Sửa lại giúp mình câu a) một chút:

\(\dfrac{-22}{25}+\left(\dfrac{3}{7}-0,12\right)\\ =\dfrac{-22}{25}+\dfrac{3}{7}-0,12\\ =\dfrac{-22}{25}+\dfrac{3}{7}-\dfrac{3}{25}\\ =\dfrac{-25}{25}+\dfrac{3}{7}\\ =-1+\dfrac{3}{7}\\ =-\dfrac{4}{7}\)

`#3107.101107`

a)

\(\dfrac{-22}{25}+\left(\dfrac{3}{7}-0,12\right)\\ =\dfrac{-22}{25}+\dfrac{3}{7}-0,12\\ =-\dfrac{79}{175}-0,12\\ =-\dfrac{4}{7}\)

b)

\(\dfrac{3}{8}-\left(1,2-\dfrac{5}{8}\right)\\ =\dfrac{3}{8}-1,2+\dfrac{5}{8}\\ =\dfrac{8}{8}-1,2\\ =1-1,2\\ =-0,2\)

c)

\(\dfrac{4}{15}-\left(2,9-\dfrac{11}{15}\right)\\ =\dfrac{4}{15}-2,9+\dfrac{11}{15}\\=\dfrac{15}{15}-2,9\\ =1-2,9\\ =-1,9\)

d)

\(\dfrac{17}{11}-\left(\dfrac{6}{5}-\dfrac{16}{11}\right)+\dfrac{26}{5}\\ =\dfrac{17}{11}-\dfrac{6}{5}+\dfrac{16}{11}+\dfrac{26}{5}\\ =\left(\dfrac{17}{11}+\dfrac{16}{11}\right)+\left(-\dfrac{6}{5}+\dfrac{26}{5}\right)\\ =\dfrac{33}{11}+\dfrac{20}{5}\\ =3+4\\ =7\)

e)

\(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{7}\right)\\ =\dfrac{39}{5}+\dfrac{9}{4}-\dfrac{9}{5}-\dfrac{5}{4}-\dfrac{6}{7}\\ =\left(\dfrac{39}{5}-\dfrac{9}{5}\right)+\left(\dfrac{9}{4}-\dfrac{5}{4}\right)-\dfrac{6}{7}\\ =\dfrac{30}{5}+\dfrac{4}{4}-\dfrac{6}{7}\\ =6+1-\dfrac{6}{7}\\=\dfrac{43}{7}\)

`#3107.101107`

a)

`2xy(x + 3y^2)`

`= 2x^2y + 6xy^3`

b)

`-7x^2(3x - 4y)`

`= -21x^3 + 28x^2y`

c)

`3x^2y(-3x^2 - y^2)`

`= -9x^4y - 3x^2y^3`

f)

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

`= -x^3y - 2x^2y^2 + 3xy`

g)

`3x^2y(x^2 - 3y + 2xy^2)`

`= 3x^4y - 9x^2y^2 + 6x^3y^3`

h)

`-xy(x^2 + xy + y^2)`

`= -x^3y - x^2y^2 - xy^3`

k)

`x^2y(2xy + x^2 - xy^2)`

`= 2x^3y^2 + x^4y - x^3y^3`

l)

`-2xy^2 (x^2 - x^3y + 3)`

`= -2x^3y^2 + 2x^4y^3 - 6xy^2`

m)

`-2x^2y (3xy^2 - y^2 + xy)`

`= -6x^3y^3 + 2x^2y^3 - 2x^3y^2`

`#3107.101107`

1.

a)

`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`

`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`

`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`

`= 1/3* (1 - 1/103)`

`= 1/3*102/103`

`= 34/103`

b)

`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`

`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`

`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`

`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`

`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`

`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`

`= -1/2 * (1 - 1/101)`

`= -1/2*100/101`

`= -50/101`

2.

`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`

`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`

`= 1-1/100`

`= 99/100`

:v Cập nhật thông tin muộn quá kh nhỉ