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a)
\(175\cdot19+38\cdot175+43\cdot175\\ =175\cdot19+175\cdot38+175\cdot43\\ =175\cdot\left(19+38+43\right)\\ =175\cdot100\\ =17500\)
b)
\(125\cdot75+125\cdot13-80\cdot125\\ =125\cdot75+125\cdot13-125\cdot80\\ =125\cdot\left(75+13-80\right)\\ =125\cdot10\\ =125\cdot8\\ =1000\)
a, 175. 19 + 38. 175 + 43. 175
= 175. 19 + 175. 38 + 175. 43
= 175.(19 + 38 + 43)
= 175. 100
= 17500
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)
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. \(3^{36}=\left(3^3\right)^{12}=27^{12}\)
\(5^{24}=\left(5^2\right)^{12}=25^{12}\)
\(27^{12}>25^{12}\Rightarrow3^{36}>5^{24}\)
\(1)2^8:2^4+3^2\cdot3=2^4+3^3=16+27=43\)
\(2)3^{24}:3^{21}+2^2\cdot2^3=3^3+2^5=27+32=59\)
\(3)5^9:5^7+12\cdot3+7^0=5^2+36+1=25+37=62\)
\(4)5^6:5^4+3^2-2021^0=5^2+3^2-1=25+9-1=33\)
\(5)3^{19}:3^{16}+5^2\cdot2^3-1^{2021}=3^3+25\cdot8-1=27+200-1=226\)
\(6)3^6:3^5+2\cdot2^3+2021^0=3^1+2^4+1=3+16+1=20\)
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) \(\)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
1: \(1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\)
\(=1026-82:41\)
=1026-2
=1024
\(2^{11}:\left\lbrace1026-\left\lbrack\left(3^4+1\right):41\right\rbrack\right\rbrace\)
\(=2^{11}:2^{10}=2\)
2: \(250:\left\lbrace1500:\left\lbrack4\cdot5^3-2^3\cdot25\right\rbrack\right\rbrace\)
\(=250:\left\lbrace1500:\left\lbrack4\cdot125-8\cdot25\right\rbrack\right\rbrace\)
\(=250:\left\lbrace1500:\left\lbrack500-200\right\rbrack\right\rbrace=250:\frac{1500}{3}=250:500=0,5\)
3: \(12+3\cdot\left\lbrace90:\left\lbrack39-\left(2^3-5\right)^2\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:\left\lbrack39-\left(8-5\right)^2\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:\left\lbrack39-3^2\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:\left\lbrack39-9\right\rbrack\right\rbrace\)
\(=12+3\cdot\left\lbrace90:30\right\rbrace=12+3\cdot3=21\)
4: \(24:\left\lbrace390:\left\lbrack500-\left(5^3+49\cdot5\right)\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:\left\lbrack500-\left(125+245\right)\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:\left\lbrack500-125-245\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:\left\lbrack375-245\right\rbrack\right\rbrace\)
\(=24:\left\lbrace390:130\right\rbrace=\frac{24}{3}=8\)
5: \(117:\left\lbrace\left\lbrack79-3\cdot\left(3^3-17\right)\right\rbrack:7+2\right\rbrace\)
\(=117:\left\lbrace\left\lbrack79-3\cdot\left(27-17\right)\right\rbrack:7+2\right\rbrace\)
\(=117:\left\lbrace\left\lbrack79-3\cdot10\right\rbrack:7+2\right\rbrace\)
\(=117:\left\lbrace49:7+2\right\rbrace=\frac{117}{9}=13\)
6: \(514-4\cdot\left\lbrace\left\lbrack40+8\left(6-3\right)^2\right\rbrack-12\right\rbrace\)
\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot3^2\right\rbrack-12\right\rbrace\)
\(=514-4\cdot\left\lbrace\left\lbrack40+8\cdot9\right\rbrack-12\right\rbrace\)
\(=514-4\cdot\left\lbrace112-12\right\rbrace\)
\(=514-4\cdot100=514-400=114\)
7: \(25\cdot\left\lbrace32:\left\lbrack\left(12-4\right)+4\cdot\left(16:2^3\right)\right\rbrack\right\rbrace\)
\(=25\cdot\left\lbrace32:\left\lbrack8+4\cdot2\right\rbrack\right\rbrace\)
\(=25\cdot\left\lbrace32:16\right\rbrace=25\cdot2=50\)
8: \(30:\left\lbrace175:\left\lbrack355-\left(135+37\cdot5\right)\right\rbrack\right\rbrace\)
\(=30:\left\lbrace175:\left\lbrack355-\left(135+185\right)\right\rbrack\right\rbrace\)
\(=30:\left\lbrace175:\left\lbrack355-320\right\rbrack\right\rbrace=30:\left\lbrace175:35\right\rbrace=\frac{30}{5}=6\)
9: \(32:\left\lbrace160:\left\lbrack300-\left(175+21\cdot5\right)\right\rbrack\right\rbrace\)
\(=32:\left\lbrace160:\left\lbrack300-\left(175+105\right)\right\rbrack\right\rbrace\)
\(=32:\left\lbrace160:\left\lbrack300-280\right\rbrack\right\rbrace\)
\(=32:\left\lbrace160:20\right\rbrace=\frac{32}{8}=4\)
10: \(750:\left\lbrace130-\left\lbrack\left(5\cdot14-65\right)^3+3\right\rbrack\right\rbrace\)
\(=750:\left\lbrace130-\left\lbrack\left(70-65\right)^3+3\right\rbrack\right\rbrace\)
\(=750:\left\lbrace130-\left\lbrack5^3+3\right\rbrack\right\rbrace\)
\(=750:\left\lbrace130-128\right\rbrace=750:2=375\)
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 \(n\in\left\{-1;1;2;4\right\}\)
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