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

\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)

\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)

\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)

\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

\(=cos^2x-sin^2x+cos^2x+3sin^2x\)

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

a/

\(\Leftrightarrow\left(sin^2\frac{x}{3}+cos^2\frac{x}{3}\right)^2-2sin^2\frac{x}{3}.cos^2\frac{x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^2\frac{2x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{4}\left(1-cos\frac{4x}{3}\right)=\frac{5}{8}\)

\(\Leftrightarrow cos\frac{4x}{3}=-\frac{1}{2}\)

\(\Leftrightarrow\frac{4x}{3}=\pm\frac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{2}+\frac{k3\pi}{2}\)

b/

\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x+\frac{1}{2}cos4x=-\frac{1}{2}\)

\(\Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)

c/

\(\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow-cos^32x+5cos^22x-7cos2x+3=0\)

\(\Leftrightarrow\left(3-cos2x\right)\left(cos2x-1\right)^2=0\)

\(\Leftrightarrow cos2x=1\)

\(\Leftrightarrow x=k\pi\)

d/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=cos4x\)

\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)=cos4x\)

\(\Leftrightarrow cos4x=1\)

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

\(P=\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)

\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)

\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)

\(=2cos^2x+1+2sin^2x+1\)

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

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

\(=cos^2x-sin^2x-2cos^2x\)

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

Lời giải:

* $x$ là biến chứ không phải tham số bạn nhé*
\(A=2[(\cos ^2x)^3+(\sin ^2x)^3]-3(\cos ^4x+\sin ^4x)\)

\(=2(\cos ^2x+\sin ^2x)(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)

\(=2(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)

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

là giá trị không phụ thuộc vào biến (đpcm)

--------------------------

\(B=\frac{\tan ^2x}{\sin ^2x\cos ^2x}-(1+\tan ^2x)^2=\frac{\sin ^2x}{\cos ^2x.\sin ^2x\cos ^2x}-(1+\frac{\sin ^2x}{\cos ^2x})^2\)

\(=\frac{1}{\cos ^4x}-(\frac{\cos ^2x+\sin ^2x}{\cos ^2x})^2=\frac{1}{\cos ^4x}-(\frac{1}{\cos ^2x})^2=\frac{1}{\cos ^4x}-\frac{1}{\cos ^4x}=0\)

là giá trị không phụ thuộc vào biến $x$ (đpcm)

Giả sử các biểu thức đều có nghĩa

\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)

\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)

\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)

\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)

b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)

\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)

c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)

\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)

\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)

\(A=27x^3-8-27x^3+3=-5\\ B=8x^3+y^3+8x^3-y^3-16x^3=0\)

phần chứng minh biểu thức không phụ thuộc \(x\)

ta có : \(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{cos^2a}{cot^2a}\)

\(=\dfrac{cot^2a-cos^2a+cos^2a}{cot^2a}=\dfrac{cot^2a}{cot^2a}=1\left(đpcm\right)\)

ý còn lại : xem lại đề nha bn

phần chứng minh đẳng thức

ta có : \(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=\dfrac{2sinacosa-2sina}{2sinacosa+2sina}+tan^2\dfrac{a}{2}\)

\(=\dfrac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}+tan^2\dfrac{a}{2}=\dfrac{cosa-1}{cosa+1}+tan^2\dfrac{a}{2}\)

\(=\dfrac{1-2sin^2\dfrac{a}{2}-1}{2cos^2\dfrac{a}{2}-1+1}+tan^2\dfrac{a}{2}=\dfrac{-2sin^2\dfrac{a}{2}}{2cos^2\dfrac{a}{2}}+tan^2\dfrac{a}{2}\)

\(=-tan^2\dfrac{a}{2}+tan^2\dfrac{a}{2}=0\left(đpcm\right)\)

ta có : \(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}\)

\(=\dfrac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}\)

\(=\dfrac{2\left(cosa+1\right)}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\left(đpcm\right)\)

còn 2 câu kia để chừng nào rảnh mk giải cho nha

mk lm 2 câu còn lại nha

ta có : \(\dfrac{sin^2x}{sinx-cosx}-\dfrac{sinx+cosx}{tan^2x-1}=\dfrac{\left(1-cos^2x\right)\left(tan^2x-1\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)

\(=\dfrac{tan^2x-sin^2x-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\dfrac{sin^4x}{cos^2x}-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2-1\right)}\)

\(=\dfrac{tan^2x\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\left(tan^2x-1\right)\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)

\(=sinx+cosx\left(đpcm\right)\)

ta có : \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-\dfrac{sin^2a.cos^2b}{cos^2a.sin^2b}}\)

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

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(\left(sina.cosb-cosa.sinb\right)\left(sina.cosb+cosa.sinb\right)\right)}\)

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-sin\left(a-b\right)sin\left(a+b\right)}=-cos^2a.sin^2b\left(đpcm\right)\)

mk lm hơi tắc ! do tối rồi , mà mk lại đang ở quán nek nên không tiện làm dài . bạn thông cảm