BÀI 3:

bài 4:

Là mấy bài này em làm được bài nào chưa?

mn giúp e với

\(\left(x^3-8\right):\left(x^2+2x+4\right)\\ =\left(x-2\right)\left(x^2+2x+4\right):\left(x^2+2x+4\right)\\ =x-2\)

bài 2

a)

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

b)

\(x^2-6xy+9y^2\\ =\left(x-3y\right)^2\)

c)

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

d)

\(x^2+4x+3\\ =x^2+3x+x+3\\ =x\left(x+3\right)+\left(x+3\right)\\ =\left(x+3\right)\left(x+1\right)\)

4,\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)

2,\(=y\left(x^2-6x+9\right)=y\left(x-3\right)^2\)

1,=\(2y\left(xy-2\right)\)

`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12

 Mik có ghi thiếu nhé mn =100/609

\(\dfrac{1}{3\times5}=\dfrac{2}{3}-\dfrac{3}{5}\)

\(\dfrac{1}{5\times7}=\dfrac{3}{5}-\dfrac{4}{7}\)

\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}\right)\)

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

\(=\dfrac{1}{2}\cdot\dfrac{2x+3-3}{3\left(2x+3\right)}\)

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

cho xin cái đề bài y

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