`A` 

`a, N = 15,32 + 27,4 - 4,25` $\times$ `2`

` N = 15,32 + 27,4 - 8,5`$\text{Kết quả của 4,25 * 2}$

` N = 42,72 - 8,5 `$\text {Kết quả của phép tính 15,32 + 27,4}$

`N = 34,22` $\text {(kết quả cuối cùng}$

`b, M = (227,45 - 142,65) : 3,2 - 25`

`M = 81,8 : 3,2 - 25`$\text{Kết quả của 227,45 - 142,65 , thực hiện nó trước vì nó là phép tính trong ngoặc}$

`M = 25,5625 - 25`$\text{Kết quả của phép tính 81,8 : 3,2}$

`M = 0,5625`$\text{Kết quả cuối cùng}$

\(\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{6};2x+3y+4z=9\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{2x+2}{4}=\dfrac{3y+9}{12}=\dfrac{4z+20}{24}\)

\(\Rightarrow\dfrac{\left(2x+3y+4z\right)+\left(2+9+20\right)}{4+12+24}=\dfrac{9+31}{40}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+1}{2}=1\\\dfrac{y+3}{4}=1\\\dfrac{z+5}{6}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+1=2\\y+3=4\\z+5=6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\)

Vậy \(x=1;y=1;z=1\)

`a, x - 75 = -26`

`x = -26 + 75` 

`x = 49`

Vậy/So...

\(b,\left(8-x\right).\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+15=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8-0\\x=0-15\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-15\end{matrix}\right.\)

Vậy \(x\in\left\{8;-15\right\}\)

\(A=3+3^2+3^3+...+3^{2024}\)

\(3A=3^2+3^3+3^4+...+3^{2025}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{2025}\right)-\left(3+3^2+3^3+...+3^{2024}\right)\)

\(2A=3^{2025}-3\)

\(A=\dfrac{3^{2025}-3}{2}\)

\(_{\left(\dfrac{13}{14}+\dfrac{1}{16}\right)-\left(\dfrac{13}{14}-\dfrac{15}{16}\right)}\)

\(=\dfrac{13}{14}+\dfrac{1}{16}-\dfrac{13}{14}+\dfrac{15}{16}\)

\(=\left(\dfrac{13}{14}-\dfrac{13}{14}\right)+\left(\dfrac{1}{16}+\dfrac{15}{16}\right)\)

\(=0+1=1\)

\(2\left(x-5\right)+3\left(x-9\right)=x-49\)

\(\Rightarrow2x-10+3x-27=x-49\)

\(\Rightarrow2x+3x-x=-49+27+10\)

\(\Rightarrow4x=-12\)

\(\Rightarrow x=-12:4\)

\(\Rightarrow x=-3\)

Vậy...

\(\left[461+\left(-78\right)+40\right]+\left(-461\right)\)

\(=\left[461-78+40\right]-461\)

\(=461-78+40-461\)

\(=\left(461-461\right)-\left(78-40\right)\)

\(=0-38=-38\)

`d, 125 . (-36) + (-36) . (-52) + 36 . (-27) + (-400)`

`= 125 . (-36) + (-36) . (-52) + (-36) . 27 + (-400)`

`= [125 + (-52) + 27] . (-36) + (-400)`

`= 100 . (-36) + (-400)`

`= (-3600) + (-400) = (-4000)`

\(1+5+5^2+...+5^{2024}\)

a,

Đặt \("1+5+5^2+...+5^{2024}"\) là `S`

Ta có :

\(S=1+5+5^2+...+5^{2024}\)

\(5S=5+5^2+5^3+...+5^{2025}\)

\(5S-S=\left(5+5^2+5^3+...+5^{2024}\right)-\left(1+5+5^2+...+5^{2024}\right)\)

\(4S=5^{2024}-1\)

\(S=\dfrac{5^{2024}-1}{4}\)

`b,`

\(4\times\dfrac{5^{2024}-1}{4}+1=5^n\)

\(\Rightarrow5^{2024}-1+1=5^n\)

\(\Rightarrow5^{2024}=5^n\)

\(\Rightarrow n=2024\)

\(2\left(x-5\right)-3\left(x+7\right)=14\)

\(\Rightarrow2x-10-3x+21=14\)

\(\Rightarrow2x-3x=14+21+10\)

\(\Rightarrow-x=45\)

\(\Rightarrow x=-45\)

Vậy...