\(27^{11}>81^8;625^5< 125^7;5^{36}>11^{24};5^{28}< 26^{14}\)

\(27^{11}>81^8;625^5< 125^7;5^{36}>11^{24};5^{28}< 26^{14}\)

Hok tốt 

27^11>81^8;625^5<125^7;5^36>11^24;5^28<26^14

Mọi giải giúp em với . Em cảm ơn ạ

a) 2711 > 818

b) 1619 > 825

c) 6255 > 1257

d) 536 < 1124

e) 32n > 23n

f) 354 > 281

1. x² - 6x + 9=(x-3)²

2. 25 +  10x + x²=(x+5)²

3. \(\dfrac{1}{4}a^2+2ab^2+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)

4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)

5.x³ + 8y³=(x+8y)(x²-8xy+64y²)

6.8y³ -125=(2y-5)(4y²+10y+25)

7.a⁶-b³=(a²-b)(a⁴+a²b+b²)

8 x² - 10x + 25=(x-2)²

1) \(x^2-6x+9=\left(x-3\right)^2\)

2) \(25+10x+x^2=\left(5+x\right)^2\)

3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)

4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)

5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)

7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)

8) \(x^2-10x+25=\left(x-5\right)^2\)

9) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

sorry nghe h tớ gửi quá 100 tin nhắn nên nó ko cho gửi

Bài 1

a)2711>818

b)6255>1257

c)536<1124

d)32n>23n

Bài 2

a)523<6.522

b)7.213>216

c)2115<275.498

bạn ơi bạn viết rõ hơn đi số mũ bạn bấm shift 6 

giúp vs

a, \(M=1+6+6^2+6^3+...+6^{99}\)

\(M=6\cdot(1+6)+6^2(1+6)+6^3(1+6)+...+6^{99}(1+6)\)

\(M=6\cdot7+6^2\cdot7+6^3\cdot7+...+6^{99}\cdot7\)

\(M=7\cdot\left[6+6^2+6^3+...+6^{99}\right]⋮7(đpcm)\)

b, \(M=1+6+6^2+6^3+...+6^{99}\)

\(M=6\cdot\left[1+6+6^2+6^3\right]+...+6^{96}\left[1+6+6^2+6^3\right]\)

\(M=6\cdot\left[7+36+216\right]+...+6^{96}\left[7+36+216\right]\)

\(M=6\cdot259+...+6^{96}\cdot259\)

\(M=259\cdot\left[6+...+6^{96}\right]⋮259\)

Vậy \(M⋮259(đpcm)\)

m : 29 =1734 ( dư 15 )

m : 29 = 1734 + 15

m : 29 =1749

m = 1749 . 29

m= 50721

1257 : m = 83 ( dư 12)

1257 : m = 83 + 12

1257 : m = 95

m = 1257 : 95

m = 1257/95