Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

Câu 3: A

Câu 4: C

Câu 5: D

Câu 6: C

Câu 7: A

Câu 8: A

Câu 9: C

Câu 10: A

Câu 11: B

Câu 12: C

Câu 13: A

Câu 1: D

Câu 1: B

Câu 2: C

Câu 3: B

Câu 4: B

Câu 5: B

Câu 6: A

Câu 7: B

Câu 8: D

Câu 9: C

Câu 10: D

Câu 11: D

Câu 12: D

b:

ĐKXĐ: \(\left\{{}\begin{matrix}cosx< >0\\sinx< >0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{2}+k\Omega\\x\ne k\Omega\end{matrix}\right.\)

=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)

 \(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left[sinx\cdot\cos\dfrac{\Omega}{3}+sin\left(\dfrac{\Omega}{3}\right)\cdot cosx\right]\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left(\dfrac{1}{2}\cdot sinx+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\)

=>\(\left(sinx+\sqrt{3}\cdot cosx\right)\left(\dfrac{1}{cosx\cdot sinx}-1\right)=0\)

=>\(2\cdot\left(sinx\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\cdot\left(\dfrac{2}{2\cdot sinx\cdot cosx}-1\right)=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\cdot\left(\dfrac{2}{sin2x}-1\right)=0\)

=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{\Omega}{3}\right)=0\\\dfrac{2}{sin2x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=k\Omega\\sin2x=2\left(loại\right)\end{matrix}\right.\)

=>\(x=-\dfrac{\Omega}{3}+k\Omega\)

Đề thiếu rồi. Bạn xem lại đề.

\(-1< =sin2x< =1\)

=>\(-3< =3\cdot sin2x< =3\)

=>\(-8< =3\cdot sin2x-5< =3-5=-2\)

=>-8<=y<=-2

Vậy: giá trị nhỏ nhất là -8 và giá trị lớn nhất là -2

\(\cot x=\dfrac{1}{\tan x}\)

chỗ nào có \(\cot x\) thì bạn thay bằng \(\dfrac{1}{\tan x}\)xong bấm máy như bình thường thôi.

mình ra kết quả = \(\sqrt{3}\), bạn bấm lại xem có ra giống không nhé.

\(\dfrac{tan225-cot81\cdot cot69}{cot261+tan201}\)

\(=\dfrac{tan\left(180+45\right)-tan\left(90-81\right)\cdot tan\left(90-69\right)}{cot\left(180+81\right)+tan\left(180+21\right)}\)

\(=\dfrac{tan45-tan9\cdot tan21}{cot81+tan21}\)

\(=\dfrac{1-tan9\cdot tan21}{tan9+tan21}\)

\(=\dfrac{1-\dfrac{sin9}{cos9}\cdot\dfrac{sin21}{cos21}}{\dfrac{sin9}{cos9}+\dfrac{sin21}{cos21}}\)

\(=\dfrac{cos9\cdot cos21-sin9\cdot sin21}{cos9\cdot cos21}:\dfrac{sin9\cdot cos21+sin21\cdot cos9}{cos9\cdot cos21}\)

\(=\dfrac{cos\left(9+21\right)}{cos9\cdot cos21}\cdot\dfrac{cos9\cdot cos21}{sin\left(9+21\right)}=\dfrac{cos30}{sin30}=cot30=\sqrt{3}\)

Đề thiếu rồi em!

  ai giải giúp mình với

Lời giải:
Do $\cos x\in [-1;1]$ với mọi $x\in\mathbb{R}$ nên $\cos^2 x\in [0;1]$

$\Rightarrow 4-3.1\leq 4-3\cos ^2x\leq 4-3.0$

$\Rightarrow 1\leq y\leq 4$
$\Rightarrow M=4; m=1$
$\Rightarrow M+m=5$

Cách 1:

Ta có: \(tan\alpha=\sqrt{2}\Rightarrow\left\{{}\begin{matrix}\dfrac{sin\alpha}{cos\alpha}=\sqrt{2}\\1+\left(\sqrt{2}\right)^2=\dfrac{1}{cos^2\alpha}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}sin\alpha=\sqrt{2}\cdot cos\alpha\\cos^2\alpha=\dfrac{1}{3}\end{matrix}\right.\)

\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}\)

    \(=\dfrac{\sqrt{2}\cdot cos\alpha-cos\alpha}{\left(\sqrt{2}\cdot cos\alpha\right)^3+3cos^3\alpha+2\cdot\sqrt{2}\cdot cos\alpha}\)

    \(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{2\sqrt{2}\cdot cos^3\alpha+3cos^3\alpha+2\sqrt{2}\cdot cos\alpha}\)

    \(=\dfrac{cos\alpha\left(\sqrt{2}-1\right)}{cos\alpha\left(2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}\right)}\)

    \(=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot cos^2\alpha+3cos^2\alpha+2\sqrt{2}}\)

Thay \(cos^2\alpha=\dfrac{1}{3}\) vào \(P\) ta có:

\(P=\dfrac{\sqrt{2}-1}{2\sqrt{2}\cdot\dfrac{1}{3}+3\cdot\dfrac{1}{3}+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{1+\dfrac{8}{3}\sqrt{2}}\)

    \(=\dfrac{3\left(\sqrt{2}-1\right)}{3\left(1+\dfrac{8}{3}\sqrt{2}\right)}=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}\)

    \(=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=5\)

Chọn đáp án A.

Cách 2:

\(P=\dfrac{sin\alpha-cos\alpha}{sin^3\alpha+3cos^3\alpha+2sin\alpha}=\dfrac{\left(sin\alpha-cos\alpha\right)\div cos^3\alpha}{\left(sin^3\alpha+3cos^3\alpha+2sin\alpha\right)\div cos^3\alpha}\)

    \(=\dfrac{\dfrac{sin\alpha}{cos^3\alpha}-\dfrac{1}{cos^2\alpha}}{\dfrac{sin^3\alpha}{cos^3\alpha}+3+2\cdot\dfrac{sin\alpha}{cos^3\alpha}}=\dfrac{\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}-\dfrac{1}{cos^2\alpha}}{tan^3\alpha+3+2\cdot\dfrac{sin\alpha}{cos\alpha}\cdot\dfrac{1}{cos^2\alpha}}\)

    \(=\dfrac{tan\alpha\cdot\left(1+tan^2\alpha\right)-\left(1+tan^2\alpha\right)}{tan^3\alpha+3+2tan\alpha\cdot\left(1+tan^2\alpha\right)}\)

Thay \(tan\alpha=\sqrt{2}\) vào ta có:

\(P=\dfrac{\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]-\left[1+\left(\sqrt{2}\right)^2\right]}{\left(\sqrt{2}\right)^3+3+2\sqrt{2}\cdot\left[1+\left(\sqrt{2}\right)^2\right]}=\dfrac{3\sqrt{2}-3}{2\sqrt{2}+3+6\sqrt{2}}\)

    \(=\dfrac{3\left(\sqrt{2}-1\right)}{3+8\sqrt{2}}=\dfrac{3\left(\sqrt{2}-1\right)}{3+2^3\sqrt{2}}=\dfrac{a\left(\sqrt{b}-1\right)}{a+b^3\sqrt{b}}\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\Rightarrow a+b=3+2=5\)

Chọn đáp án A