`(a):\  x=64(TMDK)=>A=(2+\sqrt{64})/(\sqrt{64})=(2+8)/(8)=(10)/(8)=(5)/(4)`

`(b):\  B=(\sqrt{x}-1)/(\sqrt{x})+(2\sqrt{x}+1)/(x+\sqrt{x})\   (ĐK:x>0)`

`=(\sqrt{x}-1)/(\sqrt{x})+(2\sqrt{x}+1)/(\sqrt{x}(\sqrt{x}+1))`

`=((\sqrt{x}-1)(\sqrt{x}+1)+2\sqrt{x}+1)/(\sqrt{x}(\sqrt{x}+1))`

`=(x-1+2\sqrt{x}+1)/(\sqrt{x}(\sqrt{x}+1))`

`=(\sqrt{x}(\sqrt{x}+2))/(\sqrt{x}(\sqrt{x}+1))`

`=(\sqrt{x}+2)/(\sqrt{x}+1)`

`(c):\  A/B=(2\sqrt{x}-2)/(3)\  (ĐK:x>0)`

`<=>(\sqrt{x}+2)/(\sqrt{x}):(\sqrt{x}+2)/(\sqrt{x}+1)=(2\sqrt{x}-2)/(3)`

`<=>(\sqrt{x}+1)/(\sqrt{x})=(2\sqrt{x}-2)/(3)`

`<=>\sqrt{x}(2\sqrt{x}-2)=3(\sqrt{x}+1)`

`<=>2x-2\sqrt{x}=3\sqrt{x}+3`

`<=>2x-5\sqrt{x}-3=0`

`<=>(\sqrt{x}-3)(2\sqrt{x}+1)=0`

`=>\sqrt{x}-3=0` ( Do : `2\sqrt{x}+1>0` với mọi `x\in ĐK ` )

`<=>x=9\  (TMDK)`

`(a):2\sqrt{x-2}+\sqrt{16x-32}-(1)/(5)\sqrt{25x-50}=10\   (ĐK:x\ge 2)`

`<=>2\sqrt{x-2}+\sqrt{16}.\sqrt{x-2}-(1)/(5).\sqrt{25}.\sqrt{x-2}=10`

`<=>(2+4-1).\sqrt{x-2}=10`

`<=>5\sqrt{x-2}=10`

`<=>\sqrt{x-2}=2`

`<=>x-2=4`

`<=>x=6\  (TMDK)`

`(b):\  \sqrt{2x+2\sqrt{2x-1}}+\sqrt{8x-4}=10\  (ĐK:x\ge 1/2)`

`<=>\sqrt{2x-1+2.\sqrt{2x-1}.1+1}+\sqrt{4}.\sqrt{2x-1}=10`

`<=>\sqrt{(\sqrt{2x-1}+1)^{2}}+2\sqrt{2x-1}=10`

`<=>|\sqrt{2x-1}+1|+2\sqrt{2x-1}=10`

`<=>\sqrt{2x-1}+1+2\sqrt{2x-1}=10`  ( Do : `\sqrt{2x-1}+1>0` với mọi `x\in ĐK ` )

`<=>3\sqrt{2x-1}=9`

`<=>\sqrt{2x-1}=3`

`<=>2x-1=9`

`<=>x=5\  (TMDK)`

Đổi : `6` tấn `4` tạ `=6400` kg ; `8` tạ `=800` kg

Thửa ruộng thứ nhất thu hoạch được :

    `(6400-800):2=2800` (kg)

Thửa ruộng thứ hai thu hoạch được :

    `2800+800=3600` (kg)

`A=(x^{n}+1)(x^{n}-2)-x^{n-3}(x^{n+3}-x^{3})+2007`

`=x^{n}.x^{n}+x^{n}-2x^{n}-2-x^{n-3}.x^{n+3}+x^{n-3}.x^{3}+2007`

`=x^{n+n}-x^{n}-2-x^{n-3+n+3}+x^{n-3+3}+2007`

`=x^{2n}-x^{n}-2-x^{2n}+x^{n}+2007`

`=2005` (Không phụ thuộc vào giá trị của `x`)

`3^{2}-5.[3.(4-1)-8]`

`=3.3-5.(3.3-8)`

`=9-5.(9-8)`

`=9-5.1`

`=9-5=4`

`(1)/(2):(3)/(4)+(1)/(6):(3)/(4)`

`=(1)/(2)xx(4)/(3)+(1)/(6)xx(4)/(3)`

`=(4)/(3)xx((1)/(2)+(1)/(6))`

`=(4)/(3)xx(3+1)/(6)`

`=(4)/(3)xx(4)/(6)`

`=(4)/(3)xx(2)/(3)`

`=(8)/(9)`

`(a):\  12\vdots 2;30\vdots 2`

`=>12+30\vdots 2`

Mà để : `A=12+30+x\vdots 2`

Thì : `x\vdots 2`

`(b):\  12\vdots 3;30\vdots 3`

`=>12+30\vdots 3`

Mà để : `A=12+30+x\vdots 3`

Thì : `x\vdots 3`

`(c):\ 12\vdots 6;30\vdots 6`

`=>12+30\vdots 6`

Mà để : `A=12+30+x\vdots 6`

Thì : `x\vdots 6`

`(a):\  |x|=13`

`=>x=13` hoặc `x=-13`

`(b):\  |x|=-17`

Vì : `|x|\ge 0` với mọi `x`

Mà : `-17<0`

Do vậy nên: \(x\in\varnothing\)

`x:3+x:4+x:12=6`

`x xx(1)/(3)+x xx(1)/(4)+x xx(1)/(12)=6`

`x xx((1)/(3)+(1)/(4)+(1)/(12))=6`

`x xx(4+3+1)/(12)=6`

`x xx(8)/(12)=6`

`x xx (2)/(3)=6`

`x =6:(2)/(3)=9`

`(a):\  A=((1)/(x+2\sqrt{x})-(1)/(\sqrt{x}+2)):(1-\sqrt{x})/(x+4\sqrt{x}+4)`

`(ĐK:x>0;x\ne 1)`

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

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

`=(\sqrt{x}+2)/(\sqrt{x})`

`(b):\ A=5/3<=>(\sqrt{x}+2)/(\sqrt{x})=5/3`

`<=>5\sqrt{x}=3(\sqrt{x}+2)`

`<=>5\sqrt{x}-3\sqrt{x}=6`

`<=>2\sqrt{x}=6`

`<=>\sqrt{x}=3`

`<=>x=9\  (TMDK)`