Câu 5. Cho x,y dương thỏa mãn \(x+y=\dfrac{1}{2}\).Tìm giá trị nhỏ nhất của 

\(P=\dfrac{1}{x}+\dfrac{1}{y}\)

Giải:

\(P=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{\dfrac{1}{2}}{xy}=\dfrac{2}{xy}\)

--> P nhỏ nhất khi \(xy\) lớn nhất

Ta có:

\(x^2+y^2\ge2xy\) ( BĐT AM-GM )

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow1\ge4xy\)

\(\Leftrightarrow xy\le\dfrac{1}{4}\)

\(\Rightarrow P\ge2:\dfrac{1}{4}=8\)

Vậy \(Min_P=8\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{4}\)

a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)

b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)

c/\(P=sin\left(30+60\right)=sin90=1\)

d/

\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)

\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)

e/

\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)

\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)

sina - 1 = sina - sin\(\dfrac{\pi}{2}\)

c o s 60 ° 1 + sin 60 ° + 1 t g 30 ° = 2

ta có cos60=1/2

sin 60=\(\frac{\sqrt{3}}{2}\)

tan 30=\(\frac{\sqrt{3}}{3}\)

ta thay vào biểu thức trên

=> \(\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=2\)

\(\frac{cos60^o}{1+sin60^o}+\frac{1}{tan30^o}=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=\frac{1}{2}.\frac{2}{\sqrt{3}+2}+\sqrt{3}=\frac{1}{\sqrt{3}+2}+\sqrt{3}\)

\(=\frac{2-\sqrt{3}}{4-3}+\sqrt{3}=2-\sqrt{3}+\sqrt{3}=2\)

\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)

=3/4*4/5*...*99/100

=3/100

Vì 30⁰ và 60⁰ là hai góc phụ nhau nên  sin 30 0 = cos 60 0 sin 60 0 = cos 30 0

⇒ P = cos 30 ∘ cos 60 ∘ − sin 30 ∘ sin 60 ∘ = cos 30 ∘ cos 60 ∘ − cos 60 ∘ cos 30 ∘ = 0.  

Chọn D.

Vì 30⁰ và 60⁰ là hai góc phụ nhau nên  sin 30 0 = cos 60 0 sin 60 0 = cos 30 0

⇒ P = sin 30 ∘ cos 60 ∘ + sin 60 ∘ cos 30 ∘ = cos 2 60 ∘ + sin 2 60 ∘ = 1.

 Chọn A.