\(A=cos3a+2cos\left(\pi-3a\right)sin^2\left(\dfrac{\pi}{4}-1,5a\right)\)

\(=cos3a-2cos3a\dfrac{1-cos\left(\dfrac{\pi}{2}-3a\right)}{2}\)

\(=cos3a-cos3a\left(1-sin3a\right)\)

\(=cos3a-cos3a+cos3asin3a=\dfrac{1}{2}sin6a\)

\(=\dfrac{1}{2}sin\left(6\dfrac{5\pi}{6}\right)=\dfrac{1}{2}sin\left(4\pi+\pi\right)=\dfrac{1}{2}sin\pi=0\)

\(A=\dfrac{2\left(cos\dfrac{\Pi}{2}\right)^2-1}{1+8sin^2\dfrac{\Pi}{8}\left(cos\dfrac{\Pi}{2}\right)^2}=\dfrac{2\left(0\right)^2-1}{1+8sin^2\left(0\right)^2}=\dfrac{-1}{1}=-1\)

​vậy A= -1

gọi pt đường trọng cần tìm là: \(\left(x-a\right)^2+\left(y-b\right)^2=R^2\left(C\right)\)

với I(a; b)

(C) tiếp xúc với 2 trục tọa độ \(\Rightarrow a=b=R\Rightarrow\left(C\right)\)co dang \(\left(x-a\right)^2+\left(y-a\right)^2=a^2\left(1\right)\)

lại có I(a;b) \(\in\) 4x-2y-8=0 \(\Rightarrow4a-2a-8=0\Rightarrow a=4\)

thay a = 4 vao (1) \(\Rightarrow\left(C\right)\left(x-4\right)^2+\left(y-4\right)^2=16\)

cau a: \(cos\dfrac{22\Pi}{3}=cos\dfrac{24\Pi-2\Pi}{3}=cos\left(8\Pi-\dfrac{2\Pi}{3}\right)=cos\dfrac{2\Pi}{3}=-\dfrac{1}{2}\)

câu b: \(sin\dfrac{23\Pi}{4}=sin\dfrac{24\Pi-\Pi}{4}=sin\left(6\Pi-\dfrac{\Pi}{4}\right)=-sin\dfrac{\Pi}{4}=-\dfrac{\sqrt{2}}{2}\)

cau c: \(=sin\left(8\Pi-\dfrac{\Pi}{3}\right)-tan\left(3\Pi+\dfrac{\Pi}{3}\right)=-sin\dfrac{\Pi}{3}-tan\dfrac{\Pi}{3}=-\dfrac{\sqrt{3}}{2}-\sqrt{3}=\dfrac{-3\sqrt{3}}{2}\)

cau d: \(cos^2\dfrac{\Pi}{8}-sin^2\dfrac{\Pi}{8}=cos2\left(\dfrac{\Pi}{8}\right)=cos\dfrac{\Pi}{4}=\dfrac{\sqrt{2}}{2}\)

​ta có \(sin^2a+cos^2a=1\Rightarrow sina=\pm\sqrt{1-cos^2a}=\pm\sqrt{1-\left(\dfrac{-\sqrt{5}}{3}\right)^2}=\pm\dfrac{2}{3}\)

​vì \(\Pi< a< \dfrac{3\Pi}{2}\Rightarrow sina< 0\) \(\Rightarrow sina=\dfrac{-2}{3}\)

lại có \(tana=\dfrac{sina}{cosa}=\dfrac{\dfrac{-2}{3}}{\dfrac{-\sqrt{5}}{3}}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

chia tử và mẫu của B cho sina khác 0\(B=\dfrac{4\dfrac{sina}{sina}+5\dfrac{cosa}{sina}}{2\dfrac{sina}{sina}-3\dfrac{cosa}{sina}}=\dfrac{4+5cota}{2-3cota}=\dfrac{4+5\dfrac{1}{2}}{2-3\dfrac{1}{2}}=13\)

vay B = 13

ta có : \(cos\left(\Pi-3a\right)=-cosa\)

\(sin^2\left(\dfrac{\Pi}{4}-1,5a\right)=\dfrac{1-cos\left(\dfrac{\Pi}{2}-3a\right)}{2}=\dfrac{1-cos3a}{2}\)

\(\Rightarrow cos3a+2cos\left(\Pi-3a\right)sin^2\left(\dfrac{\Pi}{4}-1,5a\right)=cos3a-2cos3a\left(\dfrac{1-cos3a}{2}\right)\) =\(cos^23a=cos^23.\dfrac{5\Pi}{6}=cos^2\dfrac{5\Pi}{2}=cos^2\dfrac{\Pi}{2}=0\)

cau a: \(y'=\dfrac{\sqrt{1-x}-\left(1+x\right)\left(\sqrt{1-x}\right)^'}{1-x}=\dfrac{\sqrt{1-x}-\left(1+x\right)\left(\dfrac{-1}{2\sqrt{1-x}}\right)}{1-x}=\dfrac{3-x}{2\left(\sqrt{1-x}\right)^2}\)câu b: \(y'=\dfrac{\sqrt{a^2-x^2}-x\left(\dfrac{-2x}{2\sqrt{a^2-x^2}}\right)}{a^2-x^2}=\dfrac{a^2}{\left(\sqrt{a^2-x^2}\right)^3}\)

vẽ hình ở ngoài rồi dán vào ko biết tại sao nó lại thụt xuống dưới

cau 12:

gọi E là trung điểm AB \(\Rightarrow\)MẸ//BC ; và EN// AC do do ME=BD/2 ;NE= AC/2

\(\Rightarrow\left[\widehat{BD;AC}\right]=\left[\widehat{ME;EN}\right]=90^0\)

\(\Delta MEN\)vuông tại E\(\Rightarrow MN^2=ME^2+NE^2=\left(\dfrac{3a}{2}\right)^2+\left(\dfrac{a}{2}\right)^2=\left(\dfrac{10a^2}{4}\right)\Rightarrow MN=\dfrac{a\sqrt{10}}{2}\)

​chọn đáp án A