Em nhập lại nội dung câu hỏi nhé!

\(3cot\left(3x+10^0\right)-3=0\)

\(\Leftrightarrow cot\left(3x+10^0\right)=1\)

\(\Leftrightarrow3x+10^0=45^0+k180^0\)

\(\Leftrightarrow x=\dfrac{35^0}{3}+k60^0\) (\(k\in Z\))

Gọi `A'(0;-3);B'(-1;-1) in d'`

`Q_{(O,-90^o )} (A)=A'<=>{(x_[A']=-y_A),(y_[A']=x_A):}<=>{(x_A=-3),(y_A=0):}`

        `=>A(-3;0) in d`

`Q_{(O,-90^o )} (B)=B'<=>{(x_[B']=-y_B),(y_[B']=x_B):}<=>{(x_B=-1),(y_B=1):}`

      `=>B(-1;1) in d`

Có: `\vec{AB}=(2;1)=>\vec{n_d}=(1;-2)`

       Mà `B(-1;1) in d`

 `=>`Ptr `d: 1(x+1)-2(y-1)=0<=>x-2y+3=0`

Cảm ơn bạn nhiều lắm:3

`a)x^3-6x^2+9x-4xy^2`

`=x(x^2-6x+9-4y^2)`

`=x[(x-3)^2-4y^2]`

`=x(x-3-2y)(x-3+2y)`

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`b)x^2-4xy+3x-12y`

`=x(x-4y)+3(x-4y)`

`=(x-4y)(x+3)`

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`c)5x^2+17x+6`

`=5x^2+15x+2x+6`

`=5x(x+3)+2(x+3)`

`=(x+3)(5x+2)`

Gọi `A(2;1)` và `B(-1;-1) in \Delta`

`Q_{(O,90^o )} (A)=A'<=>A'(-1;2)`

`Q_{(O,90^o )} (B)=B'<=>B'(1;-1)`

Gọi `\Delta'` là ảh của `\Delta=>A';B' in \Delta'`

 Có: `\vec{A'B'}=(2;-3)=>\vec{n_[\Delta']}=(3;2)`

            Mà `B'(1;-1) in \Delta'`

  `=>` Ptr `\Delta'` là: `3(x-1)+2(y+1)=0<=>3x+2y-1=0`

`1)3cos x-\sqrt{3}sin x=-3`

`<=>\sqrt{3}/2cos x-1/2sin x=-\sqrt{3}/2`

`<=>cos(x+\pi/6)=-\sqrt{3}/2`

`<=>[(x+\pi/6=[5\pi]/6+k2\pi),(x+\pi/6=[-5\pi]/6+k2\pi):}`

`<=>[(x=[2\pi]/3+k2\pi),(x=-\pi+k2\pi):}`    `(k in ZZ)`

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`2)sin 4x=\sqrt{3}(cos 4x-1)`

`<=>sin 4x-\sqrt{3}cos 4x=-\sqrt{3}`

`<=>1/2sin 4x-\sqrt{3}/2cos 4x=-\sqrt{3}/2`

`<=>sin(4x-\pi/3)=-\sqrt{3}/2`

`<=>[(4x-\pi/3=-\pi/3+k2\pi),(4x-\pi/3=[4\pi]/3+k2\pi):}`

`<=>[(x=k\pi/2),(x=[5\pi]/12+k\pi/2):}`       `(k in ZZ)`

Đặt \(q\) là công bội của cấp số nhân \(\left(u_k\right)\)

\(u_k=u_1\)\(.q^{k-1}\)

\(u_{k-1}=u_1.q^{k-2}\)

\(u_{k+1}=u_1.q^k\)

\(\Rightarrow u_{k-1}.u_{k+1}=\)\(u^2_1\)\(.q^{k-2}.q^k\)

\(=u^2_1.q^{k-2+k}\)

\(=u^2_1.q^{2k-2}\)

\(=u^2_1.\left(q^{k-1}\right)^2\)

\(=\left(uk\right)^1\)

Vậy:\(u^2_k=u_{k-1}u_{k+1}\)

thanks