`a) 1^3+ 2^3= 3^2`

`b) 1^3+ 2^3+ 3^3= 6^2`

`c) 1^3+ 2^3+ 3^3 + 4^3= 10^2`

Thay `x=4` vào P, ta được:

`P= 2.4^3+ 3.4^2+4.4+5`

`P= 128+ 48+16+5`

`P= 197`

Các số nguyên tố trong dãy trên là: 23,31,17

`a) 5/9 - (1/3)^2= 5/9- 1/9= 4/9`

`b) 1/5. -3/2 + -17/2. 1/5= 1/5. (-3/2+-17/2)= 1/5.(-10)= -2`

`c) 1+ (-2/5 + 11/13) - (3/5- 2/13)= 1+ -2/5 + 11/13 + -3/5 + 2/13= 1+ (-2/5+ -3/5) +  (11/13+2/13) =1+ (-1)+1=1`

`(3xy. 4x)(2xy^2 - 6x+4)`

`= 6x^2. y^3 - 18x^2 y+ 12xy-4x (2y^2 - 6x+4)`

`= 6x^2y^3 - 8x^2y^2 - 18x^2y + 12xy + 24x^2 - 16x`

`A=1.2+2.3+3.4+4.5+...+99.100`

`3A= 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3+ … + 99.100.3`
`3A= 1.2.3 + 2.3.(4 - 1) + 3.4.( 5 - 2) + … + 99.100. (101 - 98)`
`3A= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + … + 99.100.101 - 98.99.100`
`3A = 99.100.101`
`A = (99.100.101)/3`
`A = 333 300`

`(x^2 y + xy^2)(x-y)=xy(x-y)(x+y)`

Ta có VT `=(x^2 y + xy^2)(x-y)`

=> `VT=xy(x+y)(x-y)=xy(x-y)(x+y)=VP` (đpcm) 

Áp dụng bất đẳng thức Cauchy-Schwarz cho `x;y;z` , ta có:

`(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a})^2 \leq 3. ((a + b) + (b + c) + (c + a)) = 3.2 = 6`

-> `\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a} \leq \sqrt{6}` (đpcm)

`(4^21. (-3)^40)/(6^41) = (2^42. 3^40)/(2^41. 3^41)=2/3`

`575- (6x+70)= 445`

`6x+70=130`

`6x=60`

`x= 60/6`

`x=10`

Vậy `x=10`