giúpppppp

Đáp án:

 `a) a=\sqrt{5}`

Giải thích các bước giải:

 Bài `1:`

`a)`

`a=(\sqrt{2023}+\sqrt{2022}).(\sqrt{2023}-\sqrt{2022})+3\sqrt{5}-(2\sqrt{5}+1)`

`=[(\sqrt{2023})^{2}-(\sqrt{2022})^{2}]+3\sqrt{5}-2\sqrt{5}-1`

`=(2023-2022)+(3\sqrt{5}-2\sqrt{5})-1`

`=1-1+\sqrt{5}`

`=\sqrt{5}`

Vậy `a=\sqrt{5}`

`b)`

Ta gọi `\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}+4\sqrt{xy}` là `VT`

`(\sqrt{x}+\sqrt{y})^{2}` là `VP` ta được:

`VT=\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}+4\sqrt{xy}`

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

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

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

`=(\sqrt{x}-\sqrt{y})^{2}+4\sqrt{xy}`

`=(\sqrt{x})^{2}-2.\sqrt{x}.\sqrt{y}+(\sqrt{y})^{2}+4\sqrt{xy}`

`=x-2\sqrt{xy}+y+4\sqrt{xy}`

`=x+2\sqrt{xy}+y`

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

`=(\sqrt{x}+\sqrt{y})^{2}=VP` `(đpcm)`

Mà `VT=VP`

`=>` `\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}+4\sqrt{xy}=(\sqrt{x}+\sqrt{y})^{2}`