`96.(7^{2}+1).(7^{4}+1).(7^{8}+1).(7^{16}+1)`

`=2.(7^{2}-1).(7^{2}+1).(7^{4}+1).(7^{8}+1).(7^{16}+1)`

`=2.(7^{4}-1).(7^{4}+1).(7^{8}+1).(7^{16}+1)`

`=2.(7^{8}-1).(7^{8}+1).(7^{16}+1)`

`=2.(7^{16}-1).(7^{16}+1)`

`=2.(7^{32}-1)`

(a):

\(P=\dfrac{3x-12}{3x^2-3x-36}=\dfrac{3\left(x-4\right)}{3\left(x^2-x-12\right)}\\ =\dfrac{3\left(x-4\right)}{3\left(x-4\right)\left(x+3\right)}\\ =\dfrac{1}{x+3}\left(ĐK:x\ne\left\{4;-3\right\}\right)\)

(b):

\(x=\dfrac{1}{2}\left(TMDK\right)=>P=1:\left(\dfrac{1}{2}+3\right)=1:\dfrac{7}{2}=\dfrac{2}{7}\)

(c):

\(P=\dfrac{1}{x+3}\in Z=>1⋮\left(x+3\right)\\ =>x+3\inƯ\left(1\right)=\left\{\pm1\right\}\\ =>x\in\left\{-4;-2\right\}\left(TMDK\right)\)

\(P=\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\\ =2x^3-3x^2+4x+4x^2-6x+8-\left(2x^3+x^2-2x-1\right)\\ =2x^3+x^2-2x+8-2x^3-x^2+2x+1\\ =9\)

9

A = (1-3m)(9m² +3m +1) - (6-26m³) 

A = 9m²  + 3m + 1 - 27m³ - 9m² - 3m - 6 + 26m³

A = -m³ - 5 

với m  = 5 ⇔A = -5³ - 5 = -130

b, B = ( 2x - 3)² + ( 2x + 1)² - 2 (4x² - 9) 

B =  ( 2x - 3 - 2x - 1)² + 2(2x-3)(2x+1) - 2(4x² - 9)

B = 4² + 8x² + 4x - 12x - 6 - 8x² + 18

B =- 8x +  16 + 18  - 6 

B = -8x  + 28

với x = 3 ⇔ B = -8. 3 + 28 = 4

a/

\(\widehat{B}+\widehat{C}=200^o\) (1)

\(\widehat{B}+\widehat{D}=180^o\) (2)

\(\widehat{C}+\widehat{D}=120^o\) (3)

Cộng 2 vế của (1) (2) (3)

\(\Rightarrow2\left(\widehat{B}+\widehat{C}+\widehat{D}\right)=500^o\Rightarrow\widehat{B}+\widehat{C}+\widehat{D}=250^o\)

\(\Rightarrow\widehat{A}=360^o-250^o=110^o\) (tổng góc trong 1 tứ giác bằng \(360^o\) )

Cộng 2 vế của (1) với (2)

\(\Rightarrow\widehat{B}+\widehat{B}+\widehat{C}+\widehat{D}=380^o\Rightarrow\widehat{B}=380^o-250^o=130^o\)

Cộng 2 vế của (2) với (3)

\(\Rightarrow\widehat{B}+\widehat{C}+\widehat{D}+\widehat{D}=300^o\Rightarrow\widehat{D}=300^o-250^o=50^o\)

\(\Rightarrow\widehat{C}=360^o-\widehat{A}-\widehat{B}-\widehat{D}=360^o-110^o-130^o-50^o=70^o\)

b/

Xét tg AIB có

\(\widehat{AIB}=180^o-\widehat{IAB}-\widehat{IBA}=180^o-\dfrac{\widehat{A}}{2}-\dfrac{\widehat{B}}{2}=\)

\(=180^o-\dfrac{110^o}{2}-\dfrac{130^o}{2}=60^o\)

\(\dfrac{\widehat{C}+\widehat{D}}{2}=\dfrac{70^o+50^o}{2}=60^o\)

\(\Rightarrow\widehat{AIB}=\dfrac{\widehat{C}+\widehat{D}}{2}=60^o\)

Các điểm em cần phải viết hoa nhé. Cô đã sửa lại, em lưu ý lần sau nếu ko thì câu hỏi sẽ bị xóa

vâng

`(x+1)^{3}=1=1^{3}`

`=>x+1=1`

`=>x=1-1=0`

Vậy `x=0`

x = 0