Bài 1:

a: 2x+3=15

=>2x=15-3=12

=>\(x=\frac{12}{2}=6\)

b: 28-3x=13

=>\(3x=28-13=15\)

=>\(x=\frac{15}{3}=5\)

c: \(\left(x-1954\right)\cdot5=50\)

=>x-1954=50:5=10

=>x=10+1954=1964

d: 30(60-x)=30

=>60-x=1

=>x=60-1=59

Bài 2:

a: x+99:3=55

=>x+33=55

=>x=55-33=22

b: (x-25):15=20

=>\(x-25=20\cdot15=300\)

=>x=300+25=325

c: \(7\left(3x-15\right)=42\)

=>\(3x-15=\frac{42}{7}=6\)

=>3x=15+6=21

=>\(x=\frac{21}{3}=7\)

d: \(x\left(x+1\right)=2+4+6+\cdots+2500\)

=>\(x\left(x+1\right)=2\left(1+2+3+\cdots+1250\right)\)

=>\(x\left(x+1\right)=2\cdot1250\cdot\frac{1251}{2}=1250\cdot1251\)

=>\(x^2+x-1250\cdot1251=0\)

=>(x+1251)(x-1250)=0

=>\(\left[\begin{array}{l}x+1251=0\\ x-1250=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1251\\ x=1250\end{array}\right.\)

Bài 1:

a: \(46\cdot99=46\cdot\left(100-1\right)=46\cdot100-46\)

=4600-46

=4554

b: \(47\cdot98=47\cdot\left(100-2\right)=47\cdot100-47\cdot2\)

=4700-94

=4604

c: \(18\cdot19=18\left(20-1\right)=18\cdot20-18\)

=360-18

=342

d: \(24\cdot198=24\left(200-2\right)\)

\(=24\cdot200-24\cdot2\)

=4800-48

=4752

Bài 2:

a: \(1800\cdot5=18\cdot100\cdot5=90\cdot100=9000\)

b: \(36\cdot25=25\cdot4\cdot9=100\cdot9=900\)

c: 36600:50=3660:5=732

d: 220000:5000=220:5=44

sạc xong rồi học đk

a: \(5\cdot6\cdot7\) ⋮3

\(8\cdot9\) ⋮3

Do đó: \(5\cdot6\cdot7+8\cdot9\) ⋮3

=>Đây là hợp số

b: \(7\cdot11\cdot13\) ⋮7

\(2\cdot5\cdot7\) ⋮7

Do đó: \(7\cdot11\cdot13-2\cdot5\cdot7\) ⋮7

=>Đây là hợp số

c: \(5\cdot11\cdot13\) là số lẻ

\(17\cdot19\cdot23\) là số lẻ

Do đó: \(5\cdot11\cdot13+17\cdot19\cdot23\) là số chẵn

mà \(5\cdot11\cdot13+17\cdot19\cdot23\) <>2

nên đây là hợp số

d: \(1\cdot2\cdot3\cdot\ldots\cdot100=97\cdot1\cdot2\cdot\ldots\cdot96\cdot98\cdot99\cdot100\) ⋮97

\(97\cdot101\cdot103\) ⋮97

Do đó: \(1\cdot2\cdot3\cdot\ldots\cdot100+97\cdot101\cdot103\) ⋮97

=>Đây là hợp số

c: \(\left(x-1\right)^3=\left(-9\right)^3\)

=>x-1=-9

=>x=-9+1=-8

f: \(3x-2^3=7+\left(-9\right)\)

=>3x-8=7-9=-2

=>3x=-2+8=6

=>x=2

Gọi tổng chi phí cho chuyến đi là x(đồng)

(Điều kiện: x>0)

Số tiền mỗi bạn phải đóng ban đầu là \(\frac{x}{40}\) (đồng)

Số tiền mỗi bạn phải đóng thực tế là \(\frac{x}{36}\) (đồng)

Mỗi bạn tham gia còn lại phải đóng thêm 25000 đồng nên ta có:

\(\frac{x}{36}-\frac{x}{40}=25000\)

=>\(\frac{x}{360}=25000\)

=>\(x=25000\cdot360=9000000\) (đồng)

vậy: Tổng chi phí cho chuyến đi là 9000000(đồng)

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

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