\(b)\left\{{}\begin{matrix}x-\dfrac{y}{2}=\dfrac{1}{2}\\\dfrac{x}{3}-2y=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-6y=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\2x-12y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11y=11\\2x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\2x=1+1=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

\(c)\left\{{}\begin{matrix}5x-0,7y=1\\-10x+1,4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\5x-0,7y=1\end{matrix}\right.\) 

=> Hpt vô số nghiệm 

\(1.\left\{{}\begin{matrix}-x+3y=-10\\x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2y=6\\x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{6}{-2}=-3\\x-5\cdot\left(-3\right)=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x+15=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=16-15=1\end{matrix}\right.\\ 2.\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\-2x+8y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\9x=27\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=7\\x=\dfrac{27}{9}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+y=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=7-6=1\\x=3\end{matrix}\right.\) 

\(3.\left\{{}\begin{matrix}3x-5y=-18\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-5y=-18\\3x+6y=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x+2\cdot\left(-3\right)=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=5+6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=11\end{matrix}\right.\\ 4.\left\{{}\begin{matrix}4x+3y=-6\\2x-5y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=-6\\4x-10y=32\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=-38\\2x-5y=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x-5\cdot\dfrac{-38}{13}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x+\dfrac{190}{13}=16\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\2x=16-\dfrac{190}{13}=\dfrac{18}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-38}{13}\\x=\dfrac{18}{13}:2=\dfrac{9}{13}\end{matrix}\right.\)

20: \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+14y-2x-y=18-5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13y=13\\2x=5-y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=1\\2x=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

21: \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\9x-3y=-24\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x+3y+9x-3y=-7-24\\3x-y=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{31}{14}\\y=3\cdot\dfrac{-31}{14}+8=-\dfrac{93}{14}+\dfrac{112}{14}=\dfrac{19}{14}\end{matrix}\right.\)

22: \(\left\{{}\begin{matrix}-2x+y=-3\\3x+4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3+2x\\3x+4\left(2x-3\right)=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x-3\\3x+8x-12=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\y=2x-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=2\cdot2-3=4-3=1\end{matrix}\right.\)

23: \(\left\{{}\begin{matrix}x+y=2\\x+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y-x-y=6-2\\x+y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2y=4\\x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=2-y=2-2=0\end{matrix}\right.\)

24: \(\left\{{}\begin{matrix}x-2y=-5\\3x+4y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-10\\3x+4y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y+3x+4y=-10-5\\x-2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=-15\\2y=x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\2y=-3+5=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

25: \(\left\{{}\begin{matrix}3x-2y=12\\4x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=12\\8x+2y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-2y+8x+2y=12+10\\4x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11x=22\\y=5-4x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=5-4\cdot2=5-8=-3\end{matrix}\right.\)

26: \(\left\{{}\begin{matrix}2x-y=10\\5x+2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y=20\\5x+2y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-2y+5x+2y=20+6\\2x-y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=26\\y=2x-10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{26}{9}\\y=2x-10=2\cdot\dfrac{26}{9}-10=\dfrac{52}{9}-10=-\dfrac{38}{9}\end{matrix}\right.\)

27: \(\left\{{}\begin{matrix}5x-2y=10\\5x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-2y-5x+2y=10-6\\5x-2y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0x=4\\2y=5x-6\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)

28: \(\left\{{}\begin{matrix}3x+2y=8\\4x-3y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+8y=32\\12x-9y=-36\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x+8y-12x+9y=32+36\\3x+2y=8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}17y=68\\3x=8-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\3x=8-2\cdot4=8-8=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\)

37: \(\left\{{}\begin{matrix}2x+y=4\\2x+0y-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\y=4-2x=4-6=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

38: \(\left\{{}\begin{matrix}x-2y=2\\2x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=4\\2x-4y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y-2x+4y=4-1\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=3\\x-2y=2\end{matrix}\right.\)

=>\(\left(x;y\right)\in\varnothing\)

39: \(\left\{{}\begin{matrix}3x+2y-2=0\\9x+6y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=2\\9x+6y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x+6y=6\\9x+6y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0y=2\\3x+2y=2\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)

40: \(\left\{{}\begin{matrix}2x-y=2\\4x-2y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=2\\4x-2y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-y=2\\2x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0x=0\\y=2x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in R\\y=2x-2\end{matrix}\right.\)

41: \(\left\{{}\begin{matrix}x+2y=4\\2x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=8\\2x+9y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+9y-2x-4y=18-8\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=10\\x=4-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2\\x=4-2\cdot2=4-4=0\end{matrix}\right.\)

42: \(\left\{{}\begin{matrix}-2x+y=-3\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x+y-x-y=-3-3\\x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3x=-6\\y=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3-2=1\end{matrix}\right.\)

43: \(\left\{{}\begin{matrix}x-y=0\\2x+y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y+2x+y=0-5\\x=y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=-5\\y=x\end{matrix}\right.\Leftrightarrow y=x=-\dfrac{5}{3}\)

44: \(\left\{{}\begin{matrix}2x+y=0\\x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2x\\x-4\cdot\left(-2x\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9x=0\\y=-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-2\cdot0=0\end{matrix}\right.\)

45: \(\left\{{}\begin{matrix}-x+y=3\\x+2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y+x+2y=3+3\\x+2y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3y=6\\x=3-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=3-2\cdot2=3-4=-1\end{matrix}\right.\)

46: \(\left\{{}\begin{matrix}x-y=2\\3x-2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-3y=6\\3x-2y=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-3y-2x+2y=6-9\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y=-3\\x=y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=3+2=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x-4y=-2\\5x+2y=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-4y=-2\\2y=14-5x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-2\cdot2y=-2\\2y=14-5x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2\left(14-5x\right)=-2\\2y=14-5x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-28+10x=-2\\2y=-5x+14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=-2+28=26\\2y=-5x+14\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\2y=-5\cdot2+14=14-10=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

\(6)\left(2x+1\right)^2-\left(x+3\right)^2=0\\ \Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\\ 7)\left(x^2-4\right)+x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)+x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2+x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ 8)2\left(x+1\right)=\left(5x-1\right)\left(x+1\right)\\ \Leftrightarrow2\left(x+1\right)-\left(5x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(2-5x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(3-5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{5}\end{matrix}\right.\\ 9)\left(-4x+3\right)x=\left(2x+5\right)x\\ \Leftrightarrow\left(-4x+3\right)x-\left(2x+5\right)x=0\\ \Leftrightarrow x\left(-4x+3-2x-5\right)=0\\ \Leftrightarrow x\left(-6x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Bài 4:

\(a)x-7< 2-x\\ \Leftrightarrow x+x< 2+7\\ \Leftrightarrow2x< 9\\ \Leftrightarrow x< \dfrac{9}{2}\\ b)x+2\le2+3x\\ \Leftrightarrow3x-x\ge2-2=0\\ \Leftrightarrow2x\ge0\\ \Leftrightarrow x\ge0\\ c)4+x>5-3x\\ \Leftrightarrow x+3x>5-4\\ \Leftrightarrow4x>1\\ \Leftrightarrow x>\dfrac{1}{4}\\ d)-x+7\ge x-3\\ \Leftrightarrow x+x\le7+3\\ \Leftrightarrow2x\le10\\ \Leftrightarrow x\le\dfrac{10}{2}=5\)

\(\dfrac{3}{x}+\dfrac{4}{x-1}=\dfrac{5x-2}{x^2-x}\left(x\notin\left\{0;1\right\}\right)\\ \Leftrightarrow\dfrac{3\left(x-1\right)}{x\left(x-1\right)}+\dfrac{4x}{x\left(x-1\right)}=\dfrac{5x-1}{x\left(x-1\right)}\\ \Leftrightarrow3\left(x-1\right)+4x=5x-1\\ \Leftrightarrow3x-3+4x=5x-1\\ \Leftrightarrow7x-3=5x+1\\ \Leftrightarrow7x-5x=1+3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=\dfrac{4}{2}\\ \Leftrightarrow x=2\left(tm\right)\)

a: Ta có: ΔOCK vuông tại C

=>\(CK^2+CO^2=OK^2\)

=>\(CK=\sqrt{10^2-8^2}=6\left(cm\right)\)

Xét ΔOCK vuông tại C có CA là đường cao

nên \(OA\cdot OK=OC^2;CA\cdot OK=CO\cdot CK\)

=>\(OA=\dfrac{OC^2}{OK}=\dfrac{8^2}{10}=6,4\left(cm\right);CA=\dfrac{6\cdot8}{10}=4,8\left(cm\right)\)

Xét ΔCOK vuông tại C có \(sinCOK=\dfrac{CK}{OK}=\dfrac{6}{10}=\dfrac{3}{5}\)

nên \(\widehat{COK}=\widehat{xOy}\simeq36^052'\)

b: Xét ΔCAO vuông tại A có AH là đường cao

nên \(CH\cdot CO=CA^2\left(1\right)\)

Xét ΔCOK vuông tại O có CA là đường cao

nên \(AO\cdot AK=AC^2\left(2\right)\)

Từ (1),(2) suy ra \(CH\cdot CO=AO\cdot AK\)

Xét đường tròn (O) có tiếp tuyến MB tại B nên 

\(\widehat{MBI}=\dfrac{1}{2}sđ\stackrel\frown{IB}\)

Lại có \(\widehat{IBH}=90^o-\widehat{BIH}\)

\(=90^o-\widehat{OIB}\)

\(=90^o-\dfrac{180^o-\widehat{IOB}}{2}\)

\(=\dfrac{180^o-180^o+sđ\stackrel\frown{IB}}{2}\)

\(=\dfrac{1}{2}sđ\stackrel\frown{IB}\)

Do đó \(\widehat{MBI}=\widehat{IBH}\) hay BI là tia phân giác của \(\widehat{MBH}\)

\(\Rightarrow d\left(I,MB\right)=d\left(I,BH\right)=IH=R_I\)

Suy ra MB là tiếp tuyến của (I)

\(\left\{{}\begin{matrix}x-ay=a\\ax+y=1\end{matrix}\right.\)

Để hpt có nghiệm thì: \(\dfrac{1}{a}\ne\dfrac{-a}{1}\Leftrightarrow a^2\ne-1\) (luôn đúng) 

\(\Leftrightarrow\left\{{}\begin{matrix}x-ay=a\\a^2x+ay=a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a^2+1\right)x=2a\\ax+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2a}{a^2+1}\\\dfrac{2a^2}{a^2+1}+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2a}{a^2+1}\\y=1-\dfrac{2a^2}{a^2+1}=\dfrac{1-a^2}{a^2+1}\end{matrix}\right.\) 

Ta có: \(\left\{{}\begin{matrix}\text{x}>0\\y>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2a}{a^2+1}>0\\\dfrac{1-a^2}{a^2+1}>0\end{matrix}\right.\)

Mà: \(a^2+1>0\forall a=>\left\{{}\begin{matrix}2a>0\\1-a^2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\-1< a< 1\end{matrix}\right.\Leftrightarrow0< a< 1\)