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)

ngón tay gầy

Trả lời gúp mình với p

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 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

Gọi số tiền mỗi bạn phải đóng ban đầu là x(đồng)

(Điều kiện: x>0)

Số tiền mỗi bạn phải đóng sau đó là x+25000(đồng)

Số bạn tham gia là 40-4=36(bạn)

Tổng số tiền 40 bạn phải đóng ban đầu là 40x(đồng)

Tổng số tiền 36 bạn phải đóng lúc sau là 36(x+25000)(đồng)

Tổng số tiền không thay đổi nên ta có:

40x=36(x+25000)

=>10x=9(x+25000)

=>10x=9x+225000

=>x=225000(nhận)

Tổng chi phí cho chuyến đi là: \(225000\cdot40=9000000\) (đồng)

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 \(x=8\)