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Hì hì xin lỗi trả lời muộn thông cảm

`#040911`

`(x + 5)^3 = (2x)^3`

`\Rightarrow x + 5 = 2x`

`\Rightarrow x + 5 - 2x = 0`

`\Rightarrow 5 + (x - 2x) = 0`

`\Rightarrow 5 - x = 0`

`\Rightarrow x = 5 - 0`

`\Rightarrow x = 5`

Vậy, `x= 5.`

Cảm ơn bạn nha

Ta có : A = 3⁰ + 3¹ + 3² + 3³ + .... + 3⁵⁰

=> 3A = 3¹ + 3² + 3³ + 3⁴ + ... + 3⁵¹

Khi đó 3A - A = (3¹ + 3² + 3³ + 3⁴ + ... + 3⁵¹) - (3⁰ + 3¹ + 3² + 3³ + .... + 3⁵⁰)

=> 2A = 3⁵¹ - 3⁰ 

=> A = \(\frac{3^{51}-1}{2}\)

Khi đó A = \(\frac{3^{51}-1}{2}=\frac{3^3.3^{48}-1}{2}=\frac{27.\left(3^4\right)^{12}-1}{2}=\frac{27.\left(...1\right)^{12}-1}{2}\)

\(=\frac{\left(...7\right)-1}{2}=\frac{\left(...6\right)}{2}=\left(...3\right)\)

Vậy A tận cùng là 3

CẢM ƠN BẠN RẤT NHIỀU TvT

53−3(x+4)=32

\(^{3^2}\).\(^{3^3}\)+\(2^3\).\(2^2\)

(\(^{2^3}\).\(^{3^3}\))+(\(2^2\).​\(^{3^2}\)​​​

=275

\(A=1+3+3^2+...+3^{2002}\)

\(3A=3+3^2+3^3+...+3^{2003}\)

\(2A=3A-A=3^{2003}-1\Rightarrow A=\dfrac{3^{2003}-1}{2}\)

A = \(\dfrac{3^{2003}-1}{2}\)

\(a.x-143=57\)

\(x=200\)

\(b.\left(8x-12\right):4=3^3\)

\(8x-12=27.4\)

\(8x-12=108\)

\(8x=120\)

\(x=15\)

\(d.10+2x=4^2\)

\(2x=16-10\)

\(2x=6\)

\(x=3\)

Câu a:

2.(x + 2/5) + 1 1 /4 = 11/20

2x + 4/5 + 5/4 = 11/20

2x = 11/20 - 5/4 - 4/5

2x = 11/20 - 25/20 - 16/20

2x = -30/20

2x = -3/2

x = -3/2 : 2

x = - 3/4

Vậy x = - 3/4

Câu b:

x - 1 1/8 - 2/3x - 5/6x = 75%

(x - 2/3x - 5/6x) = 75% + 1 1/8

x (1 - 2/3 - 5/6) = 3/4 + 9/8

x (6/6 - 4/6 - 5/6) = 6/8 + 9/8

x.(-1/2) = 15/8

x = 15/8 : (-1/2)

x = 15/8 x (-2)

= -15/4