a)

`a-10>b-10`

`<=>a-10+10>b-10+10`

`<=>a>b`

c)

`-a-9≥-b-9`

`<=>-a-9+9≥-b-9+9`

`<=>-a≥-b`

`<=>-a*(-1)/1≤-b*(-1)/1`

`<=>a≤b`

e)

`-4a+9< -4b+9`

`<=>-4a+9-9< -4b+9-9`

`<=>-4a< -4b`

`<=>-4a*(-1)/4> -4b*(-1)/4`

`<=>a>b`

b)

`25+a>25+b`

`<=>25+a-25>25+b-25`

`<=>a>b`

f)

cái giữa là dấu gì vậy ạ

\(a,a-10>b-10\)

\(\Rightarrow a-10+10>b-10+10\)

\(\Leftrightarrow a>b\)

\(b,-a-9\ge-b-9\)

\(\Rightarrow-a-9+9\ge-b-9+9\)

\(\Leftrightarrow-a\ge-b\)

\(c,-4a+9< -4b+9\)

\(\Rightarrow-4a+9-9< -4b+9-9\)

\(\Leftrightarrow a< b\)

\(d,25+a>25+b\)

\(\Rightarrow25+a-25>25+b-25\)

\(\Leftrightarrow a>b\)

Câu cuối thiếu dấu bạn ơi!

\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)

\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)

Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)

Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)

Vì \(25^{10}-1>25^{10}-3\)

\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)

\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

A =25 x 33 - 10

A = 825 - 10

A = 815

B = 21 x 36 + 10

B = 756 + 10

B = 766 

Vì 815>766 => 25 x 33 - 10 > 21 x 36 + 10

Hay A>B

Ta có : 10A = 10^25 + 10/10^25 + 1 = 10^25 + 1 +9/10^25 + 1 = 10^25 + 1/10^25 + 1  +  9/10^25 + 1 

                   = 1  +  9/10^25 + 1 > 1        ( 1 )

            10B = 10^26 - 10/10^26 - 1 = 10^26 - 1 - 9/10^26 - 1 = 10^26 - 1/10^26 - 1  -  9/10^26 - 1

                   = 1  -  9/10^26 - 1  < 1         ( 2 )

Từ ( 1 ) và ( 2 ) => 1  +  9/10^25 + 1 >   1   > 1  -  9/10^26 - 1

                         => 10A > 10B

                         => A > B

Vậy PS A lớn hơn PS B.

a) \(10^{30}=2^{30}.5^{30}=2^{30}.\left(5^3\right)^{10}=2^{30}.125^{10}\)

\(2^{100}=2^{30}.2^{70}=2^{30}.\left(2^7\right)^{10}=2^{30}.128^{10}\)

mà \(125^{10}< 128^{10}\)

\(\Rightarrow10^{30}< 2^{100}\)

b) \(5^{40}=\left(5^4\right)^{10}=625^{10}>620^{10}\)

\(5^{40}>620^{10}\)

c) \(8^{25}=\left(2^3\right)^{75}=2^{75}\)

\(16^{19}=\left(2^4\right)^{19}=2^{76}>2^{75}\)

\(\Rightarrow16^{19}>8^{25}\)

a,10³⁰ và 2¹⁰⁰

10³⁰=(10³)¹⁰=1000¹⁰

2¹⁰⁰=(2¹⁰)¹⁰=1024¹⁰

Vì 1000¹⁰<1024¹⁰ nên 10³⁰<2¹⁰⁰.

b,5⁴⁰ và 620¹⁰

5⁴⁰=(5⁴)¹⁰=625¹⁰>620¹⁰

=>5⁴⁰>620¹⁰.

c,8²⁵ và 16¹⁹

Nhân 8²⁵ và 16¹⁹ với 4 , ta được 

32²⁵ và 64¹⁹

32²⁵=(32⁵)⁵=33554432⁵

64¹⁹<64²⁰=(64⁴)⁵=16777216⁵

Vì 33554432⁵>16777216⁵ nên 8²⁵>16¹⁹

\(A=10^{25}+\frac{1}{10^{26}}+1=1\cdot10^{25}\)

\(B=10^{26}+\frac{1}{10^{27}}+1=1\cdot10^{26}\)

\(1\cdot10^{25}< 1\cdot10^{26}\Rightarrow A< B\)

 ta có :

\(25^{1008}=\left(5^2\right)^{1008}=5^{2.1008}=5^{2016}\)

mà \(5^{2017}>5^{2016}\)

\(\Rightarrow\)\(5^{2017}>\left(5^2\right)^{1008}\)

\(\Rightarrow\)\(5^{2017}>25^{1008}\)

có \(5^{2017}=\left(5^2\right)^{1008}\times5\)\(=25^{1008}\times5\)

mà \(=25^{1008}\times5\)> \(25^{1008}\)

nên \(5^{2017}>25^{1008}\)

Ta có:

\(5^{2017}>5^{2016}=\text{[}5^2\text{]}^{1008}=25^{1008}\)

Suy ra: 5²⁰¹⁷ > 25¹⁰⁰⁸

Ta có:

\(1-A=1-\frac{10^{101}-1}{10^{102}-1}=\frac{10^{102}-1-\text{[}10^{101}-1\text{]}}{10^{102}-1}=\frac{10^{102}-1-10^{101}+1}{10^{102}-1}\)\(=\frac{10^{102}-10^{101}}{10^{102}-1}=\frac{10^{101}\left[10-1\right]}{10^{101}\text{[}10-\frac{1}{10^{101}}\text{]}}=\frac{10-1}{10-\frac{1}{10^{101}}}=\frac{9}{10-\frac{1}{10^{101}}}\)

\(1-B=1-\frac{10^{100}+1}{10^{101}+1}=\frac{10^{101}+1-\left[10^{100}+1\right]}{10^{101}+1}=\frac{10^{101}+1-10^{100}-1}{10^{100}+1}\)

\(=\frac{10^{101}-10^{100}}{10^{101}+1}=\frac{10^{100}\left[10-1\right]}{10^{100}\text{[}10+\frac{1}{10^{100}}\text{]}}=\frac{10-1}{10+\frac{1}{10^{100}}}=\frac{9}{10+\frac{1}{10^{100}}}\)

Vì \(\frac{9}{10-\frac{1}{10^{101}}}>\frac{9}{10+\frac{1}{10^{100}}}\Rightarrow A< B\)

Sử dụng các phương pháp so sánh đã học ( quy đồng mẫu số trung gian ...). Chú ý rút gọn phân số ( nếu cần)

a ) 11 25 > 2 5 b ) − 26 39 > − 24 32 c ) 5 − 8 < − 10 − 11