Cho cotx = -3
tính : A= \(\dfrac{2sin^2x+3sinx.cosx}{sin^2x-7}\)
K
Khách
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TM
1
Những câu hỏi liên quan
22 tháng 6 2023
b:
3/2pi<x<2pi
=>cosx>0; sin x<0
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+\left(-3\right)^2=10\)
=>cosx=1/căn 10
=>sin x=-3/căn 10
\(A=\sqrt{10}\cdot\dfrac{1}{\sqrt{10}}-2\cdot\dfrac{-3}{\sqrt{10}}+3=4+\dfrac{6}{\sqrt{10}}=\dfrac{4\sqrt{10}+6}{\sqrt{10}}\)
a: cot x=3 nên cosx/sinx=3
=>cosx=3*sinx
\(B=\dfrac{2sin^2x+3sinx\cdot3\cdot sinx}{1-2\cdot\left(3\cdot sinx\right)^2}=\dfrac{11sin^2x}{sin^2x+cos^2x-18sin^2x}\)
\(=\dfrac{11sin^2x}{-17sin^2x+9sin^2x}=\dfrac{-11}{8}\)
HQ
Hà Quang Minh
Giáo viên
31 tháng 7 2023
a, \(A=\dfrac{3sin^2\left(x\right)-cos^2\left(x\right)}{2sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\dfrac{cos^2\left(x\right)}{sin^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\dfrac{1}{tan^2\left(x\right)}=\dfrac{3}{2}-\dfrac{1}{2}\cdot\left(-\dfrac{3}{2}\right)^2=-3\)
b, \(A=\dfrac{sin^2\left(x\right)-5cos^2\left(x\right)}{2cos^2\left(x\right)}=\dfrac{1}{2}\dfrac{sin^2\left(x\right)}{cos^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\dfrac{1}{cot^2\left(x\right)}-\dfrac{5}{2}=\dfrac{1}{2}\cdot\left(\dfrac{5}{3}\right)^2-\dfrac{5}{2}=\dfrac{55}{18}\)
AH
Akai Haruma
Giáo viên
31 tháng 7 2023
Lời giải:
a.
\(A=\frac{3}{2}-2(\frac{\cos x}{\sin x})^2=\frac{3}{2}-2.(\frac{1}{\tan x})^2=\frac{3}{2}-\frac{1}{2}(\frac{-3}{2})^2=-3\)
b.
\(A=\frac{1}{2}(\frac{\sin x}{\cos x})^2-\frac{5}{2}=2(\frac{1}{\cot x})^2-\frac{5}{2}=2(\frac{5}{3})^2-\frac{5}{2}=\frac{55}{18}\)
31 tháng 7 2023
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
AH
Akai Haruma
Giáo viên
31 tháng 7 2023
Lời giải:
\(A=\frac{1}{\frac{\sin ^2x-\cos ^2x}{\sin ^2x}}=\frac{1}{1-(\frac{\cos x}{\sin x})^2}=\frac{1}{1-(\frac{1}{\tan x})^2}=\frac{1}{1-(\frac{1}{\sqrt{3}})^2}=\frac{3}{2}\)
\(A=\frac{\sin x-4\cos x}{2\sin x-\cos x}=\frac{1-4.\frac{\cos x}{\sin x}}{2-\frac{\cos x}{\sin x}}=\frac{1-4\cot x}{2-\cot x}=\frac{1-4.(-\sqrt{3})}{2-(-\sqrt{3})}=-10+7\sqrt{3}\)
17 tháng 8 2023
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
=>\(\dfrac{1}{sin^2x}=1+9=10\)
=>\(sin^2x=\dfrac{1}{10}\)
\(I=\dfrac{2\cdot sin^3x+cos^3x}{4\cdot sinx-6\cdot cosx}\)
\(=\dfrac{2\cdot sin^3x+\left(3\cdot sinx\right)^3}{4\cdot sinx-6\cdot\left(3\cdot sinx\right)}=\dfrac{2\cdot sin^3x+27\cdot sin^3x}{4\cdot sinx-18\cdot sinx}\)
\(=\dfrac{29}{-14}\cdot sin^2x=\dfrac{-29}{14}\cdot\dfrac{1}{10}=-\dfrac{29}{140}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
\(a=\int\dfrac{1}{2tan^2x+5tanx+2}.\dfrac{dx}{cos^2x}\)
Đặt \(tanx=t\Rightarrow dt=\dfrac{dx}{cos^2x}\)
\(I=\int\dfrac{dt}{2t^2+5t+2}=\int\dfrac{dt}{\left(t+2\right)\left(2t+1\right)}=\dfrac{2}{3}\int\left(\dfrac{1}{2t+1}-\dfrac{1}{2t+4}\right)dt\)
\(=\dfrac{1}{3}ln\left|\dfrac{2t+1}{2t+4}\right|+C=\dfrac{1}{3}ln\left|\dfrac{2tanx+1}{2tanx+4}\right|+C\)
Câu b hoàn toàn tương tự
NH
\(\dfrac{2sin^3x+2\sqrt{3}sin^2x.cosx-2sin^2x+cos\left(2x+\dfrac{\pi}{3}\right)}{2cosx-\sqrt{3}}=0\)
0
HP
6 tháng 8 2021
\(M=sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
\(=sinx.cosx+\dfrac{sin^2x}{\dfrac{cosx+sinx}{sinx}}+\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}\)
\(=sinx.cosx+\dfrac{sin^3x+cos^3x}{cosx+sinx}\)
\(=sinx.cosx+\dfrac{\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{cosx+sinx}\)
\(=sinx.cosx+sin^2x+cos^2x-sinx.cosx\)
\(=sin^2x+cos^2x=1\)
1 tháng 8 2023
a: tan x(cot^2x-1)
\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)
=cotx-tanx/cotx=cotx(1-tan^2x)
b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)
\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)
c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)
\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)
\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)
=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)
=-cos^2x*cos^2x=-cos^4x
=>ĐPCM
`cot x = -3 => cos x = -3 sin x`
`=> A = [ 2 sin^2 x + 3 sin x . (-3 sin x ) ] / [ sin^2 x - 7 ( sin^2 x + cos^2 x ) ]`
`<=>A = [ -7 sin^2 x ] / [ sin^2 x - 7 ( sin^2 x + 9 sin^2 x ) ]`
`<=>A = [ -7 sin^2 x ] / [ -69 sin^2 x ]`
`<=> A = 7 / 69`