Lời giải:

\(P=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\\ =\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\\ =\frac{2^{19}.3^9+2^{18}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\\ =\frac{2^{18}.3^9(2+5)}{2^{19}3^9(1+2.3)}\\ =\frac{1}{2}\)

sao nhieu qua zay

a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)

=>\(3A=2^{101}-2\)

=>\(A=\dfrac{2^{101}-2}{3}\)

b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)

c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)

\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)

\(=\dfrac{2591}{2^6\cdot3^5}\)

help

b: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{11}\cdot3^9}\)

\(=\dfrac{1}{2}\cdot\dfrac{-2}{3}=\dfrac{-1}{3}\)

Sửa đề câu a) phải là (-2020/2021)⁰

Xin lỗi mn

\(a,=\dfrac{3}{4}+\dfrac{15}{6}\cdot\dfrac{1}{5}-\dfrac{1}{36}\cdot9-\dfrac{2020}{2021}=\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{2020}{2021}=1-\dfrac{2020}{2021}=\dfrac{1}{2021}\\ b,=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^9\cdot3^8\left(2+5\right)}=\dfrac{2\left(-2\right)}{3}=-\dfrac{4}{3}\)

Hướng dẫn giải:

Rút gọn biểu thức

\(\text{a,x+(-30)-[95+(-40)+(-30)]}\)

\(=x+\left(-30\right)-95+40+30\)

\(=x-95+40=x-55\)

\(\text{a+(273-120)-(270-120)}\)

\(=a+273-120-270+120\)

\(=a+273-270=a+3\)

\(\text{b-(294+130)+(94+130)}\)

\(=b-294-130+94+130\)

\(=b-294+94=b-200\)

Cảm ơn nha

tk luôn chị Chinh ạ

a) x + (- 30) - [95 + (- 40) + 5 + (- 30)]

= x + (-60) = x - 60

b) a + ( 273 - 120 ) - ( 270 - 120 )

= a + 3

c) b - ( 294 + 130 ) + ( 94 + 130 )

= b - 648

cho tổng S=5+8+11+14+...

a,tìm số hạng thứ 100 của tổng 

b,tính tổng 100 số hạng đầu tiên

a) x + ( - 30 ) - [ 95 + ( - 40) + 5 + ( - 30) ]

   = x - 30 - 95 + 40 - 5 + 30

   = x -60

b) a + ( 273 - 120 ) - ( 270 - 120 )

   = a + 273 -120 -270 + 120

   = a+ 3

c) b - ( 294 + 130 ) + ( 94 + 130 )

    = b - 294 -130 +94 +130

    = b - 200 

Bạn nhìn xem có cặp giá trị nào có thể cho kết quả qui tắc tínbằng 0 thì ghép lại ( sử dụng qui tắc mở ngoặc và đưa vào ngoặc nha )

\(\frac{3\times21}{14\times15}=\frac{1\times21}{14\times5}=\frac{21}{70}=\frac{3}{10}\)

\(\frac{49+7\times49}{49}=\frac{392}{49}=8\)

a)36/48>12/48

=>6/8>3/12

b)135/225<180/225

=>15/25<36/45

c)1575/2835>1323/2835

=>15/27>49/105

Chúc e học tốt

a) mẫu số chung: 4

\(\dfrac{6}{8}=\dfrac{6:2}{8:2}=\dfrac{3}{4};\dfrac{3}{12}=\dfrac{3:3}{12:3}=\dfrac{1}{4}\)

vì \(\dfrac{3}{4}>\dfrac{1}{4}\)nên \(\dfrac{6}{8}>\dfrac{3}{12}\)

b) mẫu số chung: 5

\(\dfrac{15}{25}=\dfrac{15:5}{25:5}=\dfrac{3}{5};\dfrac{36}{45}=\dfrac{36:9}{45:9}=\dfrac{4}{5}\)

vì\(\dfrac{3}{5}< \dfrac{4}{5}\)nên \(\dfrac{15}{25}< \dfrac{36}{45}\)