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Giúp mik giải cho cos x=1/2.Tìm biểu thức P=3sin^2x+1
A 13/4 B 7/4 C 11/4 D 15/4
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`sin^2x+cos^2x=1`
`<=>sin^2x+(1/2)^2=1`
`<=> sinx=\pm \sqrt3/2`
• `sinx=\sqrt3/2 => P=3. (\sqrt3/2)^2 +1=13/4`
• `sinx=-\sqrt3/2 => P = 3.(-\sqrt3/2) +1=13/4`
`=>` A.
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\(P=3sin^2x+1=3\left(1-cos^2x\right)+1=3\left(1-\dfrac{1}{4}\right)+1=\dfrac{13}{4}\)
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1.thực hiện phép tính:
a,5/13 + -5/7 + -20/41 + 8/13 + -21/41
b,-5/7 . 2/11 + -5/7 . 9/11 +1
c, (4/5 + 1/2) . (3/13 - 8/13)
2. a, x + 4/7=7/11
b, -5/6 - x = 7/12 + -1/3
c,x - 1/4 = 5/8 . 2/3
d,x/126 = -5/9 . 4/7
giúp mik vs mn ơi, mik đnag cần rất gấp
Giải PT
a1) \(3.\cos4x-2^{ }\cos^23x=1\)
a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)
a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)
a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
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a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
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a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
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Tính
\(A=\dfrac{2\cos^2x-8\sin x.\cos x-sin^2x}{sin^2x+3cos^2x-4}với\cot x=2\)
Giúp mình Cho cos x=4/5. Khi đó sin(pi/2-x)+1 bằng A 9/5 B 6/5 C1/5 D 8/5
\(sin\left(\dfrac{\pi}{2}-x\right)+1=cosx+1=\dfrac{4}{5}+1=\dfrac{9}{5}\)
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1. tinh :
a) 15/8 + 3/4 - 5/12
b) 11/24 x 12/33 + 5/6
c) 15/8 + 7/24 : 5/8
2 . tim x, biet :
a) x + 5/8 : 3/4 = 17/9
b) x - 5/9 x 18/15 = 1/3
c) x x7/8 - 5/8 = 13/24
d) X x1/7 + X x 1/4 + X x 1/4 =18
giup mik nhanh nha. mik can gap lam
a)15/8+3/4-5/12
=45+18-10/24
=53/24
b)11/24.12/33+5/6
=11.12/12.2.11.3+5/6
=1/6+5/6
=6/6=1
c)15/8+7/24:5/8
=15/8+7/24.8/5
=15/8+7.8/3.8.5
=15/8+7/15
=đề sai, nếu đúng thì như này
=8/15+7/15
=15/15=1
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Tính:
a) 2^5 x 13^12 x 7^8 : 27 x 13^10 x 7^9
b) 2^n x 11 x 3 : 6^n x 3
c) 5^5 x 7 x 9^5 : 15^10
d) 7 x 8^10 x 27^9 - 8 x 9^13 x 2^27 : 7 x 6^26 x 2^2 - 2^4 x 8^3 x 27^9
e) 2^2^1^2015
f) 2015^0^2015^0^2015
72-8/3 bằng bao nhiêu số tự nhiên
A x/5=2/5
B 3/8=6/x
C 1/9=x/27
D 4/x=8/6
E x+2/3=12/4
G 14/13=28/10-x
H 3/x-5=4/x+2
K x/2=8/x
M x-2/50=2/x-2
GIẢI NHANH GIÚP MIK VỚI Ạ MIK ĐANG CẦN GẤP
A. x = 2
B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)
C. x = 3
D. \(x=\dfrac{4.6}{8}=3\)
E. \(x=\dfrac{7}{3}\)
G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)
<=>\(14\left(10-x\right)=364\)
<=> 10 - x = 26
<=> x = -16
H. \(3\left(x+2\right)=4\left(x-5\right)\)
<=> 3x + 6 = 4x - 20
<=> -x = -26
<=> x = 26
K. \(\dfrac{x}{2}=\dfrac{8}{x}\)
<=> \(x^2=16\)
<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
M. \(\left(x-2\right)^2=100\)
<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
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a=2
b=16
c=3
d=3
mik chỉ biết thế này thôi(ko chắc đúng=3)
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A, \(\dfrac{x}{5}=\dfrac{2}{5}\Rightarrow x=2\)
B,\(\dfrac{3}{8}=\dfrac{6}{x};\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16}\Rightarrow x=16\)
C,\(\dfrac{1}{9}=\dfrac{x}{27};\dfrac{1}{9}=\dfrac{1\cdot3}{9\cdot3}=\dfrac{3}{27}\Rightarrow x=3\)
D,\(\dfrac{4}{x}=\dfrac{8}{6};\dfrac{4}{x}=\dfrac{4\cdot2}{x\cdot2}=\dfrac{8}{2x}\)
\(\Rightarrow2x=6;6=x:2;x=3\)
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a, 2/9 - 7/9 x X = 1/3
b, 4/5 + 5/7 : X = 1/6
c,1/3 + 3/8 - 7/12
d, 3/14 + 5/8 - 1/2
e,1/4 - 2/3 - 11/8
g,1/4 + 5/12 - 1/13 - 7/8
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