1: cot x=-6 nên cosx/sinx=-6

=>cosx=-6*sinx

\(F=\dfrac{sinx-3\cdot cosx}{cosx+2\cdot sinx}=\dfrac{sinx+18\cdot sinx}{-6\cdot sinx+2\cdot sinx}=\dfrac{20}{-4}=-5\)

2: cotx=1

=>cosx/sinx=1

=>cosx=sinx

\(I=\dfrac{sin^3x-4\cdot sin^3x}{sinx+3sinx}=\dfrac{5\cdot sin^3x}{4\cdot sinx}=\dfrac{5}{4}\cdot sin^2x\)

\(1+cot^2x=\dfrac{1}{sin^2x}\)

=>\(\dfrac{1}{sin^2x}=1+1=2\)

=>sin^2=1/2

=>\(I=\dfrac{5}{4}\cdot\dfrac{1}{2}=\dfrac{5}{8}\)

3: cotx=3

=>cosx/sinx=3

=>cosx=3*sinx

1+cot^2x=1/sin^2x

=>\(\dfrac{1}{sin^2x}=1+9=10\)

=>\(sin^2x=\dfrac{1}{10}\)

\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)

\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)

\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)

hộ vs ae ơi

1B

2A

3A

4C

1: tan x=3 nên sin x/cosx=3

=>sin x=3*cosx

\(B=\dfrac{2\cdot sinx-3cosx}{sinx+cosx}=\dfrac{2\cdot3\cdot cosx-3cosx}{3cosx+cosx}\)

\(=\dfrac{2\cdot3-3}{3+1}=\dfrac{3}{4}\)

2: tan x=-1 nên sin x/cosx=-1

=>sinx=-cosx

\(I=\dfrac{4\cdot\left(-cosx\right)^3+\left(cosx\right)^3}{-cosx+3\cdot cosx}=\dfrac{-3\cdot cos^3x}{2cosx}=-\dfrac{3}{2}\cdot cos^2x\)

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+1=2\)

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

=>I=-3/2*1/2=-3/4

a: tan x=căn 3

=>sin x/cosx=căn 3

=>sin x=cosx*căn 3

\(A=\dfrac{\left(cosx\cdot\sqrt{3}\right)^2}{\left(cosx\cdot\sqrt{3}\right)^2-cos^2x}=\dfrac{3}{3-1}=\dfrac{3}{2}\)

b: cot x=-căn 3

=>cosx=-sinx*căn 3

\(A=\dfrac{sinx+4\cdot sinx\cdot\sqrt{3}}{2\cdot sinx+sinx\cdot\sqrt{3}}=\dfrac{1+4\sqrt{3}}{2+\sqrt{3}}=\left(4\sqrt{3}+1\right)\left(2-\sqrt{3}\right)\)

=8căn 3-12+2-căn 3

=7căn 3-10

Lời giải:

\(A=\frac{1}{\frac{\sin ^2x-\cos ^2x}{\sin ^2x}}=\frac{1}{1-(\frac{\cos x}{\sin x})^2}=\frac{1}{1-(\frac{1}{\tan x})^2}=\frac{1}{1-(\frac{1}{\sqrt{3}})^2}=\frac{3}{2}\)

\(A=\frac{\sin x-4\cos x}{2\sin x-\cos x}=\frac{1-4.\frac{\cos x}{\sin x}}{2-\frac{\cos x}{\sin x}}=\frac{1-4\cot x}{2-\cot x}=\frac{1-4.(-\sqrt{3})}{2-(-\sqrt{3})}=-10+7\sqrt{3}\)

đáp án không giống lắm 

2: \(\left(sinx+cosx\right)^2=1+2\cdot sinx\cdot cosx=1+2\cdot\dfrac{\sqrt{3}}{4}=1+\dfrac{\sqrt{3}}{2}=\dfrac{2+\sqrt{3}}{2}\)

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

mà sin x*cosx=căn 3/4

nên sinx,cosx là các nghiệm của phương trình là:

\(a^2-\dfrac{\sqrt{3}+1}{2}\cdot a+\dfrac{\sqrt{3}}{4}=0\)

=>\(\left[{}\begin{matrix}a=\dfrac{\sqrt{3}}{2}\\a=\dfrac{1}{2}\end{matrix}\right.\)

Ta sẽ có hai trường hợp:

TH1: sin x=căn 3/2; cosx=1/2

tan x=sinx/cosx=căn 3

cot x=1/căn 3

TH2: sin x=1/2; cosx=căn 3/2

tan x=sin x/cosx=1/căn 3

cot x=1:1/căn 3=căn 3

\(B=\frac{sinx+cosx}{2sinx+cosx}=\frac{\frac{sinx}{cosx}+\frac{cosx}{cosx}}{\frac{2sinx}{cosx}+\frac{cosx}{cosx}}=\frac{tanx+1}{2tanx+1}=\frac{3+1}{2.3+1}=...\)

\(C=\frac{\frac{4sin^3x}{cos^3x}+\frac{cos^3x}{cos^3x}}{\frac{sinx}{cos^3x}+\frac{3cosx}{cos^3x}}=\frac{4tan^3a+1}{tanx.\frac{1}{cos^2x}+3.\frac{1}{cos^2x}}=\frac{4tan^3x+1}{tanx\left(1+tan^2x\right)+3.\left(1+tan^2x\right)}\)

\(=\frac{4.3^3+1}{3\left(1+3^2\right)+3\left(1+3^2\right)}=...\)

a/

\(\left(\frac{sin2x}{cos2x}-\frac{sinx}{cosx}\right)cos2x=\left(\frac{sin2x.cosx-cos2x.sinx}{cos2x.cosx}\right).cos2x\)

\(=\frac{sin\left(2x-x\right)}{cosx}=\frac{sinx}{cosx}=tanx\)

b/

\(2\left(1-sinx\right)\left(1+cosx\right)=2+2cosx-2sinx-2sinxcosx\)

\(=1+sin^2x+cos^2x-2sinx+2cosx-2sinx.cosx\)

\(=\left(1-sinx+cosx\right)^2\)

c/

\(1+cotx+cot^2x+cot^3x=1+cotx+cot^2x\left(1+cotx\right)\)

\(=\left(1+cotx\right)\left(1+cot^2x\right)=\left(1+\frac{cosx}{sinx}\right)\left(1+\frac{cos^2x}{sin^2x}\right)=\frac{sinx+cosx}{sin^3x}\)

d/

\(\frac{cos3x}{sinx}+\frac{sin3x}{cosx}=\frac{cos3x.cosx+sin3x.sinx}{sinx.cosx}=\frac{cos\left(3x-x\right)}{\frac{1}{2}2sinx.cosx}=\frac{2cos2x}{sin2x}=2cot2x\)

xét vế phải

( cosa+1-sina)^2

= cos^2 +1+ sin^2+2cosa-2sina-2sinacosa

= 2( 1+ cosa-sina-sinacosa)

= 2( 1-sina) ( 1+cosa)