kẻ đường cao AH của tam giac  cân ABC ta có AH đồng thời là đường phân giác của góc BAC => \(AH\perp AM\)

mà \(AH\perp BC=>MN//BC\)

zì \(\widehat{BAH}=\widehat{HAC}=>\widehat{BAM}=\widehat{CAN}\)

do đó \(\widehat{MAC}=\widehat{NAB}\)(1)

mặt khác theo giả thiết ta có 

\(AM.AN=AB^2=>\frac{AM}{AB}=\frac{AB}{AN}\)

mà \(AB=AC\left(gt\right)\Rightarrow\frac{AM}{AC}=\frac{AB}{AM}\left(2\right)\)

từ (1) zà 2 => \(\Delta ANB~\Delta ACM\left(c.g.c\right)\)

\(M=\frac{x^2-9}{5x-10}:\frac{x^2+3x}{x-2}\)

\(M=\frac{\left(x-3\right)\left(x+3\right)}{5\left(x-2\right)}:\frac{x\left(x+3\right)}{x-2}\)

\(M=\frac{\left(x-3\right)\left(x+3\right)}{5\left(x-2\right)}.\frac{x-2}{x\left(x+3\right)}\)

không bảo rút gọn nhưng mình vẫn rút gọn cho dễ làm nhé :)))

\(M=\frac{\left(x-3\right)\left(x+3\right)\left(x-2\right)}{5\left(x-2\right).x\left(x+3\right)}=\frac{x-3}{5x}\) (1)

a) ĐKXĐ: \(x\ne0;x\ne2;x\ne-3\)

b) thay x = 1/2 vào (1), ta có: \(M=\frac{\frac{1}{2}+3}{5.\frac{1}{2}}=\frac{7}{5}\)

c) \(\frac{x-3}{5x}=\frac{1}{2}\)

<=> 2(x - 3) = 5x

<=> 2x - 6 = 5x

<=> 2x - 6 - 5x = 0

<=> -3x - 6 = 0

<=> -3x = 0 + 6

<=> -3x = 6

<=> x = -2

a, x( x - 3) - ( x + 2)( x - 1) = 3

<=> x² - 3x - x² + x - 2x + 2 = 3

<=>  x² - 3x - x² + x - 2x = 3 - 2

<=> -4x = 1

<=> x =\(-\frac{1}{4}\)

Vậy_

b,  \(x-\frac{x-1}{5}+\frac{x+2}{6}=4+\frac{x+1}{3}\)

\(\Leftrightarrow\frac{30x}{30}-\frac{\left(x-1\right)6}{30}+\frac{\left(x+2\right)5}{30}=\frac{120}{30}+\frac{\left(x+1\right)10}{30}\)

=> 30x - 6x + 6 + 5x + 10 = 120 + 10x + 10

<=> 30x - 6x + 5x - 10x = 120 + 10 - 6 - 10

<=> 19x = 114

<=> x = 6

Vậy _

=\(\left(1+\frac{4}{x-2}\right):\left(\frac{x^2-4}{2}\right)\)

=\(\left(\frac{x-2}{x-2}+\frac{4}{x-2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{2}\right)\)

=\(\frac{x-2+4}{x-2}\cdot\frac{2}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x+2}{x-2}.\frac{2}{\left(x-2\right).\left(x+2\right)}\)

=\(\frac{2}{\left(x-2\right)^2}\)

Mình quên ĐKXĐ 

Sorry bạn nha

\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x+2\right)\left(x-2\right)}\)

=> ( x + 1)( x + 2) + ( x - 1)( x - 2) = 2x² + 4

<=> x² + 2x + x + 2 + x² - 2x - x + 2 = 2x² + 4 

<=>  x² + 2x + x +  x² - 2x - x - 2x² = 4 - 2 - 2

<=> 0x = 0

Vậy phương trình vô số nghiệm