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Lời giải:

Ta có:

\(x^2=\frac{1}{4}(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}})^2=\frac{1}{4}(\frac{a}{b}+\frac{b}{a}+2)\)

\(\Rightarrow x^2-1=\frac{1}{4}(\frac{a}{b}+\frac{b}{a}-2)=\frac{1}{4}(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})^2\)

\(\Rightarrow \sqrt{x^2-1}=\frac{1}{2}|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}|=\frac{1}{2}(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})\)

Do đó:

\(2b\sqrt{x^2-1}=b.\frac{a-b}{\sqrt{ab}}=(a-b).\sqrt{\frac{b}{a}}\)

\(x-\sqrt{x^2-1}=\frac{1}{2}[\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}})]\)

\(=\sqrt{\frac{b}{a}}\)

$\Rightarrow B=a-b$

`S=[cos(a+\pi/3)+cos(a-\pi/3)]/[cot a-cot (a/2)]`

`S=[2.cos a.cos(\pi/3)]/[[cos a]/[sin a]-[cos (a/2)]/[sin (a/2)]]`

`S=[2 .cos a. 1/2]/[[cos a.sin a/2 -sin a.cos (a/2)]/[sin a.sin (a/2)]`

`S=[cos a.sin a.sin (a/2)]/[1/2[sin(3/2 a)+sin (-1/2a)-sin (3/2 a)-sin (1/2a)]]`

`S=[2.cos a.sin a.sin (a/2)]/[sin (-1/2 a)-sin(1/2 a)]`

`S=[sin 2a.sin(a/2)]/[-sin(1/2a)-sin (1/2a)]`

`S=-1/2 sin 2a`.

a) \(A=sin\left(90^0-x\right)+cos\left(180^0-x\right)+sin^2x\left(1+tan^2x\right)-tan^2x\)

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

\(=tan^2x-tan^2x\)

\(=0\)

b) \(B=\dfrac{1}{sinx}.\sqrt{\dfrac{1}{1+cosx}+\dfrac{1}{1-cosx}}-\sqrt{2}\)

\(=\dfrac{1}{sinx}.\sqrt{\dfrac{1-cosx+1+cosx}{1-cos^2x}}-\sqrt{2}\)

\(=\dfrac{1}{sinx}.\sqrt{\dfrac{2}{sin^2x}}-\sqrt{2}\)

\(=\dfrac{\sqrt{2}}{sin^2x}-\sqrt{2}\)

\(=\dfrac{\sqrt{2}\left(1-sin^2x\right)}{sin^2x}\)

\(=\dfrac{\sqrt{2}cos^2x}{sin^2x}\)

\(=\sqrt{2}tan^2x\)

\(P=\frac{1-sin^2x.cos^2x}{cos^2x}-cos^2x=\frac{1}{cos^2x}-sin^2x-cos^2x\)

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

\(M=\frac{2cos^2x-1}{sinx+cosx}=\frac{2cos^2x-\left(sin^2x+cos^2x\right)}{sinx+cosx}=\frac{cos^2x-sin^2x}{sinx+cosx}\)

\(\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}=cosx-sinx\)

Chọn B.

Ta có: 

Chọn D.

Ta có 

Chọn B.

Ta có: 

Chọn B.

Ta có 

Chọn D.

Ta có: 

Chọn A.

Ta có:

= 1 + a – (1 – b) = a + b