7 mũ9 :7 mũ7 -3 mũ2 +2 mũ3 . 25
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A = 1+2+22+...+210
=> 2A = 2+22+23+...+211
=> 2A - A = (2+22+23+...+211) - (1+2+22+...+210)
=> A = 211 - 1
B = 1+3+32+...+310
=> 3B = 3+32+33+...+311
=> 3B - B = (3+32+33+...+311) - (1+3+32+...+310)
=> 2B = 311 - 1
=> B = \(\frac{3^{11}-1}{2}\)
A = 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10
2A = 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11
2A - A = ( 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11 )
- ( 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10 )
A = 2 11 - 1
A = 2047
B = 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10
3B = 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11
3B - B= ( 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11 )
- ( 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10 )
2B = 3 11 - 1
B = \(\frac{3^{11}-1}{2}\)
B = 88573
Vì số hạng nào cũng \(⋮\) 2 nên tổng \(⋮\) 2
2 + 22 + 23 + 24 + ... + 29 + 210
= 2(1 + 2) + 22(1 + 2) + ... + 29(1 + 2)
= 3(2 + 22 + ... + 29) \(⋮\) 3
Vậy, tổng đó chia hết có 2 và 3
S = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
= (1 + 2) + (22 + 23) + (24 + 25) + (26 + 27)
= (1 + 2) + 22(1 + 2) + 24(1 + 2) + 26(1 + 2)
= (1 + 2)(1 + 22 + 24 + 26)
= 3(1 + 22 + 24 + 26) \(⋮3\)(ĐPCM)
2S = 1 + 2 + 22 + 23 + 24 + 25 + 26 + 27
S = (1+2 ) + (22 + 23 ) + (24 + 25 ) + (26 +27)
S = 3 + 22(1+2) + 24(1+2) + 26(1+2)
S = 3+22.3 + 24.3 + 26 .3
S = 3(1+22 + 24 + 26 ) \(⋮\) 3
=> đpcm
THEO ĐỀ BÀI TA CÓ
1^2+2^2+3^2+...+10^2=385
MÀ 2^2+4^2+....+20^2=2(1^2+2^2+....+10^2)=2.385=770
VẬY 2^2+2^4+....+20^2=770
1)Tính:
a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)
b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)
c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)
a
\(4^2.2=16.2=32\)
b\(36^2:6^2=36.36:6.6=36.36:36=36\)
c
Giải:
a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)
\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)
\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
Vậy ...
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)
\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)
\(\Leftrightarrow30x-37=4\)
\(\Leftrightarrow30x=41\)
\(\Leftrightarrow x=\dfrac{41}{30}\)
Vậy ...
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)
\(\Leftrightarrow x^3+8-x^3-3=14x\)
\(\Leftrightarrow5=14x\)
\(\Leftrightarrow x=\dfrac{5}{14}\)
Vậy ...
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
\(\Leftrightarrow x^3+1-x^3-3x=2\)
\(\Leftrightarrow1-3x=2\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)
=> \(x^2-2x-x-3x+7+9x=6\)
=> \(x^2-2x-x^2-3x+7+9x=6\)
=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)
=> \(4x=-1\)
Vậy \(x=\dfrac{-1}{4}\)
b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)
=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)
=> \(42x=43\)
Vậy \(x=\dfrac{43}{42}\)
c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)
=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)
=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)
=> \(5=14x\)
Vậy \(x=\dfrac{5}{14}\)
d) \(\left(x^2-x+1\right)\left(x+1\right)-x^3-3x=2\)
=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)
=>\(\left(x^3-x^3\right)+\left(-x^2+x^2\right)+\left(x-x-3x\right)=2-1\)
=> \(-3x=1\)
Vậy \(x=\dfrac{-1}{3}\)
=49-9+200=240
=49-9+200=240