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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

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

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

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

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

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

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

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

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

\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)

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

\(=\frac{\left(x+1\right)\left(x+1\right)}{x\left(x-1\right)+2\left(x-1\right)+\left(x+1\right)}\)

Ddeeff sao rồi bạn ko rút gọn được

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)`

11 tháng 11 2023

\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)

\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)

\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)

Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:

\(N=\dfrac{\left(y-1\right)\left(y+1\right)+1}{y}\)

\(=\dfrac{y^2-1+1}{y}\)

\(=\dfrac{y^2}{y}\)

\(=y\)

\(=x^2+7x+11\)

Vậy \(N=x^2+7x+11\).

\(\text{#}Toru\)

11 tháng 11 2023

À bạn ơi cho mình hỏi ngoài lề 1 chút được k ạ?

20 tháng 12 2022

Câu 4: Không có nghĩa khi x-3=0

=>x=3

Câu 5:

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

\(\dfrac{x^3+3x^2-2}{x^3+3x+4}\)

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

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

\(=\dfrac{x^2+2x-2}{x^2-x+4}\)

28 tháng 5 2022

:) bóc lột !

DD
28 tháng 5 2022

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)