Ta có : a³ + b³ = (a + b)(a - ab + b)

Thay ab = 4 và a + b = 5

=> a³ + b³ = 5(5 - 4)

=> a³ + b³ = 5

Vậy a³ + b³ = 5

Chứng minh n^6+n^4-2n^2 chia hết cho 72?

b) \(263^2+74.263+37^2\)

\(=\left(263+37\right)^2\)

\(=300^2\)

\(=90000\)

c) \(136^2-92.136+46^2\)

\(=\left(136-46\right)^2\)

\(=90^2\)

\(=8100\)

sorry,i am in grade 5 to 6,so i don't know the answer

\(a-b=8\Rightarrow\left(a-b\right)^2=8^2=\)\(64\)

\(\Rightarrow a^2-2.a.b+b^2=64\)

\(\Rightarrow a^2+b^2-2.10=64\)

\(\Rightarrow a^2+b^2=84\)

\(\Rightarrow a^2+2ab+b^2=84+2.10\)\(=84+20=104\)

\(\Rightarrow\left(a+b\right)^2=104\)

Ta có: \(a-b=5\)

\(\Rightarrow\left(a-b\right)^2=25\)

\(\Rightarrow a^2+b^2-2ab=25\)

\(\Rightarrow a^2+b^2=25+2ab=25+2\cdot4=33\)

Vậy \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)=5\left(33+4\right)=5\cdot37=185\)

1. 9

2. 25

3. 91

4. 32

5. 38^o

6. 46

7. 711

1. 9

2. 25

3. 91

4. 32

5. 38o

6. 46

7. 711

1)We have: \(a-b=8\)

\(\Rightarrow\left(a-b\right)^2=64\)

\(\Rightarrow a^2-2ab+b^2=64\)

\(\Rightarrow a^2+2ab+b^2-4ab=64\)

\(\Rightarrow\left(a+b\right)^2=64+4ab=64+4\cdot10=64+40=104\)

Hence: \(\left(a+b\right)^2=104\)

2)We have: \(a+b=8\)

\(\Rightarrow\left(a+b\right)^2=64\)

\(\Rightarrow a^2+2ab+b^2=64\)

\(\Rightarrow a^2-2ab+b^2+4ab=64\)

\(\Rightarrow\left(a-b\right)^2=64-4ab=64-4\cdot10=64-40=24\)

Hence \(\left(a-b\right)^2=24\)

Ta có :

\(6^{5x+2}=36^{3x-4}\)

\(\Rightarrow6^{5x+2}=\left(6^2\right)^{3x-4}\)

\(\Rightarrow6^{5x+2}=6^{6x-8}\)

=> 5x + 2 = 6x - 8

=> x = 10

Vậy x = 10

5x +2 = 6x -8

x = 10