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\(\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)
\(=\sqrt{sin^4x-4sin^2x+4}+\sqrt{cos^4x-4cos^2x+4}\)
\(=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}\)
\(=2-sin^2x+2-cos^2x\)
\(=4-\left(sin^2x+cos^2x\right)=3\)
\(A=cot^2x+tan^2x+2-\left(cot^2x+tan^2x-2\right)=4\)
\(B=cos^2x.cot^2x-cot^2x+cos^2x+2\left(sin^2x+cos^2x\right)\)
\(=cot^2x\left(cos^2x-1\right)+cos^2x+2\)
\(=-cot^2x.sin^2x+cos^2x+2\)
\(=-cos^2x+cos^2x+2=2\)
\(C=\left(sin^4x+cos^4x\right)^2+4sin^4x.cos^4x+4sin^2xcos^2x\left(sin^4x+cos^4x\right)+1\)
\(=\left(sin^4x+cos^4x+2sin^2x.cos^2x\right)^2+1\)
\(=\left(sin^2x+cos^2x\right)^4+1\)
\(=1^4+1=2\)
\(A=\frac{\frac{4sin^2x}{cos^2x}+\frac{5sinx.cosx}{cos^2x}+\frac{cos^2x}{cos^2x}}{\frac{sin^2x}{cos^2x}-\frac{2}{cos^2x}}=\frac{4tan^2x+5tanx+1}{tan^2x-2\left(1+tan^2x\right)}\)
\(=\frac{4.9-5.3+1}{9-2\left(1+9\right)}=...\)
\(A=2(\sin ^6x+\cos ^6x)-3(\sin ^4x+\cos ^4x)\)
\(=2(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)
\(=2(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)
\(=-(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x)=-(\sin ^2x+\cos ^2x)^2=-1^2=-1\)
là giá trị không phụ thuộc vào biến (đpcm)
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\(B=\sin ^6x+\cos ^6x-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=-\sin ^4x-\sin ^2x\cos ^2x+\sin ^2x=-\sin ^2x(\sin ^2x+\cos ^2x)+\sin ^2x\)
\(=-\sin ^2x+\sin ^2x=0\)
là giá trị không phụ thuộc vào biến (đpcm)
\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)
\(=(\sin ^4x+\cos ^4x-1).\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=(\sin ^4x+\cos ^4x-1).\frac{(\sin ^2x+\cos ^2x)^2}{\sin ^2x\cos ^2x}\)
\(=(\sin ^4x+\cos ^4x-1).\frac{1}{\sin ^2x\cos ^2x}=\frac{(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}\)
\(=\frac{(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{1-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{-2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=-2\)
là giá trị không phụ thuộc vào biến $x$
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\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)
\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)
\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)
\(=\frac{\cos ^6x+3\sin ^2x\cos ^2x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{\cos ^4x+2\sin ^2x}{\cos ^4x}\)
\(=1+\frac{2\sin ^2x}{\cos ^4x}\)
Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.
\(sina\sqrt{1+\frac{sin^2a}{cos^2a}}=sina\sqrt{\frac{cos^2a+sin^2a}{cos^2a}}=\frac{sina}{\left|cosa\right|}=\pm tana\)
\(\frac{1-cos^2x}{1-sin^2x}+tanx.cotx=\frac{sin^2x}{cos^2x}+\frac{sinx}{cosx}.\frac{cosx}{sinx}=tan^2x+1=\frac{1}{cos^2x}\)
\(\frac{1-4sin^2xcos^2x}{\left(sinx+cosx\right)^2}=\frac{\left(1-2sinx.cosx\right)\left(1+2sinx.cosx\right)}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(1-sin2x\right)\left(1+2sinx.cosx\right)}{1+2sinx.cosx}=1-2sinx\)
\(sin\left(90-x\right)+cos\left(180-x\right)+sin^2x\left(1+tan^2x\right)-tan^2x\)
\(=cosx-cosx+sin^2x.\frac{1}{cos^2x}-tan^2x=tan^2x-tan^2x=0\)
\(P=4sin^2x+\sqrt{2}\left(sin2x.cos\frac{\pi}{4}+cos2x.sin\frac{\pi}{4}\right)\)
\(P=4sin^2x+sin2x+cos2x\)
\(P=2\left(1-cos2x\right)+sin2x+cos2x\)
\(P=2+sin2x-cos2x\)
\(P=2+\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)
Do \(sin\left(2x-\frac{\pi}{4}\right)\le1\Rightarrow P\le2+\sqrt{2}\)
\(\Rightarrow P_{max}=2+\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=1\Leftrightarrow x=\frac{3\pi}{8}+k\pi\)
\(P=\dfrac{\left(cos^2x-sin^2x\right)^2}{4sin^2x.cos^2x}-\dfrac{1}{4sin^2x.cos^2x}=\dfrac{\left(cos^2x-1-sin^2x\right)\left(cos^2x+1-sin^2x\right)}{4sin^2x.cos^2x}\)
\(=\dfrac{-2sin^2x.2cos^2x}{4sin^2x.cos^2x}=-1\)
\(y^2-3y-1=0\) có \(ac=-1< 0\Leftrightarrow\) có 2 nghiệm trái dấu hay có 1 nghiệm dương
\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\left(1-cos^2x\right)}\)
\(=\sqrt{4-4sin^2x+sin^4x}+\sqrt{4-4cos^2x+cos^4x}\)
\(=\sqrt{\left(2-sin^2x\right)^2}+\sqrt{\left(2-cos^2x\right)^2}\)
\(=2-sin^2x+2-cos^2x=4-\left(sin^2x+cos^2x\right)\)
\(=3\)
Lời giải:
Ta có:
\(\sin ^2x\tan ^2x+4\sin ^2x-\tan ^2x+3\cos ^2x\)
\(=\tan ^2x(\sin ^2x-1)+4\sin ^2x+3\cos ^2x\)
\(=\tan ^2x(-\cos ^2x)+4\sin ^2x+3\cos ^2x\)
\(=\left(\frac{\sin x}{\cos x}\right)^2(-\cos ^2x)+4\sin ^2x+3\cos ^2x\)
\(=-\sin ^2x+4\sin ^2x+3\cos ^2x\)
\(=3(\sin ^2x+\cos ^2x)=3\)
Vậy giá trị của biểu thức không phụ thuộc vào $x$
Ta có đpcm.