\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

Bài 1: 

a: \(-2^{30}=-8^{10}\)

\(-3^{30}=-27^{10}\)

mà 8<27

nên \(-2^{30}>-3^{30}\)

b: \(35^5=35^5\)

\(6^{10}=36^5\)

mà 35<36

nên \(35^5< 6^{10}\)

B

C

A

A = (-25).8.2.5.(-4).13 = [(-25).(-4)].8.2.5.13 = 100.8.10.13 = 10400

B = 13.(-7) + (-13).57 + 13.(-36) = 13.(-7) - 13.57 + 13.(-36)

                                                      = 13.[(-7) + (-57) + (-36)]

                                                      = 13. (-100)

                                                      = -1300

Ta có: \(A=\left(-25\right)\cdot8\cdot2\cdot5\cdot\left(-4\right)\cdot13\)

\(=\left[\left(-25\right)\cdot\left(-4\right)\right]\cdot\left(2\cdot5\right)\cdot\left(8\cdot13\right)\)

\(=104000\)

Ta có: \(B=13\cdot\left(-7\right)+\left(-13\right)\cdot57+13\cdot\left(-36\right)\)

\(=13\cdot\left(-7\right)-13\cdot57-13\cdot36\)

\(=13\left(-7-57-36\right)\)

\(=13\cdot\left(-100\right)=-1300\)

(150 x 35 x 10) :25 = 52500 : 25 = 2100

(450 x480 x 44) : (15 x 21 x11) = 9504000 : 3465 = \(\frac{9504000}{3465}=\frac{19200}{7}\)

a: \(25496+28212:4\)

\(=25496+7053\)

=32549

b: \(\dfrac{\left(108610-87352\right)}{3}\)
\(=\dfrac{21258}{3}\)

\(=7086\)

25 – (20 - 10) = 25 - 10 = 15

80 – (30 + 25) = 80 - 55

A=x^2+6x+9+1

=(x+3)^2+1

Thay x=-103 vào A, ta được:

A=(-103+3)^2+1=10000+1=10001

\(a,A=x^2+6x+10\)

\(=\left(x^2+2\cdot x\cdot3+3^2\right)+1\)

\(=\left(x+3\right)^2+1\)

Thay \(x=-103\) vào \(A\), ta được:

\(A=\left(-103+3\right)^2+1\)

\(=\left(-100\right)^2+1\)

\(=10000+1\)

\(=10001\)

#Urushi

25 – (20 - 10) = 25 - 10 = 15