Đáp án D

Đặt \(t=\tan\frac{x}{2}\rightarrow dx=\frac{2dt}{1+t^2}\)

Khi đó : \(I=\int\frac{4\frac{dt}{1+t^2}}{\frac{4}{1+t^2}-\frac{1-t^2}{1+t^2}+1}=\int\frac{2dt}{1+2t^2}=\int\left(\frac{1}{t}-\frac{1}{t+2}\right)dt=\ln\left|\frac{1}{t+2}\right|+C=\ln\left|\frac{\tan\frac{x}{2}}{\tan\frac{x}{2}+2}\right|+C\)

tick nhé 

a) (x-2)(x+2)(x^2-10)-72=(x^2-4)(x^2-82)

b) x^8+x^6+x^4+x^2+1=x^2 (x^4+x^3+x^2+1+1/x^2)

c)(x+y)^4+x^4+y^4=(x+y)^4+(x+y)^4=2 (x+y)^4

a) (x-2)(x+2)(x^2 - 10) -72

= (x^2 - 4)(x^2 - 10) - 72

= x^4 - 4x^2 -10x^2 + 40 - 72

= x^4 - 14x^2 - 32

= x^4 - 16x^2 + 2x^2 - 32

= x^2(x^2 - 16) + 2(x^2 - 16)

= (x^2 - 16)(x^2 + 2)

= (x-4)(x+4)(x^2 + 2)

c) (x+y)⁴ + x⁴ + y⁴

= 2x⁴ + 4xy³ + 6x²y² + 4x³y + 2y³

= 2(y⁴ + 2xy³ + 3x²y² + 2x³y + x⁴)

= 2(y² + xy + y²)²

\(d,(x+1)^4+(x^2+x+1)^2\)

\(=x^4+4x^3+6x^2+4x+1+x^4+2x^3+3x^2+2x+1\)

\(=(x^4+x^4)+(4x^3+2x^3)+(6x^2+3x^2)+(4x+2x)+(1+1)\)

\(=2x^4+6x^3+9x^2+6x+2\)

\(=\left[x^2+2x+2\right]\left[2x^2+2x+1\right]\)

Chọn B

Bài 1

A= x² + 13x - 30 = x² + 15x - 2x - 30 = ( x² +15x) - ( 2x+30) = x(x+15) - 2(x+15) = (x+15)(x-2)

B = -x²+3x-2 = -(x² - 3x +2 )= -( x² - x - 2x +2) = - [(x² - x) - ( 2x - 2)] = - [x(x - 1) - 2(x - 1)] = (x - 1)(x - 2)

Bài 2

a) 4x² - 169 = (2x)² - 13² = ( 2x - 13)(2x + 13)

b) 121x³ - 27 đề sai hay sao ý

c) (x + 10)³ = x³ + 30x² + 300x + 100

d) x³ + y³ = (x + y)(x² - xy - y²)

k cho mình nhé bạn. Thanks you ! ^^

\(x^2+13x-30\)

\(=x^2+2x-15x-30\)

\(=x\left(x+2\right)-15\left(x+2\right)\)

\(=\left(x+2\right)\left(x-15\right)\)

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

i)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)

h)\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)

g)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

f)\(x^2-5x-14=x^2-7x+2x-14\)

                             \(=\left(x+2\right)\left(x-7\right)\)

i)\(x^2-7x+10=x^2-5x-2x+10\)

                              \(=\left(x-2\right)\left(x-5\right)\)

h)\(x^2-7x+12=x^2-4x-3x+12\)

                              \(=\left(x-3\right)\left(x-4\right)\)

g)\(x^2+6x+5=x^2+x+5x+5\)

                           \(=\left(x+5\right)\left(x+1\right)\)

f) \(x^2-5x-14\)

\(=x^2-7x+2x-14\)

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

\(=x\left(x-7\right)+2\left(x-7\right)\)

\(=\left(x+2\right)\left(x-7\right)\)

i) \(x^2-7x+10\)

\(=x^2-5x-2x+10\)

\(=\left(x^2-5x\right)-\left(2x-10\right)\)

\(=x\left(x-5\right)-2\left(x-5\right)\)

\(=\left(x-2\right)\left(x-5\right)\)

h) \(x^2-7x+12\)

\(=x^2-3x-4x+12\)

\(=\left(x^2-3x\right)-\left(4x-12\right)\)

\(=x\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-4\right)\left(x-3\right)\)

g) \(x^2+6x+5\)

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

\(=\left(x^2+x\right)+\left(5x+5\right)\)

\(=x\left(x+1\right)+5\left(x+1\right)\)

\(=\left(x+5\right)\left(x+1\right)\)

a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)

b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)

c)x^2-2x-5x+10=x(x-2)-5(x-2)=(x-5)(x-2)

d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(x-2)

Đặt \(t=2-3x^2\)\(\Rightarrow\begin{cases}dt=-6xdx\\x^2=\frac{2-t}{3}\end{cases}\)\(\Leftrightarrow x^2\left(2-3x^2\right)^8=\left(\frac{2-t}{3}\right)t^8=\frac{1}{3}\left(2t^8-t^9\right)\)

Vậy : 

\(I=\int x^2\left(2-3x^2\right)^8dx=\frac{1}{3}\left(2\int t^8dt-\int t^9dt\right)=\frac{2}{27}t^9-\frac{1}{30}t^{10}+C\)

  \(=\frac{2}{27}\left(2-3x^2\right)^9-\frac{1}{30}\left(2-3x^2\right)^{10}+C\)