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Xét tứ giác ABCD

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\Rightarrow\dfrac{\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}}{2}=180^o\)

Xét tg CID có

\(\widehat{ICD}+\widehat{IDC}=180^o-\widehat{CID}=180^o-115^o=65^o\)

\(\Rightarrow\dfrac{\widehat{C}+\widehat{D}}{2}=65^o\Rightarrow\dfrac{\widehat{A}+\widehat{B}}{2}=180^o-65^o=115^o\Rightarrow\widehat{A}+\widehat{B}=230^o\)

\(\Rightarrow\widehat{A}=\dfrac{230^o+50^o}{2}=140^o\Rightarrow\widehat{B}=230^o-140^o=90^o\)

A = (x -1)³- (x +1)³

A =x³ -3x² + 3x - 1 - x³ - 3x² - 3x - 1

A = -6x² - 2

B = (x-y)³ + 3xy (x-y) 

B = x³ -3x²y + 3xy² - y³ + 3x²y - 3xy²

B = x³ - y³

Bạn tham khảo nhé.

Q = (3x-5)(2x+11)-(2x+3)(3x+7)

Q = 6x² + 33x – 10x – 55 – (6x² + 14x + 9x + 21)

Q = 6x² + 33x – 10x – 55 – 6x² - 14x - 9x – 21

Q = (6x² – 6x²) + (33x – 10x - 14x - 9x) + (– 55 – 21)

Q = 0 + 0 – 76

Q = -76

P = (x + 1)(x + 2)(x + 3)(x + 4) - 8

= [(x + 1)(x + 4)][(x + 2)(x + 3] - 8

= (x² + 5x + 4)(x² + 5x + 6) - 8

= (x² + 5x + 5 - 1)(x² + 5x + 5 + 1) - 8

= (x² + 5x + 5)² - 1 - 8 = (x² + 5x + 5)² - 9

= (x² + 5x + 8)(x² + 5x + 2)

\(\dfrac{3x^2+9x-3}{x^2+x-2}-\dfrac{x+1}{x+2}+\dfrac{x-2}{1-x}\left(ĐK:x\ne\left\{1;-2\right\}\right)\\ =\dfrac{3x^2+9x-3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+1}{x+2}-\dfrac{x-2}{x-1}\\ =\dfrac{3x^2+9x-3-\left(x+1\right)\left(x-1\right)-\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{3x^2+9x-3-\left(x^2-1\right)-\left(x^2-4\right)}{\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{3x^2+9x-3-x^2+1-x^2+4}{\left(x+2\right)\left(x-1\right)}\\ =\dfrac{x^2+9x+2}{\left(x+2\right)\left(x-1\right)}\)

`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):

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