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

\(\tan x=\frac{\sin x}{\cos x}=\frac{3}{5}\Rightarrow\sin x=\frac{3}{5}\cos x\)

\(\Rightarrow N=\frac{\sin x.\cos x}{\sin^2x-\cos^2x}=\frac{\sin x.\cos x}{\left(\sin x-\cos x\right)\left(\sin x+\cos x\right)}\)

\(=\frac{\frac{3}{5}.\cos^2x}{\left(\frac{3}{5}\cos x-\cos x\right)\left(\frac{3}{5}\cos x+\cos x\right)}=\frac{\frac{3}{5}\cos^2x}{\frac{-16}{25}.\cos^2x}=\frac{-15}{16}\)

a) sin = đối / huyền => sinx < 1 => sinx - 1 < 0

b) cos = kề / huyền => cosx < 1 => 1 - cosx > 0

c) sinx - cosx = sinx - sin(90-x)

Nếu x > 90-x hay x > 45 thì sinx - sin(90-x) > 0 hay sinx - cosx > 0

Nếu x < 90-x hay x < 45 thì sinx - sin(90-x) < 0 hay sinx - cosx < 0

d) Tương tự câu c)

xin lỗi mik mới lớp 8 thui  kg jup dc j ròi

giup minh giai toan voi

1B

2A

3A

4C

a/ ĐKXĐ:

\(sin\left(\frac{\pi}{2}.sinx\right)\ne0\Rightarrow\frac{\pi}{2}.sinx\ne k\pi\)

\(\Rightarrow sinx\ne2k\)

Mà \(-1\le sinx\le1\Rightarrow sinx\ne0\Rightarrow x\ne k\pi\)

b/

\(sinx-1\ge0\Leftrightarrow sinx\ge1\Rightarrow sinx=1\)

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

c/

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow sin4x\ne0\)

\(\Rightarrow x\ne\frac{k\pi}{4}\)

d/

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\sinx+cotx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\sin^2x+cosx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x\ne k\pi\\-cos^2x+cosx+1\ne0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\cosx\ne\frac{1-\sqrt{5}}{2}\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\pm arccos\left(\frac{1-\sqrt{5}}{2}\right)+k2\pi\end{matrix}\right.\)

e/

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\Rightarrow x\ne k\pi\)

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:

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

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

Sửa đề: sin x=3/5

Ta có: \(\sin^2x+cos^2x=1\)

=>\(cos^2x=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}\)

=>\(cosx=\sqrt{\frac{16}{25}}=\frac45\)

Ta có: \(\tan x=\frac{\sin x}{cosx}\)

\(=\frac35:\frac45=\frac34\)

Ta có: \(\cot x=\frac{1}{\tan x}\)

=>\(\cot x=1:\frac34=\frac43\)

\(cosx=\sqrt{1-\dfrac{7}{16}}=\dfrac{3}{4}\)

\(tanx=\dfrac{\sqrt{7}}{4}:\dfrac{3}{4}=\dfrac{\sqrt{7}}{3}\)

\(cotx=1:\dfrac{\sqrt{7}}{3}=\dfrac{3}{\sqrt{7}}=\dfrac{3\sqrt{7}}{7}\)

\(M=\left(\dfrac{3}{7}\sqrt{7}+\dfrac{1}{3}\sqrt{7}\right):\left(\dfrac{3}{7}\sqrt{7}-\dfrac{1}{3}\sqrt{7}\right)\)

\(=\dfrac{16}{21}:\dfrac{2}{21}=8\)