\(A=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)=-8\)

\(B=\left(455-311-144\right)+789=789\)

\(C=183-150+120-183-150-20=120-20=100\)

A=-8

B=789

C=100

17/18

13/6

30/28

16/5

5/9 + 7/18 = 10/18 + 7/18 = 17/18
3 - 5/6= 18/6 - 5/6 = 13/6
5/4 x 6/7 = 15/14
2 : 5/8 = 2/1 x 8/5 = 16/5

\(\dfrac{5}{9}+\dfrac{7}{18}=\dfrac{10}{18}+\dfrac{7}{18}=\dfrac{17}{18}\)

\(\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{13}{6}\)

\(\dfrac{5}{4}\times\dfrac{6}{7}=\dfrac{30}{28}=\dfrac{15}{14}\)

\(2\times\dfrac{8}{5}=\dfrac{16}{5}\)

Câu 19: B

Câu 20: D

19, C

20, D

19. C

20. D

1) $(-81)\cdot27-3^4\cdot(-77)$

$=(-81)\cdot27-81\cdot(-77)$

$=-81\cdot[27+(-77)]$

$=-81\cdot(27-77)$

$=-81\cdot(-50)$

$=81\cdot50$

$=4050$

2) $(-16)\cdot32-(-4)^2\cdot18$

$=(-16)\cdot32-16\cdot18$

$=-16\cdot(32+18)$

$=-16\cdot50$

$=-800$

3) $(-2)^4\cdot289-16\cdot189$

$=16\cdot289+16\cdot(-189)$

$=16\cdot[289+(-189)]$

$=16\cdot(289-189)$

$=16\cdot100$

$=1600$

4) $41\cdot(-18)+41\cdot(-81)-41$

$=1\cdot(-18)+41\cdot(-81)+41\cdot(-1)$

$=41\cdot[(-18)+(-81)+(-1)]$

$=41\cdot(-18-81-1)$

$=41\cdot(-99-1)$

$=41\cdot(-100)$

$=-4100$

5) $17\cdot(-37)-23\cdot37-46\cdot(-37)$

$=17\cdot(-37)+23\cdot(-37)-46\cdot(-37)$

$=-37\cdot(17+23-46)$

$=-37\cdot(40-46)$

$=-37\cdot(-6)$

$=37\cdot6$

$=222$

6) $-48+48\cdot(-78)+48\cdot(-21)$

$=48\cdot(-1)+48\cdot(-78)+48\cdot(-21)$

$=48\cdot[(-1)+(-78)+(-21)]$

$=48\cdot(-1-78-21)$

$=48\cdot(-79-21)$

$=48\cdot(-100)$

$=-4800$

7) $29\cdot(-13)+27\cdot(-29)+(-14)\cdot(-29)$

$=-29\cdot13+27\cdot(-29)+(-14)\cdot(-29)$

$=-29\cdot[13+27+(-14)]$

$=-29\cdot(40-14)$

$=-29\cdot26$

$=-754$

$\text{#}Toru$

a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)

ta có :

 \(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)

 \(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)

\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)

Vậy \(A< 3\)

a. Ta có :

\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)

\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)

\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)

Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)

Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)

Vậy \(A< 3\)

b) \(A=\frac{1+5+5^2+5^3+...+5^{10}+5^{11}}{1+5+5^2+5^3+...+5^9+5^{10}}=5^{11}\)

bn rút gọn là dc  

\(B=\frac{1+7+7^2+7^3+...+7^{10}+7^{11}}{1+7+7^2+7^3+...+7^9+7^{10}}=7^{11}\)

\(A=5^{11},B=7^{11}\)

\(\Rightarrow7^{11}>5^{11}\Rightarrow B>A\)

hk tốt #

a)\(6450-4000:\left(610:5-114\right)\)

     \(=6450-4000:\left(122-114\right)\)

    \(=6450-4000:8\)

     \(=6450-500\)

     \(=5950\)

b)\(564:6+83x2\)

\(=94+166\)

\(=260\)

a,

\(\left(325-47\right)+\left(175-53\right)\\ =325-47+175-53\\ =325+175-47-53\\ =\left(325+175\right)-\left(47+53\right)\\ =500-100\\ =400\)

b,

\(\left(756-217\right)-\left(183-53\right)\\ =756-217-183+53\\ =756+53-217-183\\ =\left(756+53\right)-\left(217+183\right)\\ =809-400\\ =409\)

c,

\(\left(-5\right)\cdot6\cdot\left(-2\right)\cdot7\\ =\left[\left(-5\right)\cdot\left(-2\right)\right]\cdot\left(6\cdot7\right)\\ =10\cdot42\\ =420\)

d,

\(24\cdot36-\left(-24\right)\cdot64\\ =24\cdot36+24\cdot64\\ =24\cdot\left(36+64\right)\\ =24\cdot100\\ =2400\)

e,

\(24\cdot36-\left(-24\right)\cdot64\\ =24\cdot36+24\cdot64\\ =24\cdot\left(36+64\right)\\ =24\cdot100\\ =2400\)

Bài 1: Tính:

a) (325+175) - (47+53)

= 500-100=400.

b) (756-217) - (183-53)

= 539-150= 389.

c) (-5).(-2).6.7

= 10.42=420

d) 24.36- (-24).64

= 864- (-1536)

= 2400.

Tick mk nha bn!!!!!

1/Vì 179/197<1 ; 971/917>1

=>179/197<971/917

2/Vì 183/184<1 ; 184/183>1

=>183/184<184/183

3/Ta có : -3/31=-3*101/31*101=-303/3131

Vì -303>-789 =>-303/3131>-789/3131 =>-3/31>-789/3131

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