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)

b) (x – 1)(x + 1)(x² + 1)

= [x .(x + 1) – 1 .(x + 1)] . (x² + 1)

= {x.x + x.1 + (-1).x + (-1).1}. (x² + 1)

= (x² + x – x – 1) . (x² + 1)

= (x² – 1) . (x² + 1)

= x² . (x² +1) – 1.(x² + 1)

= x² . x² + x² . 1 – (1.x² + 1.1)

= x⁴ + x² – (x² + 1)

= x⁴ + x² – x² – 1

= x⁴ – 1

Ta có: 

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

________

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

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

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

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

Câu 3: 

Tuổi của chị là:

(36+8):2=22(tuổi)

Tuổi của em là:

22-8=14(tuổi)

Ta có : 1x2x3x...x9-1x2x3x...x8-1x2x3x...x8^2

=1x2x3x...x8x(9-1-8)

=1x2x3x...x8x0

=0

Nhớ  k cho mik nha !!!

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

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

\(=\dfrac{1}{x+1}\)

Ta có: 

answer

1+2=3

4x5=20

8:2=4

20-17=3

vd : 1 + 2 = One plus two equals three

giúp đc hông ạ?

#buithanhthao:

:v đầu bài ghi rõ mak

bằng Tiếng Anh -.-