giá gốc của sản phẩm là:

15 000 000 : ( 100% - 20%) = 18 750 000 (đồng)

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

\(B=\dfrac{-1}{99}+\dfrac{1}{99\cdot98}-\dfrac{1}{98\cdot97}-...-\dfrac{1}{2\cdot1}\)

\(=\dfrac{-1}{99}+\dfrac{1}{98\cdot99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{97\cdot98}\right)\)

\(=-\dfrac{2}{99}+\dfrac{1}{98}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)

\(=-\dfrac{2}{99}+\dfrac{1}{98}-\dfrac{97}{98}=\dfrac{-2}{99}-\dfrac{23}{49}=\dfrac{-2375}{4851}\)

Bổ sung đề: Cho tam giác ABC và tam giác MNP bằng nhau

a: ΔABC=ΔMNP

=>AB=MN

b: Một góc bằng góc gì vậy bạn?

bằng nhau á

ĐKXĐ: x<>0

Để F là số nguyên  thì \(x-3⋮2x\)

=>\(2x-6⋮2x\)

=>\(2x\inƯ\left(-6\right)\)

=>\(2x\in\left\{-1;1;2;-2;3;-3;6;-6\right\}\)

mà x là số nguyên

nên \(2x\in\left\{2;-2;6;-6\right\}\)

=>\(x\in\left\{1;-1;3;-3\right\}\)

ĐKXĐ: x<>0

Để E là số nguyên thì \(x-4⋮x\)

=>\(-4⋮x\)

=>\(x\inƯ\left(-4\right)\)

=>\(x\in\left\{1;-1;2;-2;4;-4\right\}\)

`a, (5^2 +20:2^2):5`

`= (25 +20:4):5`

`= ( 25+5):5`

`= 30:5`

`= 6`

`b, (-1/4)^2 * (7/9) +1/16 * (-2/9)`

`= 1/16 * (7/9) +1/16 * (-2/9)`

`=1/16 * (7/9 + (-2)/9)`

`= 1/16* 5/9`

`= 5/144`

`c, 5/12 - (-1/4+ 5/12) : 7/6 -1`

`=5/12 - ( -3/12 +5/12) xx 6/7 -1`

`= 5/12 - 2/12 xx 6/7-1`

`=  5/12 - 1/6xx 6/7-1`

`= 5/12 - 6/42-1`

`= 5/12 - 1/7 -1`

`= 23/84-1`

`= -61/84`

a: \(\dfrac{5^2+20:2^2}{5}=\dfrac{25+20:4}{5}=\dfrac{25+5}{5}=\dfrac{30}{5}=6\)

b: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{7}{9}+\dfrac{1}{16}\cdot\dfrac{-2}{9}\)

\(=\dfrac{1}{16}\cdot\dfrac{7}{9}-\dfrac{1}{16}\cdot\dfrac{2}{9}\)

\(=\dfrac{1}{16}\left(\dfrac{7}{9}-\dfrac{2}{9}\right)=\dfrac{5}{9}\cdot\dfrac{1}{16}=\dfrac{5}{144}\)

e: \(\dfrac{5}{12}-\left(-\dfrac{1}{4}+\dfrac{5}{12}\right):\dfrac{7}{6}-1\)

\(=\dfrac{-7}{12}-\left(\dfrac{-3}{12}+\dfrac{5}{12}\right)\cdot\dfrac{6}{7}\)

\(=\dfrac{-7}{12}-\dfrac{1}{6}\cdot\dfrac{6}{7}=\dfrac{-7}{12}-\dfrac{1}{7}=\dfrac{-49-12}{84}=-\dfrac{61}{84}\)

Xem lại ký hiệu ở hình 

Lần sau chụp hình rõ hơn nhé em. Hình mờ khó nhìn hình chính xác lắm

Bài 3:

a) Do c ⊥ a

a // b

⇒ c ⊥ b

b) Ta có:

∠A₃ = ∠A₁ = 115⁰ (đối đỉnh)

Do a // b

⇒ ∠B₃ = ∠A₁ = 115⁰ (so le trong)

⇒ ∠B₁ = ∠B₃ = 115⁰ (đối đỉnh)

4:

a: Ot là phân giác của \(\widehat{AOB}\)

=>\(\widehat{tOB}=\widehat{tOA}=\dfrac{\widehat{AOB}}{2}=\dfrac{120^0}{2}=60^0\)

\(\widehat{tOB}=\widehat{yBO}\left(=60^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên Ot//By

b: \(\widehat{tOA}+\widehat{xAO}=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên Ot//Ax

Ax//Ot

Ot//By

Do đó: Ax//By

3:

a: a//b

c\(\perp\)a

Do đó: c\(\perp\)b

b: a//b

=>\(\widehat{A_1}+\widehat{B_2}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{B_2}+115^0=180^0\)

=>\(\widehat{B_2}=65^0\)

a//b

=>\(\widehat{A_1}=\widehat{B_3}\)(hai góc so le trong)

mà \(\widehat{A_1}=115^0\)

nên \(\widehat{B_3}=115^0\)

\(\widehat{A_1}=\widehat{A_3}\)(hai góc đối đỉnh)

mà \(\widehat{A_1}=115^0\)

nên \(\widehat{A_3}=115^0\)

1:

a: \(\dfrac{2^{30}\cdot5^7+2^{13}\cdot5^{27}}{2^{27}\cdot5^7+2^{10}\cdot5^{27}}\)

\(=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3=8\)

b: \(\left(\dfrac{9}{15}\right)^{2022}\cdot\left(\dfrac{15}{9}\right)^{2021}\)

\(=\left(\dfrac{3}{5}\right)^{2022}\cdot\left(\dfrac{5}{3}\right)^{2021}\)

\(=\left(\dfrac{3}{5}\right)\cdot\left(\dfrac{3}{5}\right)^{2021}\cdot\left(\dfrac{5}{3}\right)^{2021}\)

\(=\dfrac{3}{5}\cdot\left(\dfrac{3}{5}\cdot\dfrac{5}{3}\right)^{2021}=\dfrac{3}{5}\)