Đáp án:
`a)` `{3\sqrt{2}}/2`
`b)` `A= -48; B=-\sqrt{3}/3`
Giải thích các bước giải:
`a)` `{\sqrt{54}+3\sqrt{14}}/{\sqrt{12}+\sqrt{28}}`
`={3\sqrt{6}+3\sqrt{14}}/{\sqrt{2}.\sqrt{6}+\sqrt{2}.\sqrt{14}}`
`={3.(\sqrt{6}+\sqrt{14})}/{\sqrt{2}.(\sqrt{6}+\sqrt{14})}=3/\sqrt{2}={3\sqrt{2}}/2`
$\\$
`b)`
`A=\sqrt{{(x-6)^4}/{(5-x)^2}}+{x^2+36}/{x-5}` `(x\ne 5)`
`A=\sqrt{{[(x-6)^2]^2}/{(5-x)^2}}+{x^2+36}/{x-5}`
`=|(x-6)^2|/|5-x|+{x^2+36}/{x-5}`
Nếu `x<5=>|5-x|=5-x`
`=>A=(x-6)^2/{5-x}-{x^2+36}/{5-x}`
`={x^2-12x+36-(x^2+36)}/{5-x}``={-12x}/{5-x}`
Với `x=4<5`
`=>A={-12.4}/{5-4}=-48`
Vậy `A=-48` khi `x=4`
$\\$
`B=3/{x^2-y^2}.\sqrt{{3x^2+6xy+3y^2}/9}` `(x\ne ±y)`
`=1/{x^2-y^2}. \sqrt{{3^2 . 3(x^2+2xy+y^2)}/9}`
`=1/{x^2-y^2}. \sqrt{3(x+y)^2}`
`={\sqrt{3}|x+y|}/{(x-y)(x+y)}`
Nếu `x+y>0=>|x+y|=x+y`
`=>B={\sqrt{3}(x+y)}/{(x-y)(x+y)}=\sqrt{3}/{x-y}`
$\\$
Với `x=-1;y=2` (thỏa mãn đk)
`=>x+y=-1+2=1>0`
`=>B=\sqrt{3}/{-1-2}=-\sqrt{3}/3`
Vậy `B=-\sqrt{3}/3` khi `x=-1;y=2`
