(-35). (-8) > (-35)

(-35). 8 < (-35)

a: B=31/39=62/78=1-16/78; A=35/37=280/296=1-16/296

16/78>16/296

=>B<A

b: C=23/27=46/54=1-8/54

D=1-8/29

8/29>8/54

=>D<C

a: \(\sqrt{12}<\sqrt{12,25}=3,5\)

\(\sqrt{20}<\sqrt{20,25}=4,5\)

\(\sqrt{30}<\sqrt{30,25}=5,5\)

\(\sqrt{42}<\sqrt{42,25}=6,5\)

Do đó: \(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\) <3,5+4,5+5,5+6,5

=>A<20

b: \(B=\sqrt{196}-\frac{1}{\sqrt3}-1=14-\frac{1}{\sqrt3}-1=13-\frac{1}{\sqrt3}\)

\(C=\sqrt{169}+\frac{1}{-\sqrt2}=13-\frac{1}{\sqrt2}\)

3>2

=>\(\frac{1}{\sqrt3}<\frac{1}{\sqrt2}\)

=>\(-\frac{1}{\sqrt3}>-\frac{1}{\sqrt2}\)

=>\(-\frac{1}{\sqrt3}+13>-\frac{1}{\sqrt2}+13\)

=>B>C

1b) có thiếu ko cau?

`1a)5^3` và `3^5`

`5^3=125`

`3^5=243`

Vì `243>125` nên `3^5>5^3`

__

`c)3^24` và `27^7`

`27^7=(3^3)^7=3^21`

Vì `3^24>3^31` nên `3^24>27^7`

`2a)x^3=216`

`=>x^3=6^3`

`=>x=6`

__

`b)3^x+15=18`

`=>3^x=18-15`

`=>3^x=3`

`=>x=1`

Ta có: \(35^2\equiv375\)( mod 425) 

\(35^3=35.35^2\equiv35.375\equiv375\)( mod 425)

\(35^4=35.35^3\equiv35.375\equiv375\)( mod 425) 

\(35^8=35^4.35^4\equiv375.375\equiv375\)( mod 425) 

\(35^{16}\equiv35^8.35^8\equiv375.375\equiv375\)( mod 425) 

\(35^{32}\equiv35^{16}.35^{16}\equiv375.375\equiv375\)( mod 425) 

=> \(35^2-35^3+35^4-35^8+35^{16}+35^{32}\equiv375-375+375-375+375+375\equiv325\)( mod 425) 

Vậy số dư cần tìm là 325

Cách hay chị Chi ơi

A

a. 35/48