72-7.(13-x)=14

=>7.(13-x)=72-14

=>7.(13-x)=58

=>13-x=58:7

=>13-x=\(\dfrac{58}{7}\)

\(=>x=13-\dfrac{58}{7}\)

\(=>x=\dfrac{33}{7}\)

Vậy ................

\(72-7(13-x)=14\)

\(7(13-x)=72-14\)

\(7(13-x)=58\)

\(13-x=58:7\)

\(13-x=\) \(\dfrac{58}{7}\)

\(x=13-\dfrac{58}{7}\)

\(x=\dfrac{33}{7}\)

Câu a:

- |- 5 + 3 - 7| - |- 5 + 7|

= -|- 2 - 7| - |2|

= - |-9| - 2

= - 9 - 2

= -11

Câu b:

24 - (72 - 13 + 24) - (72 - 13)

= 24 - 72 + 13 - 24 - 72 + 13

= (24 - 24) - (72 + 72) + (13 + 13)

= 0 - 144 + 26

= - 118

Câu c:

|4 - 9 - 5| - (4 - 9 - 5) - 15 + 9

=|- 5 - 5| - (- 5 - 5) - 15 + 9

= |-10| - (-10) - 15 + 9

= 10 + 10 - 15 + 9

= 20 - 15+ 9

= 5 +9

= 14

Câu d:

- 20 - (25 - 11 + 8) +(25 - 8 + 20)

= - 20 - 25 + 11 - 8 + 25 - 8 + 20

= (-20 + 20) + (25 - 25) + (11 - 8- 8)

= 0 + 0+ 3 - 8

= -5

Câu a:

- |- 5 + 3 - 7| - |- 5 + 7|

= -|- 2 - 7| - |2|

= - |-9| - 2

= - 9 - 2

= -11

Câu b:

24 - (72 - 13 + 24) - (72 - 13)

= 24 - 72 + 13 - 24 - 72 + 13

= (24 - 24) - (72 + 72) + (13 + 13)

= 0 - 144 + 26

= - 118

Câu c:

|4 - 9 - 5| - (4 - 9 - 5) - 15 + 9

=|- 5 - 5| - (- 5 - 5) - 15 + 9

= |-10| - (-10) - 15 + 9

= 10 + 10 - 15 + 9

= 20 - 15+ 9

= 5 +9

= 14

Câu d:

- 20 - (25 - 11 + 8) +(25 - 8 + 20)

= - 20 - 25 + 11 - 8 + 25 - 8 + 20

= (-20 + 20) + (25 - 25) + (11 - 8- 8)

= 0 + 0+ 3 - 8

= -5

Bai 2 : 

\(\hept{\begin{cases}x+y=7\\x-7=13\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=7\\x=20\end{cases}}}\)

Thay x vào phương trình đầu ta có : 

\(20+y=7\Leftrightarrow y=-13\)

Vậy \(\left\{x;y\right\}=\left\{20;-13\right\}\)

Thử \(20-13=7\); \(20-7=13\)( thỏa mãn ) 

1: Ta có: \(2155-\left(174+2155\right)+\left(-68+174\right)\)

\(=2155-174-2155-68+174\)

=-68

2) Ta có: \(-25\cdot72+25\cdot21-49\cdot25\)

\(=-25\cdot\left(72-21+49\right)\)

\(=-25\cdot100=-2500\)

a) \(\frac{3}{7}-\frac{1}{7}x=\frac{2}{3}\)

=> \(\frac{1}{7}x=\frac{3}{7}-\frac{2}{3}=-\frac{5}{21}\)

=> \(x=-\frac{5}{21}:\frac{1}{7}=-\frac{5}{21}\cdot7=-\frac{5}{3}\)

b) \(3x^2-2=72\)=> 3x² = 74 => x² = 74/3 => x không thỏa mãn

c) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)

=> \(\left(19x+2\cdot25\right):14=5^2-4^2=9\)

=> \(\left(19x+50\right):14=9\)

=> \(19x+50=126\)

=> \(19x=76\)

=> x = 4

d) \(x:\frac{1}{2}+x:\frac{1}{4}+x:\frac{1}{8}+x:\frac{1}{16}+x:\frac{1}{32}=343\)

=> \(x\cdot2+x\cdot4+x\cdot8+x\cdot16+x\cdot32=343\)

=> \(x\left(2+4+8+16+32\right)=343\)

=> x . 62 = 343

=> x = 343/62

a: Ta có: \(3^{2020}=3^{2018}\cdot3^2=3^{2018}\cdot9\)

mà 9<10

nên \(3^{2020}< 10\cdot3^{2018}\)