\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\)

\(\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

\(\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

\(x+100=0\) (do \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}>0\))

\(x=-100\)

\(144=2^4\cdot3^2;192=2^6\cdot3\)

=>\(ƯCLN\left(144;192\right)=2^4\cdot3=48\)

=>\(ƯC\left(144;192\right)=Ư\left(48\right)=\left\{1;2;3;4;6;8;12;16;24;48\right\}\)

=>Các ước chung lớn hơn 20 của 144 và 192 là 24;48

45:(3x-6)=5

=>3x-6=45:5=9

=>3x=9+6=15

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

45 : (3x - 6) = 5

3x - 6= 45 : 5

3x - 6 = 9

3x = 9 + 6

3x = 15

x = 15 : 3

x = 5

Vậy x = 5

Bài 2: Đặt *=x

Số cần tìm sẽ có dạng là \(\overline{3x5}\)

\(\overline{3x5}⋮3\)

=>\(3+x+5⋮3\)

=>\(x+8⋮3\)

=>\(x\in\left\{1;4;7\right\}\)

=>*\(\in\left\{1;4;7\right\}\)

Bài 4:

a: 720;702;270;207;762;726;627;672;276;267

b: 762; 726; 627; 672; 276; 267

a: Các góc có trong hình vẽ là \(\widehat{zMN};\widehat{tMN};\widehat{zMt}\)

b: Góc bẹt là \(\widehat{zMt}\)

Số lượng số hạng:

`(2024-1):1+1=2024` (số hạng)

Tổng của dãy số:

`(2024+1)*2024:2=2049300` 

ĐS: ... 

\(\left(x-40\right)^2-48=-12\\ =>\left(x-40\right)^2=-12+48\\ =>\left(x-40\right)^2=36\\ =>\left(x-40\right)^2=6^2\\ TH1:x-40=6\\ =>x=6+40\\ =>x=46\\ TH2:x-40=-6\\ =>x=40-6\\ =>x=34\)

Vậy: ... 

\(\left(x-40\right)^2-48=-12\)

\(\left(x-40\right)^2=\left(-12\right)+48\)

\(\left(x-40\right)^2=36\)

\(\left(x-40\right)^2=6^2\)

\(x-40=6\)

\(x=6+40\)

\(x=46\)

a: Trên tia Ox, ta có: OA<OB

nên A nằm giữa O và B

=>OA+AB=OB

=>AB+2=7

=>AB=5(cm)

b: Y là trung điểm của OB

=>\(BY=\dfrac{BO}{2}=\dfrac{7}{2}=3,5\left(cm\right)\)

Vì BY<BA

nên Y nằm giữa B và A

=>BY+YA=BA

=>YA+3,5=5

=>YA=1,5(cm)

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2003}{2004}\)

\(=\dfrac{1}{2004}\)

a: \(x-\dfrac{3}{4}=6\cdot\dfrac{3}{8}\)

=>\(x-\dfrac{3}{4}=\dfrac{3}{4}\cdot3\)

=>\(x=\dfrac{9}{4}+\dfrac{3}{4}=\dfrac{12}{4}=3\)

b: \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

=>\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=>\(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{8}\cdot\dfrac{2}{5}=\dfrac{7}{20}\)

c: \(x+\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{3}{4}\)

=>\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{9}{12}-\dfrac{2}{12}=\dfrac{7}{12}\)