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CC
2
27 tháng 8
Câu 8:
a:Sửa đề: \(4+4^2+\cdots+4^{2025}\)
Ta có: \(4+4^2+\cdots+4^{2025}\)
\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+\cdots+\left(4^{2023}+4^{2024}+4^{2025}\right)\)
\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+\cdots+4^{2023}\left(1+4+4^2\right)\)
\(=21\left(4+4^4+\cdots+4^{2023}\right)\) ⋮21
b: \(5+5^2+5^3+5^4+\cdots+5^{2024}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{2023}+5^{2024}\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+\cdots+5^{2022}\left(5+5^2\right)\)
\(=30\left(1+5^2+\cdots+5^{2022}\right)\) ⋮30
Câu 7:
a: \(A=2+2^2+2^3+\cdots+2^{99}\)
=>\(2A=2^2+2^3+\cdots+2^{100}\)
=>\(2A-A=2^2+2^3+\cdots+2^{100}-2-2^2-\cdots-2^{99}\)
=>\(A=2^{100}-2\)
b: \(B=1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(7B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}\)
=>\(7B+B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}+1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(8B=-7^{50}+1\)
=>\(B=\frac{-7^{50}+1}{8}\)
Câu 4:
a: \(x^3=125\)
=>\(x^3=5^3\)
=>x=5
b: \(11^{x+1}=121\)
=>\(11^{x+1}=11^2\)
=>x+1=2
=>x=2-1=1
c: \(\left(x-5\right)^3=27\)
=>\(\left(x-5\right)^3=3^3\)
=>x-5=3
=>x=3+5=8
d: \(4^5:4^{x}=16\)
=>\(4^{x}=4^5:16=4^5:4^2=4^3\)
=>x=3
e: \(5^{x-1}\cdot8=1000\)
=>\(5^{x-1}=1000:8=125=5^3\)
=>x-1=3
=>x=3+1=4
f: \(2^{x}+2^{x+3}=72\)
=>\(2^{x}+2^{x}\cdot8=72\)
=>\(2^{x}\cdot9=72\)
=>\(2^{x}=\frac{72}{9}=8=2^3\)
=>x=3
g: \(\left(3x+1\right)^3=343\)
=>\(\left(3x+1\right)^3=7^3\)
=>3x+1=7
=>3x=6
=>x=2
h: \(3^{x}+3^{x+2}=270\)
=>\(3^{x}+3^{x}\cdot9=270\)
=>\(10\cdot3^{x}=270\)
=>\(3^{x}=\frac{270}{10}=27=3^3\)
=>x=3
i: \(25^{2x+4}=125^{x+3}\)
=>\(\left(5^2\right)^{2x+4}=\left(5^3\right)^{x+3}\)
=>\(5^{4x+8}=5^{3x+9}\)
=>4x+8=3x+9
=>x=1
Câu 6:
1 giờ=3600 giây
Số tế bào hồng cầu được tạo ra sau mỗi giờ là:
\(25\cdot10^5\cdot3600=25\cdot36\cdot10^7=900\cdot10^7=9\cdot10^9\) =9 tỉ (tế bào)
S
subjects
CTVHS
28 tháng 8
câu 5:
a. \(16^{16}=\left(2^4\right)^{16}=2^{64}\)
\(64^{11}=\left(2^6\right)^{11}=2^{66}\)
vì \(2^{66}>2^{64}\) nên \(64^{11}>16^{16}\)
b. \(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}\)
\(5^{20}<5^{21}\Rightarrow625^5<125^7\)
c. \(3^{36}=\left(3^3\right)^{12}=27^{12}\)
\(5^{24}=\left(5^2\right)^{12}=25^{12}\)
\(27^{12}>25^{12}\Rightarrow3^{36}>5^{24}\)
16 tháng 8
Ta có: \(10A=\frac{10^{21}-60}{10^{21}-6}=\frac{10^{21}-6-54}{10^{21}-6}=1-\frac{54}{10^{21}-6}\)
\(10B=\frac{10^{22}-60}{10^{22}-6}=\frac{10^{22}-6-54}{10^{22}-6}=1-\frac{54}{10^{22}-6}\)
Ta có: \(10^{21}-6<10^{22}-6\)
=>\(\frac{54}{10^{21}-6}>\frac{54}{10^{22}-6}\)
=>\(-\frac{54}{10^{21}-6}<-\frac{54}{10^{22}-6}\)
=>\(-\frac{54}{10^{21}-6}+1<-\frac{54}{10^{22}-6}+1\)
=>10A<10B
=>A<B
16 tháng 8
Ta có: \(\frac{A}{10^{10}}=\frac{10^{20}-6}{10^{20}-6\cdot10^{10}}=\frac{10^{20}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}\)
\(\frac{B}{10^{10}}=\frac{10^{21}-6}{10^{21}-6\cdot10^{10}}=\frac{10^{21}-6\cdot10^{10}+6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}=1+\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
Ta có: \(10^{20}<10^{21}\)
=>\(10^{20}-6\cdot10^{10}<10^{21}-6\cdot10^{10}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}\)
=>\(\frac{6\left(10^{10}-1\right)}{10^{20}-6\cdot10^{10}}+1>\frac{6\left(10^{10}-1\right)}{10^{21}-6\cdot10^{10}}+1\)
=>\(\frac{A}{10^{10}}>\frac{B}{10^{10}}\)
=>A>B
23 tháng 8
bài 3:
a: \(C=5+5^2+5^3+\cdots+5^{20}\)
\(=5\left(1+5+5^2+\cdots+5^{19}\right)\) ⋮5
b: \(C=5+5^2+5^3+\cdots+5^{20}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{19}+5^{20}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{19}\left(1+5\right)\)
\(=6\left(5+5^3+\cdots+5^{19}\right)\) ⋮6
c: \(C=5+5^2+5^3+\cdots+5^{20}\)
\(=\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+\cdots+\left(5^{17}+5^{18}+5^{19}+5^{20}\right)\)
\(=5\left(1+5+5^2+5^3\right)+5^5\left(1+5+5^2+5^3\right)+\cdots+5^{17}\left(1+5+5^2+5^3\right)\)
\(=\left(1+5+5^2+5^3\right)\left(5+5^5+\cdots+5^{17}\right)=156\cdot\left(5+5^5+\cdots+5^{17}\right)\)
\(=13\cdot12\cdot\left(5+5^5+\cdots+5^{17}\right)\) ⋮13
Bài 2:
a: \(B=3+3^2+3^3+\cdots+3^{120}\)
\(=3\left(1+3+3^2+3^3+\cdots+3^{119}\right)\) ⋮3
b: \(B=3+3^2+3^3+\cdots+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+\cdots+3^{119}\left(1+3\right)\)
\(=4\left(3+3^3+\cdots+3^{119}\right)\) ⋮4
c: \(B=3+3^2+3^3+\cdots+3^{120}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+\cdots+3^{118}\left(1+3+3^2\right)\)
\(=13\left(3+3^4+\cdots+3^{118}\right)\) ⋮13
Bài 1:
a: \(A=2+2^2+2^3+\ldots+2^{20}\)
\(=2\left(1+2+2^2+\cdots+2^{19}\right)\) ⋮2
b: \(A=2+2^2+2^3+\ldots+2^{20}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{19}\left(1+2\right)\)
\(=3\left(2+2^3+\cdots+2^{19}\right)\) ⋮3
c: \(A=2+2^2+2^3+\ldots+2^{20}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{17}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+\ldots+2^{17}\right)=5\cdot3\cdot\left(2+2^5+\cdots+2^{17}\right)\) ⋮5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 8
Bài 1:
a; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)
A = 2 x (1+ 2+ 2\(^2\) + ... + 2\(^{19}\))
A ⋮ 2(đpcm)
b; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)
Xét dãy số: 1; 2;...; 20 đây là dãy số cách đều với khoảng cách là:
2 - 1 = 1
Số số hạng của dãy số trên là:
(20 - 1) : 1+ 1 = 20(số)
Vì 20 : 2 = 10
Vậy nhóm hai số hạng liên tiếp của A vào nhau khi đó ta có:
A = (2+ 2\(^2\)) + (2\(^3\) + 2\(^4\)) + ... + (2\(^{19}+\) 2\(^{20}\))
A = 2.(1 + 2) + 2\(^3\).(1+ 2) + ... + 2\(^{19}\) .(1 + 2)
A = 2.3 + 2\(^3\).3 + ... + 2\(^{19}\).3
A = 3.(2+ 2\(^3\) + ... + 2\(^{19}\))
A ⋮ 3 (đpcm)
c; A = 2 + \(2^2\) + 2\(^3\) + ... + 2\(^{20}\)
Xét dãy số: 1; 2; 3;...; 20
Dãy số trên có 20 số hạng:
Vì 20 : 4 = 5
Vậy nhóm 4 hạng tử của A thành một nhóm khi đó:
A = (2+ 2\(^2\) + 2\(^3\) + 2\(^4\)) + ... + (2\(^{17}+2^{18}+2^{19}+2^{20}\))
A = 2.(1 + 2 + 2\(^2\) + 2\(^3\)) + ... + 2\(^{17}\).(1 + 2 + 2\(^2\) + 2\(^3\))
A = (1+ 2 +2\(^2\) + 2\(^3\)).(2+ ...+ 2\(^{17}\))
A = (1 + 2 + 4 + 8).(2+ ...+ 2\(^{17}\))
A = (3+ 4 + 8).(2+ ...+ 2\(^{17}\))
A = (7 + 8)(2+ ...+ 2\(^{17}\))
A = 15.(2+ ...+ 2\(^{17}\))
A ⋮ 5(đpcm)
GD
Mình cảm ơn ạ
2
NQ
Nguyễn Quốc Đạt
CTVVIP
14 tháng 10 2023
a)
\(175\cdot19+38\cdot175+43\cdot175\\ =175\cdot19+175\cdot38+175\cdot43\\ =175\cdot\left(19+38+43\right)\\ =175\cdot100\\ =17500\)
b)
\(125\cdot75+125\cdot13-80\cdot125\\ =125\cdot75+125\cdot13-125\cdot80\\ =125\cdot\left(75+13-80\right)\\ =125\cdot10\\ =125\cdot8\\ =1000\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
14 tháng 10 2023
a, 175. 19 + 38. 175 + 43. 175
= 175. 19 + 175. 38 + 175. 43
= 175.(19 + 38 + 43)
= 175. 100
= 17500
NM
2
NQ
Nguyễn Quốc Đạt
CTVVIP
5 tháng 12 2023
2/
Xét phân số \(\dfrac{2n-3}{n+1}=\dfrac{2n+2-5}{n+1}=\dfrac{2n+2}{n+1}-\dfrac{5}{n+1}=\dfrac{2\left(n+1\right)}{n+1}-\dfrac{5}{n+1}=2-\dfrac{5}{n+1}\)
\(n\in Z\Rightarrow2n-3\inƯ\left(5\right)=\left\{-1;-5;1;5\right\}\)
Ta có bảng:
| 2n - 3 | -1 | -5 | 1 | 5 |
| n | 1 | -1 | 2 | 4 |
Vậy \(n\in\left\{-1;1;2;4\right\}\)
NQ
Nguyễn Quốc Đạt
CTVVIP
5 tháng 12 2023
1/
(x + 1) + (x + 3) + (x + 5) + ... + (x + 999) = 500
<=> (x + x + x + ... + x) + (1 + 3 + 5 + ... + 999) = 500
Xét tổng A = 1 + 3 + 5 + ... + 999
Số số hạng của A là: (999 - 1) : 2 + 1 = 500
Tổng A là: (999 + 1) x 500 : 2 = 250 000
Do A có 500 số hạng nên có 500 ẩn x.
Vậy ta có: 500x + 250 000 = 500
=> 500x = -249 500
=> x = 499
Vậy x = 499
CC
4
27 tháng 8
Câu 8:
a:Sửa đề: \(4+4^2+\cdots+4^{2025}\)
Ta có: \(4+4^2+\cdots+4^{2025}\)
\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+\cdots+\left(4^{2023}+4^{2024}+4^{2025}\right)\)
\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+\cdots+4^{2023}\left(1+4+4^2\right)\)
\(=21\left(4+4^4+\cdots+4^{2023}\right)\) ⋮21
b: \(5+5^2+5^3+5^4+\cdots+5^{2024}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{2023}+5^{2024}\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+\cdots+5^{2022}\left(5+5^2\right)\)
\(=30\left(1+5^2+\cdots+5^{2022}\right)\) ⋮30
Câu 7:
a: \(A=2+2^2+2^3+\cdots+2^{99}\)
=>\(2A=2^2+2^3+\cdots+2^{100}\)
=>\(2A-A=2^2+2^3+\cdots+2^{100}-2-2^2-\cdots-2^{99}\)
=>\(A=2^{100}-2\)
b: \(B=1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(7B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}\)
=>\(7B+B=7-7^2+7^3-7^4+\cdots+7^{49}-7^{50}+1-7+7^2-7^3+\cdots+7^{48}-7^{49}\)
=>\(8B=-7^{50}+1\)
=>\(B=\frac{-7^{50}+1}{8}\)
S
subjects
CTVHS
28 tháng 8
câu 4:
a) \(\)x³ = 125
x³ = 5³
⇒ x = 5
b. \(11^{x+1}=121\)
\(11^{x+1}=11^2\)
⇒ x + 1 = 2
⇒ x = 2 - 1 = 1
c. (x - 5)³ = 27
(x - 5)³ = 3³
⇒ x - 5 = 3
x = 3 + 5 = 8
d. \(4^5:4^{x}=16\)
\(4^{5-x}=4^2\)
⇒ 5 - x = 2
x = 5 - 2 = 3
e. \(5^{x-1}\cdot8=1000\)
\(5^{x-1}=1000:8\)
\(5^{x-1}=125\)
\(5^{x-1}=5^3\)
⇒ x - 1 = 3
x = 3 + 1 = 4
f. \(2^{x}+2^{x+3}=72\)
\(2^{x}\cdot\left(1+2^3\right)=72\)
\(2^{x}=72:9\)
\(2^{x}=8\)
\(2^{x}=2^3\)
⇒ x = 3
g. (3x + 1)³ = 343
(3x + 1)³ = 7³
⇒ 3x + 1 = 7
3x = 7 - 1
3x = 6
x = 6 : 3 = 2
h. \(3^{x}+3^{x+2}=270\)
\(3^{x}\cdot\left(1+3^2\right)=270\)
\(3^{x}=270:10\)
\(3^{x}=27\)
\(3^{x}=3^3\)
⇒ x = 3
i. \(25^{2x+4}=125^{x+3}\)
\(\left(5^2\right)^{2x+4}=\left(5^3\right)^{x+3}\)
\(5^{4x+8}=5^{3x+9}\)
=>4x + 8 = 3x + 9
4x - 3x = 9 - 8
x = 1
\(=\dfrac{2^{19}\cdot3^9\cdot5+2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\cdot5\left(2+1\right)}{2^{19}\cdot3^9\left(1-2\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{5\cdot3}{-5}=-\dfrac{3}{2}\)
\(A=\dfrac{2^{19}\cdot27^3\cdot5-15\cdot\left(-4\right)^9\cdot9^4}{6^9\cdot2^{10}-\left(-12\right)^{10}}\)
\(A=\dfrac{2^{19}\cdot\left(3^3\right)^3\cdot5-3\cdot5\cdot-\left(2^2\right)^9\cdot\left(3^2\right)^4}{2^9\cdot3^9\cdot2^{10}-\left(2^2\right)^{10}\cdot3^{10}}\)
\(A=\dfrac{2^{19}\cdot3^9\cdot5+3^9\cdot2^{18}\cdot5}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)
\(A=\dfrac{2^{18}\cdot3^9\cdot5\cdot\left(2+1\right)}{2^{19}\cdot3^9\cdot\left(1-2\cdot3\right)}\)
\(A=\dfrac{1\cdot1\cdot5\cdot3}{2\cdot1\cdot-5}\)
\(A=-\dfrac{1}{2}\cdot3\)
\(A=-\dfrac{3}{2}\)