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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

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

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

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

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

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

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

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

23 tháng 12 2023

Câu 5: B

Câu 6: 

a: ĐKXĐ: \(x-2\ne0\)

=>\(x\ne2\)

b: ĐKXĐ: \(x+1\ne0\)

=>\(x\ne-1\)

8:

\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)

\(=\dfrac{x^2+4-4x}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)

\(=\dfrac{\left(x-2\right)^2}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)

\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)

7: 

\(\dfrac{8x^3yz}{24xy^2}\)

\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)

\(=\dfrac{x^2z}{3y}\)

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 

26 tháng 2 2022

(-3).8/8.6 rút gọn

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 `

24 tháng 11 2021

\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

26 tháng 11 2021

\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

NV
26 tháng 3 2023

1.

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

\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x+4}{x-3}\)

b.

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

\(\Rightarrow x=10\) (thỏa mãn)

2.

\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)

26 tháng 3 2023

Em cảm ơn ạ