\(\frac{x^3-16x}{x^4+64x}=\frac{A}{x^2-4x+16}\)

\(\Rightarrow\frac{x\left(x^2-16\right)}{x\left(x^3+4^3\right)}=\frac{A}{x^2-4x+16}\)

\(\Rightarrow\frac{\left(x-4\right)\left(x+4\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\frac{A}{x^2-4x+16}\)

\(\Rightarrow\frac{x-4}{x^2-4x+16}=\frac{A}{x^2-4x+16}\)

\(\Rightarrow A=x-4\)

Vậy A=x-4 thì 2 biểu thức bằng nhau

Ta có: \(A=\left(x^3-16x\right).\left(x^2-4x+16\right):x^4+64x\)

\(A=\left[x^5-4x^4+16x^3-\left(-16x^3+64x^2-256x\right)\right]:\left(x^4+64x\right)\)

\(A=\left(x^5-4x^4+16x^3+16x^3-64x^2+256x\right):\left(x^4+64x\right)\)

\(A=\left(x^5-4x^4+32x^3-64x^2+256x\right):\left(x^4+64x\right)\)

\(A=x-4\) dư \(32x^3+512x\)

Chắc mình làm sai rồi nên bạn đừng tham khảo nha vì mình mới học bài 1 nên làm như thế để học mấy bài sau rồi mình tính tiếp cho nha

a: =9^8:9^3:9^2=9^3

b: =4x^2*4x*4*4x*16

=x^4*4^5

`a, 9^8 : 9^3 : 3^4`

`= 9^8 : 9^3 : 9^2`

`= 9^(8-3-2)= 9^3`.

`b, x . 4x . 16x . 64x`

`= x. 4x . 4^2x . 4^3x`

`= 4^6x^4`

a) Ta có: \(4x^4+81\)

\(=\left(2x^2\right)^2+9^2+36x^2-36x^2\)

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

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

b) Ta có: \(64x^4+y^4\)

\(=\left(8x^2\right)^2+\left(y^2\right)^2+16x^2y^2-16x^2y^2\)

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

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

a) x³ - 9x² + 27x - 27 = -8

<=> x³ - 3x².3 + 3x.3² - 3³ = -8

<=> (x - 3)³ = -2³

<=> x - 3 = -2

<=> x = 1 (T/m)

Vậy x = 1.

b) 64x³ + 48x² + 12x + 1 = 27

<=> (4x)³ + 3.(4x)².1 + 3.4x.1² + 1³ = 27

<=> (4x + 1)³ = 3³

<=> 4x + 1 = 3

<=> 4x = 2

<=> x = \(\frac{1}{2}\)(T/m)

Vậy x = \(\frac{1}{2}\).

c) (2x - 1)³ - 4x².(2x - 3) = 5

<=> (8x³ - 12x² + 6x - 1) - (8x³ - 12x²) = 5

<=> 8x³ - 12x² + 6x - 1 - 8x³ + 12x² = 5

<=> 6x - 1 = 5

<=> 6x = 6

<=> x = 1 (T/m)

Vậy x = 1.

\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)

\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)

c) \(\left(2x-1\right)^3-4x^2.\left(2x-3\right)=5\)

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-12x^2\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

d) \(\left(x+4\right)^3-x^2.\left(x+12\right)=16\)

\(\Leftrightarrow\left(x^3+12x^2+48x+64\right)-\left(x^3+12x^2\right)=16\)

\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=16\)

\(\Leftrightarrow48x+64=16\)

\(\Leftrightarrow48x=-48\)

\(\Leftrightarrow x=-1\)

#vì câu a,b có người làm rồi nên mình chỉ làm c,d thôi nhé ! :)

Học Tốt !!

<=>   \(\sqrt{64\left(x+1\right)}-\sqrt{25\left(x+1\right)}+\sqrt{4\left(x+1\right)}=20\)

<=> \(8\sqrt{\left(x+1\right)}-5\sqrt{\left(x+1\right)}+2\sqrt{\left(x+1\right)}=20\)

<=>   . \(5\sqrt{\left(x+1\right)}=20\)

<=>  \(\sqrt{\left(x+1\right)}=4\)

=> x+1=16

=> x=15