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4 tháng 7 2023

1) \(\left(x+1\right)^3-\left(x-4\right)\left(x+4\right)-x^3\)

\(=\left(x^3+3x^2+3x+1\right)-\left(x^2-16\right)-x^3\)

\(=x^3+3x^2+3x+1-x^2+16-x^3\)

\(=2x^2+3x+17\)

2) \(\left(x+2\right)^3-x\left(x+3\right)\left(x-3\right)-12x^2-8\)

\(=\left(x^3+6x^2+12x+8\right)-x\left(x^2-9\right)-12x^2-8\)

\(=x^3+6x^2+12x+8-x^3+9x-12x^2-8\)

\(=-6x^2+21x\)

`@` `\text {Ans}`

`\downarrow`

`1.`

\((x + 1) ^ 3 - (x - 4)(x + 4) - x ^ 3\)

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

`= x^3 + 3x^2 + 3x + 1 - (x^2 + 4x - 4x - 16) - x^3`

`= x^3 + 3x^2 + 3x + 1 - (x^2 - 16) - x^3`

`= x^3 + 3x^2 + 3x + 1 - x^2 + 16 - x^3`

`= (x^3 - x^3) + (3x^2 - x^2) + 3x + (1+16)`

`= 2x^2 + 3x + 17`

`2.`

\((x + 2) ^ 3 - x(x + 3)(x - 3) - 12x ^ 2 - 8\)

`= x^3 + 6x^2 + 12x + 8 - [ (x^2 + 3x)(x-3)] - 12x^2 - 8`

`= x^3 + 6x^2 + 12x + 8 - (x^3 - 9x) - 12x^2 - 8`

`= x^3 + 6x^2 + 12x +8 - x^3 + 9x - 12x^2 - 8`

`= (x^3 - x^3) + (6x^2 - 12x^2) + (12x + 9x) + (8-8)`

`= -6x^2 + 21x `

a kham khảo nha , e nhờ a e lm chứ ko phải e lm nha ! 

\(\left(x-2\right)\left(\frac{3}{x}+2-\frac{5}{2x}-4+\frac{8}{x^2}-4\right)\)

\(\left(x-2\right)\left[\left(\frac{3}{x}-\frac{5}{2x}\right)-6+\frac{8}{x^2}\right]\)

\(\left(x-2\right)\left(\frac{1}{2x}-6+\frac{8}{x^2}\right)\)

15 tháng 3 2020

\(\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)\)

\(=\left(x-2\right)\left[\frac{3}{x+2}-\frac{5}{2\left(x-2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{3.2\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{8.2}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)-5\left(x+2\right)+16}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\frac{\left(x-2\right)\left(x-6\right)}{2\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x-6}{2\left(x+2\right)}\)

16 tháng 12 2019

\(\frac{x^3+125}{x^2-3x-40}=\frac{x^3+5^3}{\left(x^2+5x\right)-\left(8x+40\right)}=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{x\left(x+5\right)-8\left(x+5\right)}\)

\(=\frac{\left(x+5\right)\left(x^2-5x+25\right)}{\left(x+5\right)\left(x-8\right)}=\frac{x^2-5x+25}{x-8}\)

22 tháng 7 2023

a) \(\dfrac{3x^2+6xy}{6x^2}=\dfrac{3x\left(x+2y\right)}{6x^2}=\dfrac{x+2y}{2x}\)

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

c) \(=\dfrac{x+1}{x^3+1}=\dfrac{x+1}{\left(x+1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

`a, (3x^2+6xy)/(6x^2) = (x+2y)/(3x)`

`b, (2x^2-x^3)/(x^2-4) = (x^2(2-x))/((x-2)(x+2))`

`= -x^2/(x+2)`

`c, (x+1)/(x^3+1) = 1/(x^2-x+1)`

20 tháng 12 2022

Câu 4: Không có nghĩa khi x-3=0

=>x=3

Câu 5:

\(A=\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)

b: Ta có: \(\left(x+5\right)^3-x^3-125\)

\(=x^3+15x^2+75x+125-x^3-125\)

\(=15x^2+75x\)

14 tháng 8 2021

1,(x+5)^3−x^3−125

=x^3+15x^2+75x+125−x^3−125

=15x(x+5)

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

25 tháng 7 2023

Mik cảm ơn ạ 

13 tháng 11 2021

2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)