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

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

sao nhieu qua zay

Lời giải:

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

a/x-115

b/x+388

tick mk đi every body

a. x-115

b.x+388

Tick mình đi!!!!!!!!!!!! Chúc bạn năm mới vui vẻ

a/ = x + 25

b/ = x + 388

a) x+ 45 - [90+(-20)+5-(-45)]+3x 

= x + 45 - [70+50]+3x

= x + 45 - 120 + 3 x 

= ( x + 3x)  + (45-120) 

= 4x + (-75)

b) x+(294+13)+(94-13)+9x

= ( x + 9x ) + (294+13)+(94-13)

= 10x + 307+81

= 10x + 388

Trả lời:

a) x + 45 - [90 + (-20) + 5 - (-45)] + 3x

= (x + 3x) + 45 - (75 + 45)

= 4x + 45 - 75 - 45

= 4x + (45 - 45) - 75

= 4x + 0 - 75

= 4x - 75

b) x + (294 + 13) + (94 - 13) + 9x

= (9x + x) + 294 + 13 + 94 - 13

= 10x + (294 + 94) + (13 - 13)

= 10x + 388 + 0

= 10x + 388

Banh Bao Tong

deo biet luon ban oi co ban free fire ko

a, x+45-[90+(-20)+5-(-45)]

=x+45-[90+(-20)+5+45]

=x+45-90+20-5-45

=x+(45-45)-(90+20)-5

=x+0-110-5

=x-115

vay x-115

b, x+(294+13)+(94-13)

=x+307+81

=x+388

vay x+388

đúng rồi

thank you Đỗ Thị Ngà.

a) x+45-[90+(-20)+5-(-45)]

=x+45-120

=x+-75

b) x+(294+13)+(94-13)

=x+307+81

=x+388

a) x+45-[90+(-20)+5-(-45)]

=x+45-[(90-20)+(5+45)]

=x+45-[70+50]

=x+45-120

=x+(45-120)

=x-75

b) x+(294+13)+(94-13)

=x+307+81

=x+(307+81)

=x+388

Bài 1 :

a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)

Mà \(x^2+1\ge1>0\)

\(\Rightarrow2x+1\ge0\)

\(\Rightarrow x\ge-\dfrac{1}{2}\)

Vậy ...

b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)

\(=-3+4-\left(-4\right)=-3+4+4=5\)

Bài 2 :

\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)

\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)

\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)

\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)

\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)

\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)

\(=3\)