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2X- 1= A.
Dãy A có số các số hạng là:
( 2112- 20): 1+ 1= 2093( số hạng)
Tổng dãy A là:
( 2112+ 20)x 2093: 2= 2231138
=> 2X- 1= 2231138.
=> 2X= 2231138+ 1.
=> 2X= 2231139.
=> X= \(\frac{2231139}{2}\).
Vậy X= \(\frac{2231139}{2}\).
a) \(x+5=20-\left(12-7\right)\)
\(\Rightarrow x+5=20-5\)
\(\Rightarrow x+5=15\)
\(\Rightarrow x=15-5\)
\(\Rightarrow x=10\)
b) \(15-\left(3+2x\right)=2^2\)
\(\Rightarrow3+2x=15-4\)
\(\Rightarrow3+2x=11\)
\(\Rightarrow2x=11-3\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=\dfrac{8}{2}\)
\(\Rightarrow x=4\)
c) \(-11-\left(19-x\right)=50\)
\(\Rightarrow19-x=-11-50\)
\(\Rightarrow19-x=-61\)
\(\Rightarrow x=61+19\)
\(\Rightarrow x=80\)
d) \(159-\left(25-x\right)=43\)
\(\Rightarrow25-x=159-43\)
\(\Rightarrow25-x=116\)
\(\Rightarrow x=25-116\)
\(\Rightarrow x=-91\)
e) \(\left(79-x\right)-43=-\left(17-52\right)\)
\(\Rightarrow\left(79-x\right)-43=52-17\)
\(\Rightarrow79-x-43=35\)
\(\Rightarrow36-x=35\)
\(\Rightarrow x=1\)
f) \(\left(7+x\right)-\left(21-13\right)=32\)
\(\Rightarrow7+x-8=32\)
\(\Rightarrow x-1=32\)
\(\Rightarrow x=32+1\)
\(\Rightarrow x=33\)
g) \(-x+20=-15+8+13\)
\(\Rightarrow-x+20=6\)
\(\Rightarrow x=20-6\)
\(\Rightarrow x=14\)
h) \(-\left(-x+13-142\right)+18=55\)
\(\Rightarrow x-13+142+18=55\)
\(\Rightarrow x+147=55\)
\(\Rightarrow x=55-147\)
\(\Rightarrow x=-92\)
x + 20 + 21 + x + 22 + 23 + x + 24 + 25 + x + 26 + 27 + x + 28 + 29 + x + 30 = 330
6x + (30 + 20) . (30 - 20 + 1) : 2 = 330
6x + 50 . 11 : 2 = 330
6x + 275 = 330
6x = 330 - 275
6x = 55
x = 55 : 6
x = 55/6
\(x+20+21+x+22+23+x+24+25+x+26+27+x+28+29+x+30=330\)
\(\Rightarrow\left(x+x+x+x+x+x\right)+\left(20+21+22+23+24+25+26+27+28+29+30\right)=330\)
\(\Rightarrow6x+\left[\left(30-20\right):1+1\right]\left(20+30\right):2=330\)
\(\Rightarrow6x+11.50:2=330\)
\(\Rightarrow6x+275=330\)
\(\Rightarrow6x=55\)
\(\Rightarrow x=\dfrac{55}{6}\)
x + 20 + 21 + x + 22 + 23 + x + 24 + 25 + x + 26 + 27 + x + 28 + 29 + x + 30 = 330
6x + (30 + 20) . (30 - 20 + 1) : 2 = 330
6x + 50 . 11 : 2 = 330
6x + 275 = 330
6x = 330 - 275
6x = 55
x = 55 : 6
x = 55/6
\(A=2^0+2^1+2^2+...+2^{111}+2^{112}\)
=> \(2A=2^1+2^2+2^3+...+2^{112}+2^{113}\)
=> \(2A-A=\left(2^1+2^2+...+2^{113}\right)-\left(2^0+2^1+....+2^{112}\right)\)
=> \(A=2^{113}-1\)
Vậy \(x=113\)
\(A=2^0+2^1+2^2+...+2^{111}+2^{112}\left(1\right)\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{112}+2^{113}\left(2\right)\)
Trừ vế với vế (2) cho (1) ta được
\(2A-A=2^{113}-2^0\)
\(\Leftrightarrow A=2^{113}-1\)
\(\Rightarrow2^x-1=2^{113}-1\)
\(\Leftrightarrow x=113\)
Vậy \(x=113\)