Câu `1:`

Ta có:

`(1+x)^10=C_(10)^0``.1^0.x^(10)+C_(10)^1.``1^1.x^9+C(10)^2``.1^2.x^8+...+C_(10)^9``.1^9.x+C_(10)^(10)``.1^0`

Cho `x=1`, ta có: `2^10=C_(10)^0+C_(10)^1+...+C_(10)^(10)`

`<=>C_(10)^0+C_(10)^1+...+C_(10)^(10)=1024`

`=>S=1024`

Câu `2:`

Ta có: `(1+x)^6=C_6^0.``x^0``.1^6+``C_6^1``.x^1``.1^5+...+C_6^5.``x^5.``1+C_6^6.``x^6.``1^0`

Cho `x=1`, ta có: `(1+1)^6=C_6^0+C_6^1+...+C_6^6`

`<=>C_6^1+C_6^2+...+C_6^5=2^6-C_6^0-C_6^6`

`<=>C_6^1+C_6^2+...+C_6^5=62`

`=>S=62`