a) M xác định \(\Leftrightarrow\hept{\begin{cases}x-3\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne3\\x\ne-2\end{cases}}}\)

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

\(M=\frac{x^2-4-5}{\left(x-3\right)\left(x+2\right)}\)

\(M=\frac{x^2-9}{\left(x-3\right)\left(x+2\right)}\)

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

\(M=\frac{x+3}{x+2}\)

neu la best toan hay tra loi cau hoi tren

x³-x+3x²y+3xy²+y³-y

=x²(x-1)+3(x²y+xy²)+y²(y-1)

=x²(x-1)+3(x².y+y².x)+y²(y-1)

=x²(x-1)+3{[x(x+1)+y(y+1)]}+y²(y-1)

=x²(x-1)+3.x(x+1)+3.y(y+1)+y²(y-1)

=x²(x-1)+2x²+3.x(x+1)+3.y(y+1)+y²(y-1)+2y²-2x²-2y²

=x²(x+1)+3.x(x+1)+3.y(y+1)+y²(y+1)-2x²-2y²

=(x²+3)(x+1)+(y²+3)(y+1)-2(x²+y²)

ta có : (x*3+3x*2y+3xy*2+y*3)-(x+y)

=(x+y)*3-(x+y)

=(x+y)((X+Y)*2-1)

(x+y)(x+y+1)(x+Y-1)

1)   (x+3)²-x(x+5)=10

x²+6x+9-x²-5x=10

x+9=10

x=1

2)xét tứ giác ABCD có

DM=AM(đối xứng)

BM=CM(gt)

suy ra ABCD là hình bình hành

b)  áp dụng định lí pytago ta có

AB²+AC²=BC²

^{=.BC =10 cm}

^{mà trong tam giác vuông đường trung tuyến ứng vs cạnh huyền và =1/2canhj huyền}

^{=>AM=1/2BC=>AM=5cm}

Bài làm

a) Ta có: 

\(P=\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

\(P=\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{\left(x^2-3x\right)-\left(2x-6\right)}\right).\left(\frac{x+1}{1-x}\right)\)

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

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

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

\(P=\left[x^2-9-\left(x^2-4\right)+x+2\right].\left(\frac{x+1}{1-x}\right)\)

\(P=\left(x^2-9-x^2+4+x+2\right)\left(\frac{x+1}{1-x}\right)\)

\(P=\frac{\left(x-3\right)\left(x+1\right)}{1-x}\)

\(P=\frac{x^2-3x+x-3}{1-x}\)

\(P=\frac{x^2-2x-3}{1-x}\)

\(P=\left(x^2-2x-3\right):\left(1-x\right)\)

b) Để P = 3P.

<=> \(P=3P=\left(x^2-2x-3\right):\left(1-x\right)=3\left(x^2-2x-3\right):\left(1-x\right)\)

<=> \(\left(x^2-2x-3\right):\left(1-x\right)=3\left(x^2-2x-3\right):\left(1-x\right)\)

<=> ( x² - 2x - 3 ) : ( 1 - x ) - 3( x² - 2x - 3 ) : ( 1 - x ) = 0

<=> ( x² - 2x - 3 ) : [ 1 - x - 3( 1 - x ) ] = 0

<=> ( x² - 2x - 3 ) = 0 . ( 1 - x - 3 + x )

<=> x² - 2x - 3 = 0

<=> x² - 3x + x - 3 = 0

<=> x( x - 3 ) + ( x - 3 ) = 0

<=> ( x + 1 )( x - 3 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

Vậy x = -1 hoặc x = 3 thì P = 3P 

\(x^3-4x^2+12x-27\)

\(=x^3-3x^2-x^2+3x+9x-27\)

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

\(2x^2y-4xy^2+6xy=2xy\left(x-2y+3\right)\)

2x²y - 4xy² + 6xy

= 2xy . ( x - 2y + 3 )