\(\dfrac{9^{29}.27^{13}}{3^{65}.81^5}=\dfrac{\left(3^2\right)^{29}.\left(3^3\right)^{13}}{3^{65}.\left(3^4\right)^5}=\dfrac{3^{58}.3^{39}}{3^{65}.3^{20}}=\dfrac{3^{97}}{3^{85}}=3^{12}\)

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

2) x(x+1)(x+2)(x+3)= 24

⇔ x.(x+3)  .   (x+2).(x+1)  = 24

⇔(\(x^2\) + 3x) . (\(x^2\) + 3x + 2) = 24

Đặt \(x^2\)+ 3x = b

⇒ b . (b+2)= 24

Hay: \(b^2\) +2b = 24

⇔\(b^2\) + 2b + 1 = 25

⇔\(\left(b+1\right)^2\)= 25

+ Xét b+1 = 5 ⇒ b=4 ⇒  \(x^2\)+ 3x = 4 ⇒ \(x^2\)+4x-x-4=0 ⇒x(x+4)-(x+4)=0

⇒(x-1)(x+4)=0⇒x=1 và x=-4

+ Xét b+1 = -5 ⇒ b=-6 ⇒ \(x^2\)+3x=-6 ⇒\(x^2\) + 3x + 6=0

⇒\(x^2\) + 2.x.\(\dfrac{3}{2}\) + (\(\dfrac{3}{2}\))² = - \(\dfrac{15}{4}\)  Hay ( \(x^2\) +\(\dfrac{3}{2}\) )²= -\(\dfrac{15}{4}\) (vô lí)

⇒x= 1 và x= 4

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

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

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

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

$=-29\cdot26$

$=-754$

$\text{#}Toru$

31.(-18) + 31.(-81) - 31

= 31.(-18 - 81 - 1)

= 31.(-100) = -3100

-48 + 48.(-78) + 48.(-21) = 48.(-1 - 78 - 21) = 48. (-100) = -4800

-75.(8 - 65) - 65.(75 - 18) = -75.18 + 75.65 - 65.75 + 65.18 = 0

29.(19 - 13) - 19.(29 - 13) = 29.19 - 29.13 - 19.29 + 19.13 = 0

13.(23 + 22) - 3(17 + 28) = 13.25 - 3.45 = (10 + 3).25 - 3.45

= 10 . 25 + 3.25 - 3.45 = 10.25 - 3.(45 - 25)

= 250 - 3.20 = 250 - 60 = 190

(-25).21 . (-2)².(|3|). (-1)^{2n + 2} (n \(\in\)N*)

= (-25. 4) . (21.3).1

= -100. 63 = -6300

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

a: \(A=21\cdot100-11\cdot100+90\cdot100+100\cdot125\cdot16\)

\(=100\left(21-11+90\right)+100\cdot2000\)

\(=100\left(10+90+2000\right)=2100\cdot100=210000\)

b: \(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\)

a: \(=\dfrac{2^9\cdot5^9\cdot3^{40}}{2^{12}\cdot5^{10}\cdot3^{20}}=\dfrac{3^{20}}{5\cdot2^3}\)

b: \(=\dfrac{-3^8\cdot2^{10}\cdot5^6}{2^9\cdot\left(-1\right)\cdot3^6\cdot5^7}=\dfrac{-2}{5}\cdot3^2=-\dfrac{18}{5}\)

c: \(=\dfrac{3^{186}\cdot5^{100}}{5^{100}\cdot3^{187}}=\dfrac{1}{3}\)

lung tung dã man bài toán này bịa

bài này đến cả giáo viên con không nghĩ ra nói gì tụi này

hại não qá bà con ơi 

Em nến gõ bằng công thức toán học em nhé. Như vậy mọi người nới hiểu đúng đề bài để trợ giúp cho em một cách tốt nhất em ạ!