Ôn tập cuối năm môn Đại số - Hỏi đáp

sina.cosa=1 => sina,cosa≠0 => sina+cosa≠0

\(P=\frac{\sin^3a+\cos^3a}{\sin a+\cos a}=\frac{\left(\sin a+\cos a\right).\left(\sin^2a-\sin a.\cos a+\cos^2a\right)}{\sin a+\cos a}\)

\(=\sin^2a+\cos^2a-\sin a.\cos a=1-1=0\)

Nhận thấy \(cos\frac{x}{2}=0\) ko phải nghiệm, chia 2 vế cho \(cos^3\frac{x}{2}\) ta được:

\(3tan^3\frac{x}{2}+3tan^2\frac{x}{2}=tan\frac{x}{2}+1\)

\(\Leftrightarrow3tan^2\frac{x}{2}\left(tan\frac{x}{2}+1\right)-\left(tan\frac{x}{2}+1\right)=0\)

\(\Leftrightarrow\left(3tan^2\frac{x}{2}-1\right)\left(tan\frac{x}{2}+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tan\frac{x}{2}=-1\\tan\frac{x}{2}=\frac{1}{\sqrt{3}}\\tan\frac{x}{2}=-\frac{1}{\sqrt{3}}\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\x+y+2xy=2\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)

\(\Rightarrow\left\{{}\begin{matrix}a^3-3ab=8\\a+2b=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=8\\a=2-2b\end{matrix}\right.\)

\(\Rightarrow\left(2-2b\right)^3-3b\left(2-2b\right)-8=0\)

\(\Leftrightarrow2b\left(4b^2-15b+15\right)=0\)

\(\Leftrightarrow b=0\Rightarrow a=2\Rightarrow\left(x;y\right)=\left(0;2\right);\left(2;0\right)\)

Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\):

\(x^2+\frac{1}{x^2}-4\left(x+\frac{1}{x}\right)+5=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\) pt trở thành:

\(t^2-2-4t+5=0\)

\(\Leftrightarrow t^2-4t+3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=1\\x+\frac{1}{x}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x^2-3x+1=0\end{matrix}\right.\)

mọi người ơi giúp mình với

Câu đầu ko dịch được đề, lỗi kí tự rồi bạn

b/

\(\Leftrightarrow2cos^6x+sin^4x+2cos^2x-1=0\)

\(\Leftrightarrow2cos^2x\left(cos^4x+1\right)+\left(sin^2x-1\right)\left(sin^2x+1\right)=0\)

\(\Leftrightarrow cos^2x\left(2cos^4x+2\right)-cos^2x\left(sin^2x+1\right)=0\)

\(\Leftrightarrow cos^2x\left(2cos^4x+1-sin^2x=0\right)\)

\(\Leftrightarrow cos^2x\left(2cos^4x+cos^2x\right)=0\)

\(\Leftrightarrow cos^4x\left(2cos^2x+1\right)=0\)

\(\Leftrightarrow cos^4x=0\Leftrightarrow cosx=0\)

\(\Leftrightarrow x=\frac{\pi}{2}+k\pi\)

\(A=2sin2x.cos2x.cos4x=sin4x.cos4x=\frac{1}{2}sin8x\)

\(B=sin^4x+cos^6x-6sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x\)

\(=1-2\left(2sinx.cosx\right)^2=1-2sin^22x=cos4x\)

\(C=\frac{cos2a+1-2cos^22a}{2sin2a.cos2a+sin2a}=\frac{\left(1-cos2a\right)\left(2cos2a+1\right)}{sin2a\left(2cos2a+1\right)}=\frac{1-cos2a}{sin2a}\)

\(=\frac{1-\left(1-2sin^2a\right)}{2sina.cosa}=\frac{2sin^2a}{2sina.cosa}=\frac{sina}{cosa}=tana\)

\(D=\frac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}=\frac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\frac{cos3a}{sin3a}=cot3a\)

\(E=\frac{1}{2}-\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)-\frac{1}{2}+\frac{1}{2}cos\left(\frac{\pi}{4}+x\right)\)

\(=\frac{1}{2}\left[cos\left(\frac{\pi}{4}+x\right)-cos\left(\frac{\pi}{4}-x\right)\right]=-sin\frac{\pi}{4}.sinx=-\frac{\sqrt{2}}{2}sinx\)

Đặt t=\(\left(\frac{x}{x+1}\right)^2\) (t>0)

Có: t²-t-\(\frac{3}{2}\)=0

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{1+\sqrt{7}}{2}\left(TM\right)\\t=\frac{1-\sqrt{7}}{2}\left(KTM\right)\end{matrix}\right.\)

\(\Rightarrow\left(\frac{x}{x+1}\right)^2=\frac{1+\sqrt{7}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{x+1}=\sqrt{\frac{1+\sqrt{7}}{2}}\\\frac{x}{x+1}=-\sqrt{\frac{1+\sqrt{7}}{2}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{\frac{1+\sqrt{7}}{2}}}{1-\sqrt{\frac{1+\sqrt{7}}{2}}}\\x=\frac{-\sqrt{\frac{1+\sqrt{7}}{2}}}{1+\sqrt{\frac{1+\sqrt{7}}{2}}}\end{matrix}\right.\)

Bạn tham khảo nha

Lời giải:

$B=\cos a+\cos b+\cos (a+b)+1$

$=2\cos \frac{a+b}{2}.\cos \frac{a-b}{2}+\cos ^2.\frac{a+b}{2}-\sin ^2\frac{a+b}{2}+1$

$=2\cos \frac{a+b}{2}.\cos \frac{a-b}{2}+2\cos ^2\frac{a+b}{2}-1+1$

$=2\cos \frac{a+b}{2}.\cos \frac{a-b}{2}+2\cos ^2\frac{a+b}{2}=2\cos \frac{a+b}{2}\left(\cos \frac{a-b}{2}+\cos \frac{a+b}{2}\right)$