = 20189.(2018 + 1) = 20189. 2019
201710 = 20179.2017
=> 201810 + 20189 > 201710
2. A = 1 + 2 + 22 + 23 + ... + 2100
2A = 2(1 + 2 + 22 + 23 + ... + 2100)
2A = 2 + 22 + 23 + ... + 2101
2A - A = (2 + 22 + 23 + ... + 2101) - (1 + 2 + 22 +. ... + 2100)
A = 2101 - 1
B = 1 + 6 + 11 + 16 + ... + 51
B = (51 + 1)[(51 - 1) : 5 + 1] : 2
B = 52. 11 : 2
B = 286