Giúp tớ với ạ cảm nhìuuu ạ

`#Sky`

`"Câu 6:D"`

`sin∝+cos∝=a`

`->(sin∝+cos∝)^2=a^2`

`->sin^2∝+cos^2∝+2.sin∝cos∝=a^2`

`->1+2.sin∝cos∝=a^2`

`->sin∝cos∝=(a^2-1)/2`

`->` Chọn `D`

_____________________

`"Câu 7:B"`

Ta có: `cos∝=-2/3` `=>90^o<∝<180^o`

`sin^2∝+cos^2∝=1`

`->sin^2∝=1-cos^2∝=5/9`

`->sin∝=sqrt5/3`

`-> cot∝=(cos∝)/(sin∝)=-(2sqrt5)/5`

`->tan∝=-sqrt5/2`

Thay `cot∝=-(2sqrt5)/5;tan∝=-sqrt5/2` vào `B` ta được:

`B=(19)/(13)`

`->` Chọn `B`

_____________________

`"Câu 8:D"`

`cot∝=5`

`->(cos∝)/(sin∝)=5`

`->` `cos∝=5sin∝`

Lại có: `sin^2∝+cos^2∝=1`

Giải hpt trên ta được:
`sin∝=(sqrt26)/(26)`

`cos∝=(5sqrt26)/(26)`

Thay `sin∝=(sqrt26)/(26); cos∝=(5sqrt26)/(26)` vào `E` ta được:

`E=(101)/(26)`

`->` Chọn `D`

_____________________

`"Câu 9:D"`

Ta có: `cot∝=1/3`

`->(cos∝)/(sin∝)=1/3`

Ta có: `A=(3sin∝+4cos∝)/(2sin∝-5cos∝)`

`=(3+4.(cos∝)/(sin∝))/(2-5.(cos∝)/(sin∝))`

`=13`

`->` Chọn `D`

_____________________

`"Câu 10:C"`

Ta có: `cos∝=-2/3` `=>90^o<∝<180^o`

`sin^2∝+cos^2∝=1`

`->sin^2∝=1-cos^2∝=5/9`

`->sin∝=sqrt5/3`

`-> cot∝=(cos∝)/(sin∝)=-(2sqrt5)/5`

`->tan∝=-sqrt5/2`

Thay `cot∝=-(2sqrt5)/5;tan∝=-sqrt5/2` vào `B` ta được:

`E=-(11)/3`

`->` Chọn `C`

Đáp án:

 `6)` `bbD`

`7)`  `bbB`

`8)`  `bbD`

`9)` `bbD`

`10)` `bbC`

Giải thích các bước giải:

 Câu `6:`

Ta có:

`sin\alpha+cos\alpha=a`

`<=>(sin\alpha+cos\alpha)^{2}=a^{2}`

`<=>sin^{2}\alpha+cos^{2}\alpha+2sin\alpha.cos\alpha=a^{2}`

`<=>1+2sin\alpha.cos\alpha=a^{2}`

`<=>2sin\alpha.cos\alpha=a^{2}-1`

`<=>sin\alpha.cos\alpha=\frac{a^{2}-1}{2}`

`->`  Chọn `bbD`

Câu `7:`

`E=\frac{cot\alpha+3tan\alpha}{2cot\alpha+tan\alpha}`

`=\frac{\frac{1}{tan\alpha}+3tan\alpha}{2.\frac{1}{tan\alpha}+tan\alpha}`

`=(\frac{1+3tan^{2}\alpha}{tan\alpha}):(\frac{2+tan^{2}\apha}{tan\alpha})`

`=\frac{1+3.(\frac{1}{cos^{2}\alpha}-1)}{2+\frac{1}{cos^{2}\alpha}-1}`

`={1+3.[1:(-2/3)^{2}-1]}:{2+[1:(-2/3)^{2}-1]}`

`=[1+3.(9/4-1)]:[2+(9/4-1)]`

`=(1+3. 5/4):(2+5/4)`

`=19/4:13/4`

`=19/13`

`->` Chọn `bbB`

Câu `8:`

`E=2cos^{2}\alpha+5sin\alpha cos\alpha+1`

Chia cả hai vế cho `sin^{2}\alpha` ta được:

`->E: sin^{2}\alpha=2.\frac{cos^{2}\alpha}{sin^{2}\alpha}+\frac{5sin\alpha.cos\alpha}{sin^{2}\alpha}+\frac{1}{sin^{2}\alpha}`

`=2.cot^{2}\alpha+5. cot\alpha+1+cot^{2}\alpha`

`=3 cot^{2}\alpha+5. 5+1`

`=3.5^{2}+25+1`

`=75+26`

`=101`

`->E=101:(1+cot^{2}\alpha)`

`->E=\frac{101}{1+5^{2}}`

`->E=\frac{101}{26}`

`->` Chọn `bbD`

Câu `9:`

`A=\frac{3sin\alpha+4cos\alpha}{2sin\alpha-5cos\alpha}`

Chia cả hai vế cho `sin\alpha` ta được:

`->A=\frac{3sin\alpha:sin\alpha+4.\frac{cos\alpha}{sin\alpha}}{2sin\alpha-5.\frac{cos\alpha}{sin\alpha}}`

`=\frac{3+4.cot\alpha}{2-5.cot\alpha}`

`=\frac{3+4. 1/3}{2-5. 1/3}`

`=(3+4/3):(2-5/3)`

`=13/3:1/3`

`=13`

`->` Chọn `bbD`

Câu `10:`

`E=\frac{cot\alpha-3tan\alpha}{2cot\alpha-tan\alpha}`

`=\frac{\frac{1}{tan\alpha}-\frac{3tan^{2}\alpha}{tan\alpha}}{2. \frac{1}{tan\alpha}-\frac{tan^{2}\alpha}{tan\alpha}}`

`=\frac{\frac{1-3tan^{2}\alpha}{tan\alpha}}{\frac{2-tan^{2}\alpha}{tan\alpha}}`

`=[1-3. (\frac{1}{cos^{2}\alpha}-1)]:[2-(\frac{1}{cos^{2}\alpha}-1)]`

`={1-3.[1:(-2/3)^{2}-1]}:{2-[1:(-2/3)^{2}-1]}`

`=[1-3.(9/4-1)]:[2-(4/9-1)]`

`=(1-3. 5/4):(2-5/4)`

`=(-11/4):3/4`

`=-11/3`

`->` Chọn `bbC`