a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)

\(=\left(1-sin^2a\right)-sin^2a=1\)

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

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

\(=2-sin^2a-cos^2a=2-1=1\)

1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)

 \(y=2-\left(-cosx\right).\left(-sinx\right)\)

y = 2 - sinx.cosx

y = \(2-\dfrac{1}{2}sin2x\)

Max = 2 + \(\dfrac{1}{2}\) = 2,5

Min = \(2-\dfrac{1}{2}\) = 1,5

2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)

Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)

Max = \(\sqrt{5}\)

\(A=\cos^4x+2\sin^2x.\cos^2x\left(\sin^2x+\cos^2x\right)+\sin^4x+1\)

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

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

1: Ta có: \(-1<=\sin\left(2x+\frac{\pi}{4}\right)\le1\)

=>\(-3\le3\cdot\sin\left(2x+\frac{\pi}{4}\right)\le3\)

=>\(-3-1\le3\cdot\sin\left(2x+\frac{\pi}{4}\right)-1\le3-1\)

=>-4<=y<=2

=>Tập giá trị là T=[-4;2]

\(y_{\min}=-4\) khi \(\sin\left(2x+\frac{\pi}{4}\right)=-1\)

=>\(2x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\)

=>\(2x=-\frac34\pi+k2\pi\)

=>\(x=-\frac38\pi+k\pi\)

2: \(0\le cos^2x\le1\)

=>\(0\ge-5\cdot cos^2x\ge-5\)

=>\(0+3\ge-5\cdot cos^2x+3\ge-5+3\)

=>3>=y>=-2

=>Tập giá trị là T=[-2;3]

\(y_{\max}=3\) khi \(cos^2x=1\)

=>\(\sin^2x=0\)

=>sin x=0

=>\(x=k\pi\)

\(y_{\min}=-2\) khi \(cos^2x=0\)

=>cosx=0

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

3: \(-1\le cosx\le1\)

=>\(-3\le3\cdot cosx\le3\)

=>\(-3+4\le3\cdot cosx+4\le3+4\)

=>\(1\le3\cdot cosx+4\le7\)

=>\(\frac51\ge\frac{5}{3\cdot cosx+4}\ge\frac57\)

=>\(\frac57\le y\le5\)

=>Tập giá trị là \(T=\left\lbrack\frac57;5\right\rbrack\)

\(y_{\min}=\frac57\) khi cosx=1

=>\(x=k2\pi\)

\(y_{\max}=5\) khi cosx=-1

=>\(x=\pi+k2\pi\)

4: \(y=\sin^2x-4\cdot\sin x+8\)

\(=\sin^2x-4\cdot\sin x+4+4\)

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

Ta có: \(-1\le\sin x\le1\)

=>\(-1-2\le\sin x-2\le1-2\)

=>\(-3\le\sin x-2\le-1\)

=>\(1\le\left(\sin x-2\right)^2\le9\)

=>\(5\le\left(\sin x-2\right)^2+4\le13\)

=>5<=y<=13

=>Tập giá trị là T=[5;13]

\(y_{\min}=5\) khi sin x-2=-1

=>sin x=1

=>\(x=\frac{\pi}{2}+k2\pi\)

\(y_{\max}\) =13 khi sin x-2=-3

=>sin x=-1

=>\(x=-\frac{\pi}{2}+k2\pi\)

heo me tim gtnn gtln cua bieu thuc:asinx + bcosx (a,b la hang so,a^2+b^2=/o)? | Yahoo Hỏi & Đáp

cám ơn bn nhìu nha 

\(B=cos^2x+sin^2x+tan^2x\)

\(=1+tan^2x\)

\(=\dfrac{1}{cos^2x}=1:\dfrac{1}{4}=4\)

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