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

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

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

@Chu Gia Hân

bạn hỏi gì vậy???

\(250=5^3\cdot2\)

=>\(Ư\left(250\right)=\left\{1;-1;2;-2;5;-5;10;-10;25;-25;50;-50;125;-125;250;-250\right\}\)

\(Ư\left(250\right)\)\(=\left\{1;2;5;10;25;50;125;250\right\}\)

\(42\cdot40+6\cdot8\cdot21-22\cdot96\)

\(=42\cdot40+24\cdot42-22\cdot96\)

\(=42\left(40+24\right)-22\cdot1,5\cdot64\)

\(=64\cdot42-64\cdot33=64\cdot9=576\)

\(C=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}\cdot4\dfrac{1}{2}-2\cdot2\dfrac{1}{3}\right):\dfrac{7}{4}\)

\(=\dfrac{59}{10}\cdot\dfrac{2}{3}-\left(\dfrac{7}{3}\cdot\dfrac{9}{2}-2\cdot\dfrac{7}{3}\right)\cdot\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{7}{3}\cdot\dfrac{5}{2}\cdot\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{10}{3}=\dfrac{9}{15}=\dfrac{3}{5}\)