Đưa 1 thừa số vào trong dấu căn

Đáp án+Giải thích các bước giải:

 `b)`

`1)`

`xy\sqrt{y/x}` `(xy>0;x\ne0)`

`=\sqrt{(xy)^{2}. y/x}`

`=\sqrt{\frac{x^{2}y^{2}.y}{x}}`

`=\sqrt{xy^{3}}`

`=y\sqrt{xy}`

`2)`

`(4-x).\sqrt{\frac{x}{x-4}}` `(đk:x>4)`

`=-(x-4).\sqrt{\frac{x}{x-4}}` (vì `x>4=>4-x<0=>|4-x|=-(4-x)=x-4`)

`=-(x-4).\sqrt{\frac{x}{x-4}}`

`=-\sqrt{(x-4)^{2}. \frac{x}{x-4}}`

`=-\sqrt{\frac{(x-4)^{2}.x}{x-4}}`

`=-\sqrt{(x-4).x}`

`=-\sqrt{x^{2}-4x}`

`3)`

`x\sqrt{2/x}(x>0)`

`=\sqrt{x^{2}. 2/x}`

`=\sqrt{\frac{2x^{2}}{x}}`

`=\sqrt{2x}`

`c)` 

`1)`

`x\sqrt{2/5}` `(1)`

Ta xét `2TH:`

`+)TH1:` Nếu `x>0=>|x|=x` từ đó `(1)` trở thành:

`x\sqrt{2/5}`

`=\sqrt{x^{2}. 2/5}`

`=\sqrt{\frac{2x^{2}}{5}}`

`=\frac{x\sqrt{10}}{5}`

`+)TH2:` Nếu `x<0=>|x|=-x` từ đó `(1)` có dạng:

`x\sqrt{2/5}`

`=-\sqrt{x^{2}. 2/5}`

`=-\sqrt{\frac{2x^{2}}{5}}`

`=-\frac{x\sqrt{10}}{5}`

`2)`

`(x+5).\sqrt{\frac{x}{25-x^{2}}` `(đk:x< -5)`

`=-\sqrt{(x+5)^{2}.\frac{x}{25-x^{2}}}` (vì `x< -5=>x+5<0=>|x+5|=-(x+5)`)

`=-\sqrt{\frac{x.(x+5)^{2}}{(5-x).(5+x)}}`

`=-\sqrt{\frac{x.(x+5)}{5-x}}`

`c)`

`(\sqrt{28}-2\sqrt{14}+\sqrt{7}).\sqrt{7}+7\sqrt{8}`

`=(\sqrt{2^{2}.7}-2.\sqrt{2.7}+\sqrt{7}).\sqrt{7}+7\sqrt{8}`

`=2\sqrt{7}.\sqrt{7}-2\sqrt{2}.\sqrt{7}.\sqrt{7}+\sqrt{7^{2}}+7.\sqrt{2^{2}.2}`

`=2\sqrt{7^{2}}-2\sqrt{2}.7+7+7.2\sqrt{2}`

`=2.7+7-14\sqrt{2}+14\sqrt{2}`

`=14+7`

`=21`

`d)`

`(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}):\sqrt{10}`

`=15\sqrt{50}:\sqrt{10}+5\sqrt{200}:\sqrt{10}-3\sqrt{450}:\sqrt{10}`

`=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}`

`=15\sqrt{5}+5\sqrt{2^{2}.5}-3\sqrt{3^{2}.5}`

`=15\sqrt{5}+5.2\sqrt{5}-3.3\sqrt{5}`

`=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}`

`=16\sqrt{5}`