Đề bài là: Tính cos2x 

Cảm ơn mn nhiều ạ!

`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`

`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`

`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`

`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`

`<=> cos2x - cos4x - cos2x = 1/2`

`<=> cos4x = cos(2π)/3`

`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)

\(sin3x.sinx+sin\left(x-\dfrac{\pi}{3}\right)cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x+\dfrac{1}{2}sin\left(2x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin\left(-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}cos2x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos4x+\dfrac{1}{2}=0\)

\(\Leftrightarrow2cos^22x-1+\dfrac{1}{2}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

\(\Rightarrow cos2x=\pm\dfrac{1}{2}\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

(3x)^2=3^2.x^2=9x^2

bạn tự vẽ hình giúp mik nha

a. xét \(\Delta ADN\) và \(\Delta BAM\) có

AB=AD(gt)

\(\widehat{ADN}=\widehat{BAM}=90^o\)

DN=MA(N,M là trung điểm của cạnh DC,AD)

\(\Rightarrow\Delta ADN\sim\Delta BAM\left(c.g.c\right)\)

\(\Rightarrow\widehat{DNA}=\widehat{AMB}\)

mà:\(\widehat{DNA}+\widehat{DAN}=90^o\Rightarrow\widehat{BMA}+\widehat{DAN}=90^o\)

\(\Rightarrow\Delta MAI\) vuông tại I

\(\Rightarrow AI\perp MI\) hay \(MB\perp AN\)

b.ta có M là trung điểm của AD\(\Rightarrow AM=\dfrac{1}{2}AD=\sqrt{5}\)

trong \(\Delta MAB\) vuông tại A có

\(MB=\sqrt{AM^2+AB^2}=\sqrt{\sqrt{5^2}+\left(2\sqrt{5}\right)^2}=5\)

\(AM^2=MB.MI\Rightarrow MI=\dfrac{AM^2}{MB}=\dfrac{\sqrt{5^2}}{5^5}=0,2\)

\(AI.MB=AM.AB\Rightarrow AI=\dfrac{AM.AB}{MB}=\dfrac{\sqrt{5}.2\sqrt{5}}{5}\)=2

c.IB=MB-MI=5-0,2=4,8

\(S_{\Delta AIB}=\dfrac{AI.IB}{2}=\)\(\dfrac{2.4,8}{2}=4,8\)

\(S_{\Delta ADN}=\dfrac{AD.DN}{2}=\dfrac{2\sqrt{5}.\sqrt{5}}{2}=5\)

\(S_{\Delta ABCD}=\left(2\sqrt{5}\right)^2=20\)

\(S_{BINC}=S_{ABCD}-S_{\Delta AIB}-S_{\Delta DAN}\)=20-4,8-5=10,2