giúp mình câu c câu d với

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(x+\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\pi+k2\pi\\2x+\dfrac{\pi}{6}=-x-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

c.

\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)

\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

2.

\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)

\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)

\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)

Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)

\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)

Help

a: Xét ΔABH vuông tại H và ΔACK vuông tại K có 

AB=AC

\(\widehat{KAC}\) chung

Do đó: ΔABH=ΔACK

Suy ra: AH=AK

b: Xét ΔKIB vuông tại K và ΔHIC vuông tại H có 

KB=HC

\(\widehat{KBI}=\widehat{HCI}\)

Do đó: ΔKIB=ΔHIC

Suy ra: IK=IH

Xét ΔAKI vuông tại K và ΔAHI vuông tại H có 

AI chung

KI=HI

Do đó: ΔAKI=ΔAHI

Suy ra: \(\widehat{KAI}=\widehat{HAI}\)

Câu 7:

b: Tọa độ của C là:

\(\left\{{}\begin{matrix}2x-3=x+1\\y=x+1\end{matrix}\right.\Leftrightarrow C\left(4;5\right)\)

d. \(\dfrac{\pi}{2}< a;b< \pi\Rightarrow sina>0;sinb>0\)

\(sina=\sqrt{1-cos^2a}=\dfrac{4}{5}\Rightarrow tana=\dfrac{sina}{cosa}=-\dfrac{4}{3}\)

\(sinb=\sqrt{1-cos^2b}=\dfrac{5}{13}\Rightarrow tanb=-\dfrac{5}{12}\)

Vậy:

\(sin\left(a-b\right)=sina.cosb-cosa.sinb=\dfrac{4}{5}.\left(-\dfrac{12}{13}\right)-\left(-\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)=...\)

\(cos\left(a-b\right)=cosa.cosb-sina.sinb=...\) (bạn tự thay số bấm máy)

\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=...\)

\(cot\left(a+b\right)=\dfrac{1}{tan\left(a+b\right)}=\dfrac{1-tana.tanb}{tana+tanb}=...\)

e.

\(0< y< \dfrac{\pi}{2}\Rightarrow cosy>0\Rightarrow cosy=\sqrt{1-sin^2y}=\dfrac{4}{5}\)

\(\Rightarrow tany=\dfrac{siny}{cosy}=\dfrac{3}{4}\)

Vậy: \(tan\left(x+y\right)=\dfrac{tanx+tany}{1-tanx.tany}=...\)

\(cot\left(x-y\right)=\dfrac{1}{tan\left(x-y\right)}=\dfrac{1+tanx.tany}{tanx-tany}=...\)

a: Xet ΔBAE vuông tại A và ΔBDE vuông tại D có

BE chung

góc ABE=góc DBE

=>ΔBAE=ΔBDE

=>BA=BD

=>ΔBAD cân tại B

BA=BD

EA=ED

=>BE là trung trực của AD

b: AE=ED

ED<EC

=>AE<EC

b) Ta có: \(\left(-\dfrac{1}{2}\right)^2\cdot2\dfrac{6}{7}-\dfrac{14}{15}:2\dfrac{1}{3}+\left(-1.21\right)^0\)

\(=\dfrac{1}{4}\cdot\dfrac{20}{7}-\dfrac{14}{15}:\dfrac{7}{3}+1\)

\(=\dfrac{5}{7}-\dfrac{14}{15}\cdot\dfrac{3}{7}+1\)

\(=\dfrac{5}{7}-\dfrac{2}{5}+1\)

\(=\dfrac{25-14-35}{35}=\dfrac{-24}{35}\)