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A=2^0+2^1+2^2+2^3+...+2^2002
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\(\Rightarrow\left(x-4\right)\left(2x+x-4\right)=0\\ \Rightarrow\left(x-4\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)
S=1+2+2^2+2^3+...+2^2002
2S=2+2^2+2^3+2^4+...+2^2003
2S-S=(2+2^2+2^3+2^4+...+2^2003)-(1+2+2^2+2^3+...+2^2002)
2S-S=(2-2)+(2^2-2^2)+(2^3-2^3)+...+(2^2002-2^2002)+(2^2003-1)
S=0+0+0+...+0+(2^2003-1)
S=2^2003-1 (rút gọn)
tick nha
S=1+2+2^2+2^3+...+2^2002
=>2S=2+22+23+24+...+22013
=>2S-S=2+22+23+24+...+22013-1-2-22-23-...-22012
=>S=22013-1
\(a,=\left(x+2-3x\right)\left(x+2+3x\right)=4\left(1-x\right)\left(2x+1\right)\\ b,=25-\left(x+y\right)^2=\left(5-x-y\right)\left(5+x+y\right)\)
\(c,=x^4+2x^2+1-x^2=\left(x^2+1\right)-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
S=30+32+34+36+...+3200
6S=32+34+36+...+3202
6S-S=(32+34+36+...+3202)-(1+32+34+...+3200)
5S=1+(32-32)+(34-34)+...+(3200-3200)+3202
S=(3200+1):5\(\frac{ }{ }\)
\(\text{C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006}\)
\(\text{C=1+(2-3-4+5)+(6-7-8+9)+...+(2002-2003-2004+2005)+2006}\)
\(\text{C=1+0+0+...+0+2006}\)
\(\text{C=1+2006}\)
\(C=2007\)
HỌC TỐT!!!
a: =>\(4\cdot3^x\cdot\dfrac{1}{3}+2\cdot3^x\cdot9=4\cdot3^6+2\cdot3^9\)
=>3^x(4*1/3+2*9)=3^6(4+2*3^3)
=>3^x*58/3=3^6*58
=>3^x/3^6=3
=>x-6=1
=>x=7
b: =>\(2^x\cdot\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)
=>2^x=2^7
=>x=7
A = 20 + 21 + 22 + 23 +....+ 22002
2A = 21 + 22 + 23 + 24 +....+ 22003
2A - A = ( 21 + 22 + 23 + 24 +....+ 22003 ) - ( 20 + 21 + 22 + 23 + .... + 22002 )
A = 22003 - 1
Hk tốt
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