a) A = (2/7 * 1/4 -1/3 * 2/7) : (2/7 * 3/9 - 2/7 * 2/5)

A = [2/7 * (1/4 - 1/3)] : [2/7 * (1/3 - 2/5)]

A = (1/4 - 1/3) : (1/3 - 2/5)

A = 1 1/2

b)  B = (1/5 - 2/7) * 3/4 - 3/4 * (1/3 - 2/7)

1/5 * 2/7 - 1/3 * (2/7 + 3/9) + 3/9 * 1/5

3/4 * (1/5 - 2/7 - 1/3 + 2/7)

1/5 * (2/7 + 1/3) - 1/3 * (2/7 + 1/3)

3/4 * (1/5 - 1/3)

(1/5 - 1/3) * (3/7 + 1/3)

= 1 11/52

a) A = 3/5 * 6/7 + 3/7 : 5/3 - 2/7 : 1 2/3

A = (3/5 * 6/7) + (3/7 : 5/3) - (2/7 : 5/3)

A = 18/35 + 9/35 - 6/35

A = 3/5

b) B = (-13 * 2/5 + -2/9 : 2 1/2 + 2/5 * 11/9) * 2 1/2

B = (-13 + 11/9 + -2/9) * 2/5 * 5/2

B = -13 (11/9 - 2/9)

B = -12

c) C = (-4/5 + 5/7) : 2/3 + (-1/5 + 2/7) : 2/3

C = [(-4/5 + 5/7) + (-1/5 + 2/7)] : 2/3

C = [(-4/5 + -1/5) + (5/7 + 2/7)] : 2/3

C = (-1 + 1 ) : 2/3

d) D = 4/9 : (1/15 - 2/3) + 4/9 : (1/11 - 5/22)

D = 4/9 : [(1/15 - 2/3) + (1/11 - 5/22)]

D = 4/9 : [-9/15 + (-3/22)]

D = 4/9 : [-9/15 - 3/22]

D = 4/9 : -9/15 - 4/9 : -3/22

D = 60/-81 - 88/-27 

D = 20/-27 - 88/-27

D = -4

a) P = 2/3 - (-1/4) + 3/5 - 7/45 - (-5/9) + 1/12 + 1/35

P = (2/3 + 1/4 + 1/12) + (3/5 - 7/45 + 5/9) + 1/35

P = (8/12 + 3/12 + 1/12) + (27/45 - 7/45 + 25/45) + 1/35

P = 1 + 1 + 1/35

P = 2 1/35

b) Q = (5 - 3/4 + 1/5) - (6 + 7/4 - 8/5) - (2 - 5/4 + 16/5)

Q = (5 - 6 - 2) + (-3/4 - 7/4 + 5/4) + (1/5 + 8/5 - 16/5)

Q = -(3 + 5/4 + 7/5)

Q = - 111/20

a) A = (1/3 - 8/15 - 1/7) + (2/3 + -7/15 + 1 1/7)

A = (1/3 + 1/2) + (-8/15 + -7/15) + (-1/7 + 1 1/7)

A = 1 + 5/3 + 1

A = 3/3 + 5/3 + 3/3

A = 11/3

b) B = 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)

B = 1/4 + 3/5 - (1/8 - 2/5 + 1 1/4)

B = (1/4 - 5/4) + (3/5 + 2/5) - 1/8

B = -1 + 1 - 1/8

B = -1/8

a)Ta có:

$\genfrac{}{}{0ex}{}{-3}{8}=\genfrac{}{}{0ex}{}{\left(-3\right).3}{8.3}=\genfrac{}{}{0ex}{}{-9}{24}$

$\genfrac{}{}{0ex}{}{5}{-12}=\genfrac{}{}{0ex}{}{-5}{12}=\genfrac{}{}{0ex}{}{-5.2}{12.2}=\genfrac{}{}{0ex}{}{-10}{24}$

Vì $\genfrac{}{}{0ex}{}{-9}{24}>\genfrac{}{}{0ex}{}{-10}{24}$ nên $\genfrac{}{}{0ex}{}{-3}{8}>\genfrac{}{}{0ex}{}{5}{-12}$

b)Ta có:

$\genfrac{}{}{0ex}{}{3131}{5252/}=\genfrac{}{}{0ex}{}{3131:101}{5252:101}=\genfrac{}{}{0ex}{}{31}{52}$

$\genfrac{}{}{0ex}{}{31}{52}=\genfrac{}{}{0ex}{}{31}{52}$

Vì $\genfrac{}{}{0ex}{}{31}{52}=\genfrac{}{}{0ex}{}{31}{52}$ nên $\genfrac{}{}{0ex}{}{3131}{5252/}=\genfrac{}{}{0ex}{}{31}{52}$

$-127+208-73+92$

$=(-127-73)+(208+92)$

$=-200+300$

$=100$

__

$2353-(473+2153)+(-55+373)$

$=2353-473-2153-55+373$

$=(2353-2153)+(-473+373)-55$

$=200-100-55$

$=100-55$

$=45$

a) (454−�)+43=116=>454−�+64=116=>518−�=116=>�=518−116=402(454−x)+4³3=116

=>454−x+64=116

=>518−x=116

=>x=518−116=402

b) �∈N=>�+1∈N∗x∈N=>x+1∈N^*

Theo bài ra : $x+1\in Ư\left(15\right)=\left\{1;3;5;15\right\}$

$=>x\in \left\{0;2;4;14\right\}$

Gọi số hàng dọc chia được là $x$ (hàng), $x\in \mathbb{N}$ và $x\ge 5$.

Theo bài ra ta có: $48⋮x$; $18⋮x$

Suy ra $x\in$ ƯC$(18,48)$.

Ta có: $18=2.3^2$; $48=2^4.3$

Suy ra ƯCLN$(18,48)=2.3=6$

Do đó, $x\in$ ƯC$(18,48)$ = Ư$(6)=\{1;2;3;6\}$.

Mà $x\ge 5$ nên $x=6$.

Vậy có thể xếp được thành $6$ hàng dọc.

Chiều cao hình bình hành : 

       $24:3=8\left(m\right)$

Diện tích hình bình hành ban đầu :

        $20\times 8=160\left(m^2\right)$

$-9\le x\le 7$

Mà $x$ là số nguyên

   $=>x\in${$-9;-8;-7;-6;....;6;7$}

Tổng các số nguyên $x$ là:

    $17(7-9):2=-17$