Đáp án+Giải thích các bước giải:
`1)`
`A=\frac{tan\alpha+3 cot\alpha}{tan \alpha+cot\alpha}`
`=\frac{tan\alpha+3. \frac{1}{tan\alpha}}{tan\alpha+\frac{1}{tan\alpha}}`
Nhân cả tử và mẫu cho `tan\alpha` ta được:
`->A=\frac{tan^{2}\alpha+3}{tan^{2}\alpha+1}`
`=\frac{\frac{1}{cos^{2}\alpha}+(-1)+3}{\frac{1}{cos^{2}}+(-1)+1}`
Mà `cos \alpha =3/4`
`->A=\frac{\frac{1}{(3/4)^{2}}+2}{\frac{1}{(3/4)^{2}}`
`=\frac{16/9+2}{16/9}`
`=(16/9+18/9): 16/9`
`=34/9:16/9`
`=17/8`
Vậy `A=17/8`

 `2)`
`B=\frac{sin\alpha-cos\alpha}{sin^{3}\alpha+3cos^{3}\alpha+2sin\alpha}`
Ta có: `tan=\sqrt{2}->cos\alpha\ne0`
Chia cả tử và mẫu cho `cos^{3}\alpha` ta được:
`->B=\frac{\frac{sin\alpha}{cos^{3}\alpha}-\frac{1}{cos^{2}\alpha}}{tan^{3}\alpha+3+2. \frac{sin\alpha}{cos\alpha}.\frac{1}{cos^{2}\alpha}}`
`=\frac{tan\alpha. \frac{1}{cos^{2}\alpha}-(1+tan^{2}\alpha)}{tan^{3}\alpha+3+2tan\alpha.(1+tan^{2}\alpha)}`
`=\frac{tan\alpha. (1+tan^{2}\alpha)-1-tan^{2}\alpha}{tan^{3}\alpha+3+2. (tan\alpha+tan^{3}\alpha)}`
`=\frac{\sqrt{2}.[1+(\sqrt{2})^{2}]-1-(\sqrt{2})^{2}}{(\sqrt{2})^{3}+3+2.(\sqrt{2}+(\sqrt{2})^{3})}`
`=\frac{3\sqrt{2}-1-2}{2\sqrt{2}+3+2.(\sqrt{2}+2\sqrt{2})}`
`=\frac{3\sqrt{2}-3}{2\sqrt{2}+3+2.3\sqrt{2}}`
`=\frac{3\sqrt{2}-3}{8\sqrt{2}+3}`
`=\frac{(3\sqrt{2}-3).(8\sqrt{2}-3)}{(8\sqrt{2})^{2}-3^{2}}`
`=\frac{24.2-9\sqrt{2}-24\sqrt{2}+9}{119}`
`=\frac{57-33\sqrt{2}}{119}`
Vậy `B=\frac{58-33\sqrt{2}}{119}.`