\(\sin^3\frac{x}{3} 3\sin^3\frac{x}{3^2} ... 3^{n-1}\sin^3\frac{x}{3}=\frac{1}{4}\left(3^n\sin^3\frac{x}{3^n}-\sin x\right)\)\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{2n 1}{2n 2}

#Toán lớp 9

0

NA

Nguyen ANhh

16 tháng 7 2020

Giải các pt lượng giác sau

1) \(cos^2\left(x-\frac{\pi}{6}\right)-sin^2\left(x-\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\)

2) \(sin^4-sin^4\left(x+\frac{\pi}{2}\right)=sin\left(x+\frac{\pi}{3}\right)\)

3) \(8cos^3\left(x-\frac{\pi}{3}\right)-1=0\)

3

NV

Nguyễn Việt Lâm

Giáo viên

16 tháng 7 2020

c/

\(\Leftrightarrow cos^3\left(x-\frac{\pi}{3}\right)=\frac{1}{8}\)

\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(x-\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=k2\pi\end{matrix}\right.\)

NV

Nguyễn Việt Lâm

Giáo viên

16 tháng 7 2020

a/

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{6}-x\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{6}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{6}+\frac{k2\pi}{3}\)

b/

\(\Rightarrow sin^4x-cos^4x=sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow-cos2x=sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow cos2x=-sin\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{5\pi}{6}\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x=x+\frac{5\pi}{6}+k2\pi\\2x=-x-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

giải phương trình \(\sin x\sqrt{1+2\sin x}=\cos2x\) \(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\) \(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\) \(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\) \(\sin2x\sin4x+2\left(3\sin...

QN

Quỳnh Nguyễn Thị Ngọc

11 tháng 2 2020

Giải các phương trình : 1) \(\frac{\sin^4x+\cos^4x}{\sin2x}=\frac{1}{2}\left(\tan x+\cot2x\right)\) 2) \(\frac{1}{\sin...

1

NV

Nguyễn Việt Lâm

Giáo viên

11 tháng 2 2020

a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp

b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)

\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)

\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)

\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)

c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:

\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)

Đặt \(\sqrt{tanx+1}=t\ge0\)

\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)

\(\Leftrightarrow3t^3-5t^2+3t-10=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)

d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)

Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)

\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)

\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)

Đúng(1)

TT

thai thai

21 tháng 5 2019

Giải phương trình:a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)d) \(2\cos 3x + 5 = 3\)e) \(3\tan x = - \sqrt 3 \)g) \(\cot x - 3 = \sqrt 3 \left( {1 - \cot x}...

0

QT

Quoc Tran Anh Le

Giáo viên

21 tháng 9 2023

3

HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)

b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)

\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

d) \(2\cos 3x + 5 = 3\)

\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

LN

lu nguyễn

15 tháng 4 2019

rút gọn biểu thức:

A= cosa.sin( b-c )+ cosb. sin(c-a) + cosc.sin( a-b)

B= \(sin^2x+cos\left(\frac{\pi}{3}-x\right).cos\left(\frac{\pi}{3}+x\right)\)

C=\(sin^2x+sin^2\left(\frac{2\pi}{3}+x\right)+sin^2\left(\frac{2\pi}{3}-x\right)\)

D=\(sin^2\left(\frac{\pi}{4}+x\right)-sin^2x-2sinx.sin\frac{\pi}{4}.cos\left(\frac{\pi}{4}+x\right)\)

#Toán lớp 10

2

NV

Nguyễn Việt Lâm

Giáo viên

17 tháng 4 2019

\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)

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

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

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

NA

Ngoc Anh Nguyen

30 tháng 4 2019

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là

JE

Julian Edward

16 tháng 9 2020

GPT

a) \(sin\left(2x+1\right)+cos\left(3x-1\right)=0\)

b) \(sin\left(2x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{4}\right)\)

c) \(sin\left(3x+\frac{2\pi}{3}\right)+sin\left(x-\frac{7\pi}{5}\right)=0\)

d) \(cos\left(4x+\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{4}\right)=0\)

2

NV

Nguyễn Việt Lâm

Giáo viên

16 tháng 9 2020

c.

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)

\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)

d.

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)

NV

Nguyễn Việt Lâm

Giáo viên

16 tháng 9 2020

a.

\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)

\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

b.

\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)

QT

Quoc Tran Anh Le

Giáo viên

21 tháng 9 2023

Cho \(\cos 2x = \frac{1}{4}\).

Tính: \(A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right)\); \(B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right)\)

1

HQ

Hà Quang Minh

Giáo viên

21 tháng 9 2023

\(\begin{array}{l}A = \cos \left( {x + \frac{\pi }{6}} \right)\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{6} + x - \frac{\pi }{6}} \right) + \cos \left( {x + \frac{\pi }{6} - x + \frac{\pi }{6}} \right)} \right]\\A = \frac{1}{2}\left[ {\cos 2x + \cos \frac{\pi }{3}} \right] = \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = \frac{3}{8}\end{array}\)

\(\begin{array}{l}B = \sin \left( {x + \frac{\pi }{3}} \right)\sin \left( {x - \frac{\pi }{3}} \right) = - \frac{1}{2}\left[ {\cos \left( {x + \frac{\pi }{3} + x - \frac{\pi }{3}} \right) - \cos \left( {x + \frac{\pi }{3} - x + \frac{\pi }{3}} \right)} \right]\\B = - \frac{1}{2}\left( {\cos 2x - \cos \frac{{2\pi }}{3}} \right) = - \frac{1}{2}\left( {\frac{1}{4} + \frac{1}{2}} \right) = - \frac{3}{8}\end{array}\)

Giair các pt lượng giác sau: 1) \(sin\left(x-\frac{\pi}{4}\right)\left(2cos+\sqrt{2}\right)tan2x=0\) 2) \(tan2x.sinx+3\left(sin-\sqrt{3}tan2x\right)-3\sqrt{3}=0\) 3) \(\frac{cos2x}{sin\left(x+\frac{3\pi}{4}\right)}=\frac{sin\left(x+\frac{3\pi}{4}\right)}{cos2x}\) 4) \(\left(\frac{tanx-1}{tanx+1}+cot2x\right)\left(3tan-\sqrt{3}\right)=0;0 x...

NA

Nguyen ANhh

16 tháng 7 2020

4

TQ

Trần Quốc Lộc

16 tháng 7 2020

\(\frac{tanx-1}{tanx+1}+cot2x=0\\ \Leftrightarrow cot2x-\frac{1-tanx\cdot tan\frac{\pi}{4}}{tanx+tan\frac{\pi}{4}}=0\\ \Leftrightarrow cot2x-cot\left(x+\frac{\pi}{4}\right)=0\)

NV

Nguyễn Việt Lâm

Giáo viên

16 tháng 7 2020

d/

ĐKXĐ: \(\left\{{}\begin{matrix}sin2x\ne0\\tanx\ne-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}+cot2x=0\\3tanx-\sqrt{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{tanx-1}{tanx+1}-\frac{tan^2x-1}{2tanx}=0\\tanx=\frac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(tanx-1\right)\left(\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}\right)=0\left(1\right)\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Xét (1): \(\Leftrightarrow\left[{}\begin{matrix}tanx=1\Rightarrow x=\frac{\pi}{4}+k\pi\\\frac{1}{tanx+1}-\frac{tanx+1}{2tanx}=0\left(2\right)\end{matrix}\right.\)

Xét (2)

\(\Leftrightarrow\left(tanx+1\right)^2-2tanx=0\)

\(\Leftrightarrow tan^2x+1=0\left(vn\right)\)

QN

Quỳnh Nguyễn Thị Ngọc

23 tháng 9 2019

giải phương trình

a, \(2\sin\frac{x}{2}\left(\sin\frac{3x}{2}+\cos\frac{3x}{2}\right)=3-4\cos x\)

b, \(\frac{2\cos^2x+\sqrt{3}\sin2x+3}{2\cos^2x.\sin\left(x+\frac{\pi}{3}\right)}=3\left(\tan^2x+1\right)\)

1

NV

Nguyễn Việt Lâm

Giáo viên

24 tháng 9 2019

a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)

\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)

\(\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0\)

\(\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)

\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)

\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)

\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)