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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. \(3^{36}=\left(3^3\right)^{12}=27^{12}\)
\(5^{24}=\left(5^2\right)^{12}=25^{12}\)
\(27^{12}>25^{12}\Rightarrow3^{36}>5^{24}\)
Bài 3:
4; 45 + 5\(x\) = 10\(^3\): 10
45 + 5\(x\) = 100
5\(x\) = 100 - 45
5\(x\) = 55
\(x\) = 55 : 5
\(x\) = 11
Vậy \(x=11\)
5; 4\(x\) - 20 = 2\(^5\) : 2\(^2\)
4\(x\) - 20 = 2\(^3\)
4\(x\) = 8 + 20
4\(x\) = 28
\(x\) = 28 : 4
\(x=7\)
Vậy \(x=7\)
Bài 4:
1; 82 - (25 + 4\(x^{}\)) = 17
25 + 4\(x\) \(^{}\) = 82 - 17
4\(x^{}\) = 65 - 25
4\(x^{}\) = 40
\(x=40:4\)
\(x\) = 10
Vậy \(x=10\)
2; 71 - (24 + 3\(x\)) = 24
24 + 3\(x\) = 71 - 24
24 + 3\(x\) = 47
3\(x\) = 47 - 24
3\(x\) = 23
\(x\) = 23 : 3
Vậy \(x=\frac{23}{3}\)
3; 145 - (125 + \(x\)) = 12
125 + \(x\) = 145 - 12
125 + \(x\) = 133
\(x\) = 133 - 125
\(x\) = 8
Vậy \(x=8\)
Bài 5:
a: \(B=2009\cdot2011\)
\(=\left(2010-1\right)\left(2010+1\right)\)
\(=2010\cdot2010-1=A-1\)
=>B<A
b: \(B=2019\cdot2021\)
\(=\left(2020-1\right)\left(2020+1\right)\)
\(=2020\cdot2020-1\)
=A-1
=>B<A
c: \(A=234234\cdot233=234\cdot233\cdot1001\)
\(B=233233\cdot234=233\cdot234\cdot1001\)
Do đó: A=B
d: \(A=123\cdot137137=123\cdot137\cdot1001\)
\(B=137\cdot123123=137\cdot123\cdot1001\)
Do đó: A=B
Bài 4:
a: \(391-125<\overline{26x}<184+84\)
=>\(266<\overline{26x}<268\)
=>\(\overline{26x}=267\)
=>x=7
b: \(935+167<\overline{110x}<1240-135\)
=>\(1102<\overline{110x}<1105\)
=>x∈{3;4}
c: \(135\cdot12<\overline{162x}<4869:3\)
=>\(1620<\overline{162x}<1623\)
=>x∈{1;2}
d: \(11268:3<\overline{375x}<235\cdot16\)
=>\(3756<\overline{375x}<3760\)
=>x∈{7;8;9}
Bài 3:
l: x+125=492
=>x=492-125=367
m: 327-x=129
=>x=327-129=198
n: 124+(118-x)=217
=>118-x=217-124=93
=>x=118-93=25
o: 89-(73-x)=20
=>73-x=89-20=69
=>x=73-69=4
p: 198-(x+4)=120
=>x+4=198-120=78
=>x=78-4=74
q: (x+7)-25=23
=>x+7=25+23=48
=>x=48-7=41
r: 140:(x-8)=7
=>x-8=140:7=20
=>x=20+8=28
s: 4(x+41)=400
=>x+41=400:4=100
=>x=100-41=59
t: 4(3x-4)-2=18
=>4(3x-4)=2+18=20
=>3x-4=5
=>3x=9
=>x=3
u: 123-5(x+4)=38
=>5(x+4)=123-38=85
=>x+4=85:5=17
=>x=17-4=13
v: 231-(x-6)=1339:13
=>231-(x-6)=103
=>x-6=231-103=128
=>x=128+6=134
w: (x-36):18+12=14
=>(x-36):18=2
=>\(x-36=2\cdot18=36\)
=>x=36+36=72
Bài 5:
a: \(B=2009\cdot2011\)
\(=\left(2010-1\right)\left(2010+1\right)\)
\(=2010\cdot2010-1=A-1\)
=>B<A
b: \(B=2019\cdot2021\)
\(=\left(2020-1\right)\left(2020+1\right)\)
\(=2020\cdot2020-1\)
=A-1
=>B<A
c: \(A=234234\cdot233=234\cdot233\cdot1001\)
\(B=233233\cdot234=233\cdot234\cdot1001\)
Do đó: A=B
d: \(A=123\cdot137137=123\cdot137\cdot1001\)
\(B=137\cdot123123=137\cdot123\cdot1001\)
Do đó: A=B
Bài 4:
a: \(391-125<\overline{26x}<184+84\)
=>\(266<\overline{26x}<268\)
=>\(\overline{26x}=267\)
=>x=7
b: \(935+167<\overline{110x}<1240-135\)
=>\(1102<\overline{110x}<1105\)
=>x∈{3;4}
c: \(135\cdot12<\overline{162x}<4869:3\)
=>\(1620<\overline{162x}<1623\)
=>x∈{1;2}
d: \(11268:3<\overline{375x}<235\cdot16\)
=>\(3756<\overline{375x}<3760\)
=>x∈{7;8;9}
Bài 3:
l: x+125=492
=>x=492-125=367
m: 327-x=129
=>x=327-129=198
n: 124+(118-x)=217
=>118-x=217-124=93
=>x=118-93=25
o: 89-(73-x)=20
=>73-x=89-20=69
=>x=73-69=4
p: 198-(x+4)=120
=>x+4=198-120=78
=>x=78-4=74
q: (x+7)-25=23
=>x+7=25+23=48
=>x=48-7=41
r: 140:(x-8)=7
=>x-8=140:7=20
=>x=20+8=28
s: 4(x+41)=400
=>x+41=400:4=100
=>x=100-41=59
t: 4(3x-4)-2=18
=>4(3x-4)=2+18=20
=>3x-4=5
=>3x=9
=>x=3
u: 123-5(x+4)=38
=>5(x+4)=123-38=85
=>x+4=85:5=17
=>x=17-4=13
v: 231-(x-6)=1339:13
=>231-(x-6)=103
=>x-6=231-103=128
=>x=128+6=134
w: (x-36):18+12=14
=>(x-36):18=2
=>\(x-36=2\cdot18=36\)
=>x=36+36=72
Bài 5:
a: \(B=2009\cdot2011\)
\(=\left(2010-1\right)\left(2010+1\right)\)
\(=2010\cdot2010-1=A-1\)
=>B<A
b: \(B=2019\cdot2021\)
\(=\left(2020-1\right)\left(2020+1\right)\)
\(=2020\cdot2020-1\)
=A-1
=>B<A
c: \(A=234234\cdot233=234\cdot233\cdot1001\)
\(B=233233\cdot234=233\cdot234\cdot1001\)
Do đó: A=B
d: \(A=123\cdot137137=123\cdot137\cdot1001\)
\(B=137\cdot123123=137\cdot123\cdot1001\)
Do đó: A=B
Bài 4:
a: \(391-125<\overline{26x}<184+84\)
=>\(266<\overline{26x}<268\)
=>\(\overline{26x}=267\)
=>x=7
b: \(935+167<\overline{110x}<1240-135\)
=>\(1102<\overline{110x}<1105\)
=>x∈{3;4}
c: \(135\cdot12<\overline{162x}<4869:3\)
=>\(1620<\overline{162x}<1623\)
=>x∈{1;2}
d: \(11268:3<\overline{375x}<235\cdot16\)
=>\(3756<\overline{375x}<3760\)
=>x∈{7;8;9}
Bài 3:
l: x+125=492
=>x=492-125=367
m: 327-x=129
=>x=327-129=198
n: 124+(118-x)=217
=>118-x=217-124=93
=>x=118-93=25
o: 89-(73-x)=20
=>73-x=89-20=69
=>x=73-69=4
p: 198-(x+4)=120
=>x+4=198-120=78
=>x=78-4=74
q: (x+7)-25=23
=>x+7=25+23=48
=>x=48-7=41
r: 140:(x-8)=7
=>x-8=140:7=20
=>x=20+8=28
s: 4(x+41)=400
=>x+41=400:4=100
=>x=100-41=59
t: 4(3x-4)-2=18
=>4(3x-4)=2+18=20
=>3x-4=5
=>3x=9
=>x=3
u: 123-5(x+4)=38
=>5(x+4)=123-38=85
=>x+4=85:5=17
=>x=17-4=13
v: 231-(x-6)=1339:13
=>231-(x-6)=103
=>x-6=231-103=128
=>x=128+6=134
w: (x-36):18+12=14
=>(x-36):18=2
=>\(x-36=2\cdot18=36\)
=>x=36+36=72
d: \(48\cdot26+24\cdot148\)
\(=48\cdot26+48\cdot74\)
\(=48\cdot\left(26+74\right)=48\cdot100=4800\)
e: \(23\cdot48+92\cdot88\)
\(=23\cdot4\cdot12+92\cdot88\)
\(=92\cdot12+92\cdot88=92\cdot100=9200\)
b: \(89\cdot25+89\cdot74+89\)
\(=89\cdot\left(25+74+1\right)\)
\(=89\cdot100=8900\)
Bài 7: Ta có: \(a\cdot b=ƯCLN\left(a;b\right)\cdot BCN\mathbb{N}\left(a;b\right)\)
=>\(a\cdot b=10\cdot900=9000\)
ƯCLN(a;b)=10
=>a⋮10; b⋮10
ab=9000
mà a⋮10 và b⋮10 và a<b
nên (a;b)∈{(10;900);(20;450);(30;300);(50;180);(60;150);(90;100)}
mà ƯCLN(a;b)=10
nên (a;b)∈{(10;900);(20;450);(50;180);(90;100)}
Bài 5:
ƯCLN(a;b)=6
=>a⋮6; b⋮6
ab=216
mà a⋮6; b⋮6
nên (a;b)∈{(6;36);(12;18);(18;12);(36;6)}
Bài 4: Gọi hai số tự nhiên cần tìm là a,b
ƯCLN(a;b)=5
=>a⋮5; b⋮5
Ta có: ab=300
mà a⋮5; b⋮5
nên (a;b)∈{(5;60);(60;5);(10;30);(30;10);(15;20);(20;15)}
Bài 1:
a: ƯCLN(a;b)=6
=>a⋮6 và b⋮6
a+b=96
mà a⋮6 và b⋮6
nên (a;b)∈{(6;90);(90;6);(12;84);(84;12);(18;78);(78;18);(24;72);(72;24);(30;66);(66;30);(36;60);(60;36);(42;54);(54;42);(48;48)}
mà ƯCLN(a;b)=6
nên (a;b)∈{(6;90);(90;6);(18;78);(78;18);(30;66);(66;30);(42;54);(54;42)}
b: ƯCLN(a;b)=4
=>a⋮4 và b⋮4
a+b=16
mà a⋮4; b⋮4 và a>b
nên (a;b)∈{(12;4);(8;8)}
mà ƯCLN(a;b)=4
nên (a;b)=(12;4)








xét phân số thứ nhất:
tử số: \(2^{12}\cdot3^5-4^6\cdot9^2=2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2\)
=\(2^{12}\cdot3^5-2^{12}\cdot3^4=2^{12}\cdot3^4\cdot\left(3-1\right)=2^{13}\cdot3^4\)
mẫu số: \(\left(2^2\cdot3\right)^6+8^4\cdot3^5=\left(2^2\right)^6\cdot3^6+\left(2^3\right)^4\cdot3^5\)
=\(2^{12}\cdot3^6+2^{12}\cdot3^5=2^{12}\cdot3^5\cdot\left(3+1\right)=2^{14}\cdot3^5\)
rút gọn phân số thứ 1:
\(\frac{2^{13}\cdot3^4}{2^{14}\cdot3^5}=\frac{1}{2\cdot3}=\frac16\)
xét phân số thứ 2:
tử số: \(5^{10}\cdot7^3-25^5\cdot49^2=5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2\)
=\(5^{10}\cdot7^3-5^{10}\cdot7^4=5^{10}\cdot7^3\cdot\left(1-7\right)=5^{10}\cdot7^3\cdot\left(-6\right)\)
mẫu số: \(\left(125\cdot7\right)^3+5^9\cdot14^3=\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3\)
=\(\left(5^3\right)^3\cdot7^3+2^3\cdot7^3=5^9\cdot7^3+5^9\cdot2^3\cdot7^3\)
=\(5^9\cdot7^3\cdot\left(1+2^3\right)=5^9\cdot7^3\cdot9\)
rút gọn phân số thứ 2:
\(\frac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}=\frac{5\cdot\left(-6\right)}{9}=-\frac{10}{3}\)
cộng hai phân số lại:
A=\(\frac16-\frac{-10}{3}=\frac16+\frac{10}{3}=\frac{21}{6}=\frac72\)
b) gọi biểu thức cần cm là:
B=\(3^{n+2}-2^{n+2}+3^{n}-2^{n}\)
=\(\left(3^{n+2}+3^{n}\right)-\left(2^{n+2}+2^{n}\right)\)
=\(\left(3^{n}\cdot9+3^{n}\right)-\left(2^{n}\cdot4+2^{n}\right)\)
\(=3^{n}\left(9+1\right)-2^{n}\left(4+1\right)\)
=\(3^{n}\cdot10-2^{n}\cdot5\)
mà \(3^{n}\cdot10\) ⋮10
\(2^{n}\cdot5\) ( vì \(n\ge1)\)
=> \(2^{n}\cdot5=2^{n-1}\cdot2\cdot5=2^{n-1}\cdot10\) ⋮10
=> B⋮10 với mọi số nguyên dương n
cảm ơn bn nha
a: \(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)
\(=\frac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=\frac13\cdot\frac24=\frac13\cdot\frac12=\frac16\)
Ta có: \(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\frac{5^{10}\cdot7^3\cdot\left(1-7\right)}{5^9\cdot7^3\cdot\left(1+2^3\right)}=5\cdot\frac{-6}{9}=\frac{-30}{9}=\frac{-10}{3}\)
TA có: \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(=\frac16+\frac{10}{3}=\frac{21}{6}=\frac72\)
b: \(3^{n+2}-2^{n+2}+3^{n}-2^{n}\)
\(=3^{n}\cdot9+3^{n}-2^{n}\cdot4-2^{n}\)
\(=3^{n}\cdot\left(9+1\right)-2^{n}\left(4+1\right)=3^{n}\cdot10-2^{n}\cdot5\)
\(=3^{n}\cdot10-2^{n-1}\cdot10=10\left(3^{n}-2^{n-1}\right)\) ⋮10
\(\frac{2^{12}\cdot 3^{5}-(2^{2})^{6}\cdot (3^{2})^{2}}{(2^{12}\cdot 3^{6})+(2^{3})^{4}\cdot 3^{5}}=\frac{2^{12}\cdot 3^{5}-2^{12}\cdot 3^{4}}{2^{12}\cdot 3^{6}+2^{12}\cdot 3^{5}}\)
\(=\frac{2^{12}\cdot 3^{4}(3-1)}{2^{12}\cdot 3^{5}(3+1)}=\frac{3^{4}\cdot 2}{3^{5}\cdot 4}=\frac{1\cdot 2}{3\cdot 4}=\frac{2}{12}=\frac{\mathbf{1}}{\mathbf{6}}\) 2. Tính cụm phân số thứ hai:
\(\frac{5^{10}\cdot 7^{3}-(5^{2})^{5}\cdot (7^{2})^{2}}{(5^{3}\cdot 7)^{3}+5^{9}\cdot (2\cdot 7)^{3}}=\frac{5^{10}\cdot 7^{3}-5^{10}\cdot 7^{4}}{5^{9}\cdot 7^{3}+5^{9}\cdot 2^{3}\cdot 7^{3}}\)
\(=\frac{5^{10}\cdot 7^{3}(1-7)}{5^{9}\cdot 7^{3}(1+2^{3})}=\frac{5\cdot (-6)}{1+8}=\frac{-30}{9}=\mathbf{-}\frac{\mathbf{10}}{\mathbf{3}}\) 3. Kết quả biểu thức A:
\(A=\frac{1}{6}-\left(-\frac{10}{3}\right)=\frac{1}{6}+\frac{20}{6}=\frac{\mathbf{21}}{\mathbf{6}}\mathbf{=}\frac{\mathbf{7}}{\mathbf{2}}\) b) Chứng minh chia hết Chứng minh rằng với mọi số nguyên dương \(n\) thì: \(B = 3^{n+2} - 2^{n+2} + 3^n - 2^n\) chia hết cho \(10\). Lời giải: Ta nhóm các hạng tử cùng cơ số với nhau:
\(B=(3^{n+2}+3^{n})-(2^{n+2}+2^{n})\)
\(B=3^{n}(3^{2}+1)-2^{n}(2^{2}+1)\)
\(B=3^{n}\cdot 10-2^{n}\cdot 5\) Xét số hạng thứ hai \(2^n \cdot 5\):
- Vì \(n\) là số nguyên dương (\(n \ge 1\)), nên \(2^{n}\) luôn chia hết cho \(2\).
- Do đó, \(2^n \cdot 5 = 2^{n-1} \cdot (2 \cdot 5) = 2^{n-1} \cdot 10\).
Thay lại vào biểu thức \(B\):\(B=3^{n}\cdot 10-2^{n-1}\cdot 10\)
\(B=10\cdot (3^{n}-2^{n-1})\) Vì \(10 \cdot (3^n - 2^{n-1})\) luôn chia hết cho \(10\) với mọi số nguyên dương \(n\).
Vậy \(3^{n+2} - 2^{n+2} + 3^n - 2^n\) chia hết cho \(10\) (Đpcm).