`36xx17+12xx34+6xx30`

`=12xx3xx17+12xx34+6xx2xx15`

`=12xx51+12xx34+12xx15`

`=12xx(51+34+15)`

`=12xx100`

`=1200`

`(8)/(3)-(1)/(2):(5)/(6)`

`=(8)/(3)-(1)/(2)xx(6)/(5)`

`=(8)/(3)-(1xx3xx2)/(2xx5)`

`=(8)/(3)-(3)/(5)`

`=(8xx5)/(15)-(3xx3)/(15)`

`=(40-9)/(15)=31/15`

`x-50xx25=8`

`x-1250=8`

`x=8+1250`

`x=1258`

`(4):`

`2x(x^{2}+x-1)`

`=2x.x^{2}+2x.x+2x.(-1)`

`=2x^{3}+2x^{2}-2x`

`(5):`

`((z)/(3)+2t-3t^{2}).(-zt)`

`=(z)/(3).(-zt)+2t.(-zt)+(-3t^{2}).(-zt)`

`=(-z^{2}t)/(3)-2t^{2}z+3t^{3}z`

`3^{n}.((1)/(9))^{4}=27`

`=>3^{n}.(1)/(9^{4})=27`

`=>3^{n}=27:(1)/(9^{4})`

`=>3^{n}=27.9^{4}`

`=>3^{n}=3^{3}.(3^{2})^{4}`

`=>3^{n}=3^{3}.3^{2.4}=3^{3}.3^{8}`

`=>3^{n}=3^{3+8}`

`=>n=3+8=11`

`(505)/(11)=(495+10)/(11)=(495)/(11)+(10)/(11)`

`=(45xx11)/(11)+(10)/(11)`

`=45(10)/(11)`

`(3)/(2)\sqrt{6}+2\sqrt{2/3}-4\sqrt{3/2}`

`=(3)/(2)\sqrt{6}+2.\sqrt{2}/\sqrt{3}-4.\sqrt{3}/\sqrt{2}`

`=(3)/(2)\sqrt{6}+2.(\sqrt{2}.\sqrt{3})/(\sqrt{3}^{2})-4.(\sqrt{3}.\sqrt{2})/(\sqrt{2}^{2})`

`=(3)/(2)\sqrt{6}+2.(\sqrt{6})/(3)-4.(\sqrt{6})/(2)`

`=(3)/(2)\sqrt{6}+(2)/(3)\sqrt{6}-2\sqrt{6}`

`=((3)/(2)+(2)/(3)-2).\sqrt{6}`

`=(1)/(6)\sqrt{6}`

`=(\sqrt{6})/(6)`

`(1):`

`A=(x+y)^{2}-x^{3}-y^{3}`

`=(x+y)^{2}-(x^{3}+y^{3})`

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

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

``

`(2):`

`A=x^{3}+1-x^{2}-x`

`=(x^{3}-x^{2})-(x-1)`

`=x^{2}(x-1)-(x-1)`

`=(x-1)(x^{2}-1)`

`=(x-1)^{2}(x+1)`

`(4):`

`A=x^{3}-4x^{2}+8x-8`

`=(x^{3}-2^{3})-(4x^{2}-8x)`

`=(x-2)(x^{2}+2x+4)-4x(x-2)`

`=(x-2)(x^{2}+2x+4-4x)`

`=(x-2)(x^{2}-2x+4)`

`(1)/(2)+(3)/(4)xx(-2)/(18)+(-3)/(7):(9)/(-14)`

`=(1)/(2)+(3)/(4)xx(-1)/(9)+(-3)/(7)xx(-14)/(9)`

`=(1)/(2)+(3xx(-1))/(4xx3xx3)+(3xx7xx2)/(7xx3xx3)`

`=(1)/(2)-(1)/(4xx3)+(2)/(3)`

`=(1)/(2)-(1)/(12)+(2)/(3)`

`=(6)/(12)-(1)/(12)+(8)/(12)`

`=(6-1+8)/(12)=13/12`

`(a):\  (6x^{4}-15x^{3}+x^{2}):3x^{2}`

`=(6x^{4})/(3x^{2})-(15x^{3})/(3x^{2})+(x^{2})/(3x^{2})`

`=2x^{2}-5x+(1)/(3)`   `(ĐK:x\ne 0)`

`(b):`  `A(x)=2x^{3}-3x^{2}+x+a`

`=(2x^{3}+4x^{2})-(7x^{2}+14x)+(15x+30)+a-30`

`=2x^{2}(x+2)-7x(x+2)+15(x+2)+a-30`

`=(x+2)(2x^{2}-7x+15)+a-30`

Do : `(x+2)(2x^{2}-7x+15)\vdots (x+2)`

Nên để : `A(x)` chia hết cho `B(x)`

Thì : `a-30=0`

`<=>a=30`