\(3^2.5+2^3.10-3^4:3\)

\(=5\left(3^2+2^3.2\right)-3^{4-1}\)

\(=5\left(9+16\right)-3^3\)

\(=5.25-27\)

\(=125-27=98\)

\(3^2\times5+2^3\times10-3^4:3\\ =9\times5+8\times10-27\\ =45+80-27\\ =98.\)

\(D=-\dfrac{4}{5}+\dfrac{4}{5^2}-\dfrac{4}{5^3}+...+\dfrac{4}{5^{200}}\)

\(\Rightarrow D=4\left(-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{200}}\right)\)

\(5D=4\cdot\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{199}}\right)\)

\(\Rightarrow5D+D=4\cdot\left(-1+\dfrac{1}{5}-\dfrac{1}{5^2}+...+\dfrac{1}{5^{199}}-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+...+\dfrac{1}{5^{200}}\right)\)

\(\Rightarrow6D=4\cdot\left(\dfrac{1}{5^{200}}-1\right)\)

\(\Rightarrow D=\dfrac{2}{3}\cdot\left(\dfrac{1}{5^{200}}-1\right)\)

oki

(x + 1)³ - 4 = 60

(x + 1)³ = 60 + 4

(x + 1)³ = 64

(x + 1)³ = 4³

x + 1 = 4

x = 4 - 1

x = 3

\(B=7-7^4+7^7-7^{10}+...+7^{295}-7^{298}+7^{301}\)

\(=7\left(1-7^3\right)+7^7\left(1-7^3\right)+...+7^{295}\left(1-7^3\right)+7^{301}\)

\(=\left(1-7^3\right)\left(7+7^7+...+7^{295}\right)+7^{301}\)

\(=\left(1-7^3\right)\left(\dfrac{7^{296}-7}{6}\right)+7^{301}\)

\(=-57\left(7^{296}-7\right)+7^{301}\)

Thời gian xe đi từ B đi đến chỗ gặp nhau:

9-7=2(giờ)

Thời gian xe đi từ A đi đến chỗ gặp nhau:

2-1=1(giờ(

Quãng đường xe đi từ B đi được tới chỗ gặp nhau:

50 x 2= 100(km)

Quãng đường xe đi từ A đi được tới chỗ gặp nhau:

40 x 1 = 40 (km)

Độ dài quãng đường AB:

100+40=140(km)

Đ.số: 140km

Thời gian xe A đi là:

\(9-7-1=1\left(h\right)\)

Thời gian xe B đi là:
\(9-7=2\left(h\right)\)

Độ dài quãng đường AB là:

\(2\cdot50+40\cdot1=140\left(km\right)\)

Đáp số: 140 km 

(2x + 1) : 7 = 2² + 3²

(2x + 1) : 7 = 4 + 9

(2x + 1) : 7 = 13

2x + 1 = 13 . 7

2x + 1 = 91

2x = 91 - 1

2x = 90

x = 90 : 2

x = 45

\((2x+1):7=2^2+3^2\\\Rightarrow (2x+1):7=4+9\\\Rightarrow(2x+1):7=13\\\Rightarrow2x+1=13\cdot7\\\Rightarrow2x+1=91\\\Rightarrow2x=91-1\\\Rightarrow2x=90\\\Rightarrow x=90:2\\\Rightarrow x=45\\Vậy:x=45\)

\(2xy-6x+y=13\)

\(2x\left(y-3\right)+y-3=10\)

\(\left(2x+1\right)\left(y-3\right)=10\)

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+1=10\\y-3=1\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=1\\y-3=10\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=2\\y-3=5\end{matrix}\right.\\\left\{{}\begin{matrix}2x+1=5\\y-3=2\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left\{\left(0,13\right);\left(2,5\right)\right\}\)

\((5x-39)\cdot7+3=80\\\Rightarrow (5x-39)\cdot7=80-3\\\Rightarrow (5x-39)\cdot7=77\\\Rightarrow 5x-39=77:7\\\Rightarrow 5x-39=11\\\Rightarrow5x=11+39\\\Rightarrow5x=50\\\Rightarrow x=50:5=10\\Vậy:x=10\)

(5x - 39).7 + 3 = 80

(5x - 39).7 = 80 - 3

(5x - 39).7 = 77

5x - 39 = 77 : 7

5x - 39 = 11

5x = 11 + 39

5x = 50

x = 50 : 5

x = 10

\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+140\right)=5041\)

\(x+x+...+x+2+4+...+140=5041\)

Có tất cả số hạng là:

\(\dfrac{\left(140-2\right)}{2}+1=70\left(số\right)\)

=> \(71x+\dfrac{\left(140+2\right).70}{2}=5041\)

=> \(71x=71\)

=> \(x=1\)

x + (x + 2) + (x + 4) + ... + (x + 140) = 5041

x + 70x + (140 + 2) . 70 : 2 = 5041

71x + 4970 = 5041

71x = 5041 - 4970

71x = 71

x = 71 : 71

x = 1

\(\left(6^{2024}-6^{2023}\right):6^{2023}\)

\(=6^{2024}:6^{2023}-6^{2023}:6^{2023}\)

\(=6-1\)

\(=5\)

\(\left(6^{2024}-6^{2023}\right):6^{2023}\)

\(=\dfrac{6^{2024}}{6^{2023}}-\dfrac{6^{2023}}{6^{2023}}\)

\(=6^{2024-2023}-6^{2023-2023}\)

\(=6^1-6^0\)

\(=6-1=5\)