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\(2^{x+3}-3.2^{x+1}=32\)
<=> 2^x.2^3-3.2^x.2=32
<=> 2^x.(2^3-3.2)=32
<=> 2^x.2=32
<=>2^x=16
<=>2^x=2^4
=>x=4
Vậy...
2x+3-3.2x+1=32
= 2x . 23 - 3. 21 .2x = 32
2x . 8 - 6 . 2x = 32
2x . (8 - 6 ) = 32
2x . 2 = 32
2x+1 = 32
vì 32 = 25 => x + 1=5
x = 5-1
x = 4
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\(7\times2^{x+1}-3\times2^{x+1}=32\)
\(\Rightarrow2^{x+1}\left(7-3\right)=32\)
\(\Rightarrow2^{x+1}\times4=32\)
\(\Rightarrow2^{x+1}=32\div4\)
\(\Rightarrow2^{x+1}=8\)
\(\Rightarrow2^{x+1}=2^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1\)
\(\Rightarrow x=2\)
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80 - (4.52 - 3.23) = 210 - (x - 4)
<=> 80 - (100 - 24) = 210 - (x - 4)
<=> 80 - 76 = 210 - (x - 4)
<=> 4 + x - 4 = 210
<=> x = 210
@Đức Đạt Đỗ (Đạt 301 Channel)
80-[2^2(5^2-6)]=2^10-(x-4)
80-(2^2.19)=2^10-x+4
80-76=2^10-x+4
\(\Rightarrow\)2^10-x+4=4
x+4=1024-4=1020
x=1020-4=1016
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2.22 + 3.23 + 4.24 + ... + n.2n = 2n+11
Đặt vế trái là A ta có:
A = 2.22 + 3.23 + 4.24 + ... (n -1).2n-1+ n.2n
2A = 2.23 + 3.24 + 4.25 +....+ (n- 1).2n + n.2n+1
2A-A = [2.23+3.24 + 4.25 +...+(n-1).2n+n.2n+1] - [2.22 + 3.23+...+n.2n]
A = -2.22+ (2.33-3.23) + (3.24 - 4.24) +...+ [(n-1).2n - n.2n ] + n.2n+1
A = -2.22 - 23 - 24 -...- 2n + n.2n+1
Đặt B = -2.22 - 23 - 24 - ... - 2n
2B = -2.23 - 24 - 25 -...-2n+1
2B - B = (-2.23 - 24 - 25 -..-2n+1) - (-2.22-23-24-..-2n)
B = -24 -24 - 25 -..2n-2n+1 + 23 + 23 + 24+ 25+ 2n
B = (-24 + 23) + (- 2n+1 + 23) +(-24+24)+(-25+25)+(-2n+2n)
B = -16 + 8 - 2n+1 + 8
B = (-16 + 8 + 8 ) - 2n+1
B = - 2n+1
A = n.2n+1 - 2n+1
Theo bài ra ta có:
n.2n+1 - 2n+1 = 2n+11
n.2n+1 - 2n+1 - 2n+11 = 0
2n+1.(n - 1 - 210) = 0
Vì n là số tự nhiên nên 2n+1 > 0
Vậy 2n+1.(n - 1- 210) = 0 ⇔ n - 1 - 210 = 0 ⇒ n = 1 + 210 ⇒ n = 1025
Vậy n = 1025
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a ) sai đầu bài
b) 15. { 32 : [ 6 - 5 + 5 ( 9: 3 )]}
= 15 . { 32 : [ 6 - 5 + 5 . 3 ]}
= 15 . { 32 : [ 6 - 5 + 15 ] }
= 15 . { 32 : [ 1 +15 ] }
= 15 . { 32 : 16 }
= 15 . 2
= 30
bài 2:
a) 14
b)1024
Bài1:
b)15.{32:[6-5+5(9:3)]}
=15.{32:[6-5+5.3]}
=15.{32:[6-5+15]}
=15.{32:[1+15]}
=15.{32:16}
=15.2
=30
\(2^m:2+3.2^{m-1}=32\)
\(\Leftrightarrow2^{m-1}+3.2^{m-1}=32\)
\(\Leftrightarrow2^{m-1}.\left(1+3\right)=32\)
\(\Leftrightarrow2.^{m-1}.4=32\)
\(\Leftrightarrow2^{m-1}=32:4=8=2^3\)
\(\Leftrightarrow m-1=3\Leftrightarrow m=4\)