a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)

\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)

=(x-3)^2/(x-3)

=x-3

b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)

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

\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)

=-7(x+2)/(x-2)(x+2)

=-7/(x-2)

Hình như đâu có câu nào đúng đâu bạn!

`!`

`(x^3-3x^2):(x-3)`         

`= x^2(x-3)  :(x-3)` 

`=x^2`

`----`

`(2x^2+2x-4):(x+2)`

`=2(x^2+x-2) :(x+2)`

`= 2 (x^2+2x-x-2) :(x+2)`

\(=2\left[x\left(x+2\right)-\left(x+2\right)\right]:\left(x+2\right)\)

`= 2(x+2)(x-1) :(x+2)`

`=2(x-1)`

`-------`

`(x^4-x-14):(x-2)`

https://hoc24.vn/cau-hoi/x4-x-14-x-2-giup-minh.204306769717

bn tham khảo ở đây nha

`----`

`(x^3-3x^2+x-3):(x-3)`

`= x^2(x-3) +(x-3) :(x-3)`

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

`=x^2+1`

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)

\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)

\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)

b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)

c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)

phần dưới là tìm x

\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

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

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

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

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

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)

\(=2x^3-x+x^2-y\)

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349

\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{6.2}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}\\ =\dfrac{3\left(4+x\right)}{2x\left(x+4\right)}\\ =\dfrac{3}{2x}\)

________

\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\\ \left(\text{đ}k\text{x}\text{đ}:x\ne\pm2\right)\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{x^2-4}\\ =2\)

a: \(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)

\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)

b: \(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-2\right)^2+x-14}{x^2-4}\)

\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{x^2-4}=\dfrac{2x^2-8}{x^2-4}=2\)

a. \(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{12}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)

b. \(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\left(đk:x\ne\pm2\right)\\ =\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-2x-2x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}=2\)

`a)3x(2x^2-3x+4)`

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

______________________________________________

`b)(x+3)^2+(3x-2)(x+4)`

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

`=4x^2+16x+1`

______________________________________________

`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]`       `ĐK: x \ne +-1`

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

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

`=[2x^2-2]/[x^2-1]`

`=2`

hếp