67:

a: \(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac83\)

=>\(\frac{-6x}{12}+\frac{8x}{12}+\frac{3\left(x+1\right)}{12}+\frac{2\left(2x+1\right)}{12}=\frac{32}{12}\)

=>-6x+8x+3(x+1)+2(2x+1)=32

=>2x+3x+3+4x+2=32

=>9x=32-5=27

=>x=3

b: \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

=>\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{12}{26}=\frac{6}{13}\)

=>\(\frac{6}{2x+1}=\frac{6}{13}\)

=>2x+1=13

=>2x=12

=>x=6

Bài 68:

a: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};...;\frac{1}{100}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{50}{50}=1\) (1)

Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};\ldots;\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+..+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}\)

=>\(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}>\frac{50}{100}=\frac12\) (2)

Từ (1),(2) suy ra \(\frac12<\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}<1\)

b: Ta có: \(\frac{1}{21}<\frac{1}{20};\frac{1}{22}<\frac{1}{20};\ldots;\frac{1}{30}<\frac{1}{20}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}<\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}=\frac{10}{20}=\frac12\) (3)

Ta có: \(\frac{1}{31}<\frac{1}{30};\frac{1}{32}<\frac{1}{30};\ldots;\frac{1}{40}<\frac{1}{30}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}<\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (4)

Từ (3),(4) suy ra \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{40}<\frac12+\frac13=\frac56\left(5\right)\)

Ta có: \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\ldots;\frac{1}{30}=\frac{1}{30}\)

Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (6)

Ta có: \(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)

Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (7)

Từ (6),(7) suy ra \(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}>\frac13+\frac14=\frac{7}{12}\) (8)

Từ (5),(8) suy ra \(\frac{7}{12}<\frac{1}{21}+\ldots+\frac{1}{40}<\frac56\)

a: A={x∈N|x⋮3 và x<=15}

b: B={x∈N|x⋮5 và x<=30}

c: C={x∈N|x⋮10 và x<=90}

d: D={x∈N|x=4k+1; 0<=k<=4}

20:

a: \(4^{n}=256\)

=>\(4^{n}=4^4\)

=>n=4

b: \(9^{5n-8}=81\)

=>\(9^{5n-8}=9^2\)

=>5n-8=2

=>5n=10

=>n=2

c: \(3^{n+2}:27=3\)

=>\(3^{n+2}=27\cdot3=81=3^4\)

=>n+2=4

=>n=2

d: \(8^{n+2}\cdot2^3=8^5\)

=>\(8^{n+2}=8^5:8=8^4\)

=>n+2=4

=>n=2

Bài 21:

a: \(30-2x^2=12\)

=>\(2x^2=30-12=18\)

=>\(x^2=9\)

mà x>=0(do x là số tự nhiên)

nên x=3

b: \(\left(9-2x\right)^3=125\)

=>9-2x=5

=>2x=4

=>x=2

c: \(\left(2x-2\right)^4=0\)

=>2x-2=0

=>2x=2

=>x=1

d: \(\left(x+5\right)^3=\left(2x\right)^3\)

=>2x=x+5

=>2x-x=5

=>x=5

bài 20:

\(a.4^{n}=256\)

\(4^{n}=4^4\)

⇒ n = 4

b . \(9^{5n-8}=81\)

\(9^{5n-8}=9^2\)

⇒ 5n - 8 = 2

5n = 2 + 8

5n = 10

n = 10 : 5 = 2

c. \(3^{n+2}:27=3\)

\(3^{n+2}=3\cdot27\)

\(3^{n+2}=81\)

\(3^{n+2}=3^4\)

⇒ n + 2 = 4

⇒ n = 4 - 2 = 2

d. \(8^{n+2}\cdot2^3=8^5\)

\(8^{n+2}=8^5:2^3\)

\(8^{n+2}=8^4\)

⇒ n + 2 = 4

⇒ n = 4 - 2 = 2

bài 21 :

\(a.30-2x^2=12\)

\(2x^2=30-12\)

\(2x^2=18\)

\(x^2=18:2\)

\(x^2=9\)

⇒ x = 3 hoặc x = -3

b. \(\left(9-2x\right)^3=125\)

\(\left(9-2x\right)^3=5^3\)

⇒ 9 - 2x = 5

2x = 9 - 5

2x = 4

x = 4 : 2 = 2

c. \(\left(2x-2\right)^4=0\)

⇒ 2x - 2 = 0

2x = 2

x = 2 : 2 = 1

d. \(\left(x+5\right)^3=\left(2x\right)^3\)

⇒ x + 5 = 2x

⇒ 2x - x = 5

x = 5

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

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: \(89\cdot25+89\cdot74+89\)

\(=89\cdot\left(25+74+1\right)\)

\(=89\cdot100=8900\)