Ta có: `sin^2 a+cos^2 a=1`

    `=>cos a=+- 4/5`   Mà `90^o < a < 180^o`

    `=>cos a=-4/5`

    `=>{(tan a=[sin a]/[cos a]=-3/4),(cot a=1/[tan a]=-4/3):}`

Có: `E=[cot a-2tan a]/[tan a+3cot a]`

      `E=[-4/3+2. 3/4]/[-3/4- 3. 4/3]=-2/57`.

Xét `\triangle MNP` vuông tại `M` có:

 `@cos \hat{N}=[MN]/[NP]=3/5`

     `=>\hat{N}~~53^o`

`@\hat{P}=90^o -\hat{N}=37^o`

`@MP=\sqrt{NP^2 -MN^2}=16`.

:))

Gọi đường thẳng cần tìm là `y=ax+b`   `(1)`

`a)` Thay `A(4;0);B(-1;2)` vào `(1)` có hệ:

   `{(4a+b=0),(-a+b=2):}<=>{(a=2/5),(b=-8/5):}`

 `=>` Ptr đường thẳng `(1)` là: `y=2/5x-8/5`

`b)-2x+y=3<=>y=2x-3`

`(1) \bot y=2x-3<=>a.2=-1<=>a=-1/2`

Thay `a=-1/2; M(-1;2)` vào `(1)` có:

      `2=-1/2 .(-1)+b<=>b=3/2`

  `=>` Ptr đường thẳng `(1)` là: `y=-1/2x+3/2`.

`A=[sin x+sin 2x+sin 3x]/[cos x+cos 2x+cos 3x]`

`A=[(sin x+sin 3x)+sin 2x]/[(cos x+cos 3x)+cos 2x]`

`A=[2sin 2x.cos (-x)+sin 2x]/[2cos 2x.cos (-x)+cos 2x]`

`A=[sin 2x(2cos(-x)+1)]/[cos 2x(2cos(-x)+1)]`

`A=[sin 2x]/[cos 2x]=tan 2x`.

`@R_[tđ]=[R_1(R_2+R_3)]/[R_1+R_2+R_3]=20/3(\Omega)`

`=>I=U/[R_[tđ]]=1,5(A)`

`@R_1 //// R_[23]=>U=U_1=U_[23]=10(V)`

  `=>{(I_1=10/10=1(A)),(I_[23]=10/[5+15]=0,5(A)=I_1 =I_2):}`

`x+y-6=0<=>y=-x+6`

Hàm số `y=(4m-5)x+3m //// y=-x+6`

    `<=>{(4m-5=-1),(3m ne 6):}`

   `<=>{(m=1),(m ne 2):}=>m=1`.

`a)\root[3]{135}/\root[3]{5}-\root[3]{54}.\root[3]{4}`

`=\root[3]{135/5}-\root[3]{54.4}`

`=\root[3]{27}-\root[3]{216}`

`=3-6=-3`

`b)(\root[3]{25}-\root[3]{10}+\root[3]{4})(\root[3]{5}+\root[3]{2})`

`=5+\root[3]{50}-\root[3]{50}-\root[3]{20}+\root[3]{20}+2`

`=7`.

`ĐK: x >= 1`

`\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}`

`=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}`

`=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}`

`=\sqrt{x-1}+1+|\sqrt{x-1}-1|`

`a)\sqrt{45x}-2\sqrt{20x}+2\sqrt{80x}=21`    `ĐK: x >= 0`

`<=>3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21`

`<=>7\sqrt{5x}=21`

`<=>\sqrt{5x}=3`

`<=>5x=9<=>x=9/5` (t/m).

`b)\sqrt{x^2-10x+25}=4`

`<=>\sqrt{(x-5)^2}=4`

`<=>|x-5|=4`

`<=>[(x-5=4),(x-5=-4):}`

`<=>[(x=9),(x=1):}`