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

10 cách

1: 2⋮x

mà x là số tự nhiên

nên x∈{1;2}

2: 2⋮x+1

=>x+1∈{1;-1;2;-2}

=>x∈{0;-2;1;-3}

mà x>=0

nên x∈{0;1}

3: 2⋮x+2

mà x+2>=2(Do x là số tự nhiên)

nên x+2=2

=>x=0

4: 2⋮x-1

=>x-1∈{1;-1;2;-2}

=>x∈{2;0;3;-1}

mà x>=0

nên x∈{0;2;3}

5: 2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

6: 2⋮2-x

=>2⋮x-2

=>x-2∈{1;-1;2;-2}

=>x∈{3;1;4;0}

Bài 1:

2 ⋮ \(x\)(\(x\) ∈ N*)

2 ⋮ \(x\)

⇒ \(x\) ∈ Ư(2) = {-2; -1; 1; 2}

Vì \(x\) ∈ N* nên \(x\) ∈ {1; 2}

Vậy \(x\) ∈ {1; 2}

Giúp mình với mng ơi!!

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: \(750:\left\lbrace130-\left\lbrack\left(5\cdot14-65\right)^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-\left\lbrack\left(70-65\right)^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-\left\lbrack5^3+3\right\rbrack\right\rbrace\)

\(=750:\left\lbrace130-128\right\rbrace=750:2=375\)

Bài 8:

a: \(5^3=125;3^5=243\)

mà 125<243

nên \(5^3<3^5\)

b: \(7\cdot2^{13}<8\cdot2^{13}=2^3\cdot2^{13}=2^{16}\)

c: \(27^5=\left(3^3\right)^5=3^{3\cdot5}=3^{15}\)

\(243^3=\left(3^5\right)^3=3^{5\cdot3}=3^{15}\)

Do đó: \(27^5=243^5\)

d: \(625^5=\left(5^4\right)^5=5^{4\cdot5}=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{3\cdot7}=5^{21}\)

mà 20<21

nên \(625^5<125^7\)

Bài 9:

a: \(3^{x}\cdot5=135\)

=>\(3^{x}=\frac{135}{5}=27=3^3\)

=>x=3(nhận)

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

=>\(\left(x-3\right)^3-\left(x-3\right)^2=0\)

=>\(\left(x-3\right)^2\cdot\left\lbrack\left(x-3\right)-1\right\rbrack=0\)

=>\(\left(x-3\right)^2\cdot\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-3=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\left(nhận\right)\\ x=4\left(nhận\right)\end{array}\right.\)

c: \(\left(2x-1\right)^4=81\)

=>\(\left[\begin{array}{l}2x-1=3\\ 2x-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-1\left(loại\right)\end{array}\right.\)

d: \(\left(5x+1\right)^2=3^2\cdot5+76\)

=>\(\left(5x+1\right)^2=9\cdot5+76=45+76=121\)

=>\(\left[\begin{array}{l}5x+1=11\\ 5x+1=-11\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=10\\ 5x=-12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=-\frac{12}{5}\left(loại\right)\end{array}\right.\)

e: \(5+2^{x-3}=29-\left\lbrack4^2-\left(3^2-1\right)\right\rbrack\)

=>\(2^{x-3}+5=29-\left\lbrack16-9+1\right\rbrack\)

=>\(2^{x-3}+5=29-8=21\)

=>\(2^{x-3}=16=2^4\)

=>x-3=4

=>x=4+3=7(nhận)

f: \(3+2^{x-1}=24-\left\lbrack4^2-\left(2^2-1\right)\right\rbrack\)

=>\(2^{x-1}+3=24-\left\lbrack16-4+1\right\rbrack=24-13=11\)

=>\(2^{x-1}=11-3=8=2^3\)

=>x-1=3

=>x=4(nhận)

Bài 6:

a: \(5\cdot5\cdot5\cdot5\cdot5\cdot5=5^6\)

b: \(27\cdot14\cdot7\cdot2=27\cdot14\cdot14=3^3\cdot14^2\)

c: \(x\cdot x\cdot x\cdot y=x^3\cdot y\)

d: \(5^3\cdot5^4=5^{3+4}=5^7\)

e: \(7^8:7^2=7^{8-2}=7^6\)

f: \(42^7:6^7\cdot49=7^7\cdot49=7^7\cdot7^2=7^{7+2}=7^9\)

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)