a.    x=(x-1)^2

b.    câu hỏi chưa xong kìa

(x-1)³\(=\)(x-5)³

\(\Leftrightarrow\)(x-1)³-(x-1)⁵\(=\)0

\(\Leftrightarrow\)(x-1)³\([\)1-(x-1)²\(]\)\(=\)0

\(\Leftrightarrow\)(x-1)³\(=\)0 hoặc 1-(x-1)²\(=\)0

\(\Leftrightarrow\)x-1\(=\)0 hoặc x-1\(=\pm\)1

\(\Leftrightarrow\)x\(=\)1 hoặc x\(=\)2; x\(=\)0

Vậy x\(\in\){1;2;0}

b) (x-1)ⁿ\(=\)(x-1)ⁿ⁺²

\(\Leftrightarrow\)(x-1)ⁿ-(x-1)ⁿ⁺²\(=\)0

\(\Leftrightarrow\)(x-1)ⁿ\([\)1-(x-1)²\(]\)\(=\)0

\(\Leftrightarrow\)(x-1)ⁿ\(=\)0 hoặc (x-1)²\(=\)1

\(\Leftrightarrow\)x\(=\)1 hoặc x\(=\)2; x\(=\)0

Vậy x\(\in\){1;2;0}

a) Đặt A = \(6^5.5-3^5\)

\(=\left(2.3\right)^5.5-3^5\)

\(=2^5.3^5.5-3^5\)

\(=3^5.\left(2^5.5-1\right)\)

\(=3^5.\left(32.5-1\right)\)

\(=3^5.159\)

\(=3^5.3.53⋮53\)

Vậy \(A⋮53\)

b) Đặt \(B=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{119}.3\)

\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(B⋮3\)

\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7\)

\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)

Vậy \(B⋮7\)

\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)

\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{116}.31\)

\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)

Vậy \(B⋮31\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)

\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(=2.255+2^9.255+...+2^{113}.255\)

\(=255.\left(2+2^9+...+2^{113}\right)\)

\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)

Vậy \(B⋮17\)

c) Đặt C = \(3^{4n+1}+2^{4n+1}\)

Ta có:

\(3^{4n+1}=\left(3^4\right)^n.3\)

\(2^{4n}=\left(2^4\right)^n.2\)

\(3^4\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)

\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)

\(2^4\equiv6\left(mod10\right)\)

\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)

\(\Rightarrow\) Chữ số tận cùng của C là 5

\(\Rightarrow C⋮5\)

d) Đặt \(D=75+\left(4^{2006}+4^{2005}+4^{2004}+...+1\right).25\)

Đặt \(E=4^{2006}+4^{2005}+4^{2004}+...+1\)

\(\Rightarrow4E=4^{2007}+4^{2006}+4^{2005}+...+4\)

\(\Rightarrow3E=4E-E\)

\(=\left(4^{2007}+4^{2006}+4^{2005}+...+4\right)-\left(4^{2006}+4^{2005}+4^{2004}+...+1\right)\)

\(=4^{2007}-1\)

\(\Rightarrow E=\dfrac{\left(4^{2007}-1\right)}{3}\)

\(\Rightarrow D=75+\dfrac{4^{2007}-1}{3}.25\)

Ta có:

\(4^{2007}=\left(4^2\right)^{1003}.4\)

\(4^2\equiv6\left(mod10\right)\)

\(\left(4^2\right)^{1003}\equiv6^{1003}\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow4^{2007}\equiv\left(4^2\right)^{1003}.4\left(mod10\right)\equiv6.4\left(mod10\right)\equiv4\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(4^{2007}\) là 4

`a, 8x=64`

`=>x= 64:8`

`=> x=8`

`b, (-5)x=25`

`=>x=25:(-5)`

`=>x=-5`

`c,4x+1=21`

`=>4x=21-1`

`=>4x=20`

`=>x=20:4`

`=>x=5`

`d, (-3)x-1=8`

`=>(-3)x=8+1`

`=>(-3)x=9`

`=>x=9:(-3)`

`=>x=(-3)`

a) \(\Leftrightarrow x^2-36=64\)

\(\Leftrightarrow x^2=100\)

\(\Leftrightarrow x=\pm10\)

Vậy \(x=\pm10\)

b) \(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{1;3\right\}\)

a) Ta có: \(x^4+64\)

\(=x^4+16x^2+64-16x^2\)

\(=\left(x^2+8\right)^2-\left(4x\right)^2\)

\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)

b) Ta có: \(81x^4+4y^4\)

\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)

\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)

\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)

c) Ta có: \(x^5+x+1\)

\(=x^5+x^2-x^2+x-1\)

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

\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

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

a,= (x+4)\(^3\)

b,= (x-2)\(^3\)

c,= x\(^3\)+8

d,=x\(^3\)-27

a,= x\(^3\)+3.4x\(^2\)+3.4\(^2\).x+4\(^3\)=(x+4)\(^3\)

-3.2.x\(^2\)+3.2\(^2\).x-2\(^3\)= (x-2)\(^3\)

còn c,d áp dụng HĐT là ra! ( đc chx bà nội)

`a)(x+5)^3=-64`

`(x+5)^3=(-4)^3`

`x+5=-4`

`x=-4-5=-9`

Vậy `x=-9`

`2)(2x-3)^3=8`(9 không được)

`(2x-3)^3=2^3`

`2x-3=2`

`2x=5`

`x=5/2`

Vậy `x=5/2`

a)\(\left(x+5\right)^3=64\)

\(\left(x+5\right)^3=4^3\)

\(x+5=4\)

\(x=4-5\)

\(x=-1\)

b) \(\left(2x-3\right)^3=9\)

\(\left(2x-3\right)^3=3^3\)

\(2x-3=3\)

\(2x=3+3\)

\(2x=6\)

\(x=\dfrac{6}{2}\)

\(x=3\)

Giải:

a) \(\left(x+5\right)^3=-64\) 

\(\Rightarrow\left(x+5\right)^3=\left(-4\right)^3\) 

\(\Rightarrow x+5=-4\) 

\(\Rightarrow x=-4-5\) 

\(\Rightarrow x=-9\) 

b) \(\left(2x-3\right)^3=8\) 

\(\Rightarrow\left(2x-3\right)^3=2^3\) 

\(\Rightarrow2x-3=2\) 

\(\Rightarrow2x=2+3\) 

\(\Rightarrow2x=5\) 

\(\Rightarrow x=5:2\) 

\(\Rightarrow x=\dfrac{5}{2}\)

A = 3 + 3² + 3³ + ...+3¹⁰⁰ 

3A = 3² + 3³ + 3⁴ + ...+ 3¹⁰¹

3A - A = ( 3² + 3³ + 3⁴ + ...+ 3¹⁰¹ )  - ( 3 + 3² + 3³ + ...+3¹⁰⁰  ) 

 2A = 3¹⁰¹ - 3 

Thay vào 2A + 3 = 3ⁿ ta có 

 3¹⁰¹ - 3 + 3 = 3ⁿ

3¹⁰¹ = 3ⁿ

=> n = 101

A = 3 + 3² + 3³ +....+ 3¹⁰⁰

\(\Rightarrow\) 3A= 3.(3 + 3² + 3³ +....+ 3¹⁰⁰)

\(\Rightarrow\) 3A= 3² + 3³ + 3⁴ +.....+ 3¹⁰¹

\(\Rightarrow\)3A - A= (3² + 3³ + 3⁴ +.....+ 3¹⁰¹) - (3 + 3² + 3³ +....+ 3¹⁰⁰)

\(\Rightarrow\)2A= 3¹⁰¹ - 3

mà 2A + 3 = 3ⁿ

\(\Rightarrow\)3¹⁰¹ - 3 + 3 = 3ⁿ

\(\Rightarrow\)3¹⁰¹ = 3ⁿ

\(\Rightarrow\)n=101

\(A=3+3^2+3^3+3^{100}\)

\(3A=3^2+3^3+3^4+...+3^{101}\)

\(3A-A=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)

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

Mà \(2A+3=3^n\)

\(3^{101}-3+3=3^n\)

\(3^{101}=3^n\)

\(n=101\)