a: Xét ΔACD và ΔBDC có

AC=BD

CD chung

AD=BC

Do đó: ΔACD=ΔBDC

Suy ra: ˆACD=ˆBDCACD^=BDC^

hay ˆODC=ˆOCDODC^=OCD^

Xét ΔOCD có ˆODC=ˆOCDODC^=OCD^

nên ΔOCD cân tại O

Suy ra: OC=OD

Ta có: AO+OC=AC

OB+OD=BD

mà AC=BD

và OC=OD

nên OA=OB

x²-y²+2yz-x²

=x²-(y²-2yz+x²)

=x²-(y-x)²

=(x-y+x)(x+y-x)

HT

x^2-2xy+y^2-xz+yz =(x-y)²-xz+yz

                               =(x-y)²-(xz-yz)

                                =(x-y)²-z(x-y)

                                 =(x-y)(x-y-z)

\(x^2-2xy+y^2-xz+yz\)   

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y-z\right)\left(x-y\right)\)

bài 3

 a, x² + 4x + 3

= x² + x + 3x + 3

= x ( x +1 ) + 3 ( x +1 )

= ( x + 1 ) ( x +3 )

b, x² + 7x + 10

= x² + 2x + 5x + 10

x ( x + 2 ) + 5( x + 2)

= ( x +2 ) ( x +5 )

bài 4

a, 3x( x -1 ) - 5( x -1 ) =0

( x -1 ) ( 3x -5 ) =0

=> \(\orbr{\begin{cases}x-1=0\\3x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=\frac{5}{3}\end{cases}}\)

vậy x =1; x=5/3

b, 4x²( x -2 ) + 9(2-x) =0

4x²( x-2 ) -9( x-2) =0

( x-2 ) ( 4x² -9 ) =0

=> \(\orbr{\begin{cases}x-2=0\\4x^2-9=0\end{cases}}\) => \(\orbr{\begin{cases}x=2\\\left(2x\right)^2=9\end{cases}}\) => \(\orbr{\begin{cases}x=2\\\left(2x\right)^2=\left(\pm3\right)^2\end{cases}}\) => \(\orbr{\begin{cases}x=2\\2x=\pm3\end{cases}}\)

=>  ( chỗ này mk tạm ẩn vế x=2 đi ) \(\orbr{\begin{cases}2x=3\Rightarrow x=\frac{3}{2}\\2x=-3\Rightarrow x=-\frac{3}{2}\end{cases}}\)

vậy x = 2; x= \(\pm\frac{3}{2}\)

c, 8x³ -4x² + 2x -1 =0

8x³-1 - ( 4x² -2x ) =0

( 2x-1)( 4x² +2x +1 ) - 2x( 2x-1 )=0

( 2x -1 ) ( 4x² +2x +1 -2x ) =0

( 2x -1 ) ( 4x² +1 )=0

=> \(\orbr{\begin{cases}2x-1=0\\4x^2+1=0\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{1}{2}\\4x^2=-1\left(loại\right)\end{cases}}\) => \(x=\frac{1}{2}\)

vậy x = 1/2

kết quả 3*(y+x)*(z+x)*(z+y)

ta có :

\(\left(x+y+z\right)^3=x^3+\left(y+z\right)^3+3x\left(y+z\right)\left(x+y+z\right)\)

\(=x^3+y^3+z^3+3yz\left(y+z\right)+3x\left(y+z\right)\left(x+y+z\right)\)

Vậy \(\left(x+y+z\right)^3-x^3-y^3-z^3=3yz\left(y+z\right)+3x\left(y+z\right)\left(x+y+z\right)=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

x² + 2xz + 2xy + 4yz

= ( x² + 2xy ) + ( 2xz +4yz )

= x( x +2y ) + 2z( x +2y )

= (x +2y ) ( x +2z )

2xy + 3z +6y + xz

= 2xy + 6y + xz + 3z

= 2y( x +3 ) + z( x +3)

= ( x +3 ) ( 2y+ z)

2xy+3z+6y+xz

=2xy+6y+xz+3z

=2y.(x+3)+z.(x+3)

=(x+3).(2y+z)