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

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)

Bài 4:

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

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

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

\(\dfrac{3}{4}x-6< =0\)

=>\(\dfrac{3}{4}x< =6\)

=>\(x< =6:\dfrac{3}{4}=6\cdot\dfrac{4}{3}=8\)

không hiểu chỗ nào cứ hỏi nha

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

mọi người giúp e với ạ

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)

Khi x=5 thì \(11x-52=11\cdot5-52=55-52=3>0\)

=>Đúng

Khi x=5 thì \(6x-29=6\cdot5-29=30-29=1>0\)

=>6x-29>0 đúng

Khi x=5 thì 5-2=3<=0(sai)

=>x-2<=0 là đáp án sai duy nhất, hai cái còn lại đúng

a: \(\left\{{}\begin{matrix}4x+3y=6\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\15x-3y=33\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+3y+15x-3y=6+33\\5x-y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19x=39\\y=5x-11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{39}{19}\\y=5\cdot\dfrac{39}{19}-11=-\dfrac{14}{19}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{1}{5}x-\dfrac{1}{6}y=0\\5x-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=\dfrac{y}{6}\\5x-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\5\cdot\dfrac{5}{6}y-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{25}{6}y-4y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{5}{6}y\\\dfrac{1}{6}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=\dfrac{5}{6}\cdot12=10\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{8}y=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}-\dfrac{y}{8}=3\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{8x-3y}{24}=3\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x-3y=72\\7x+9y=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}24x-9y=216\\7x+9y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y+7x+9y=216-2\\8x-3y=72\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}31x=214\\3y=8x-72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{214}{31}\\y=\dfrac{8x-72}{3}=\dfrac{-520}{93}\end{matrix}\right.\)