1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

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

=> \(\frac{4-x}{x+5}=\frac{4}{9}\)

=> 9.(4-x)=4.(x+5)

=> 36-9x=4x+20

=> -9x-4x=20-36

=> -13x=-16

=> x=\(\frac{-16}{-13}=\frac{16}{13}\)

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

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

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

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

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

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

(-100):(x+5)=-5

x+5=-100:(-5)

x+5=20

x=20-5

x=15

=> -5(x+5)=-100

=>x+5=20

=>x=20-5=15

Vậy x=15

thấy đúng thì k hộ chị nha,chúc e học tốt

`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