Tam giác \(BKC\)vuông tại \(K\)có \(M\)là trung điểm của cạnh huyền \(BC\)nên \(KM=\frac{1}{2}BC\).

Tương tự ta cũng có \(HM=\frac{1}{2}BC\)

Suy ra \(KM=HM\)

\(\Rightarrow\Delta MKH\)cân tại \(M\).

Kẻ \(MN\)vuông góc với \(DE\).

Suy ra \(MN//BD//CE\)mà \(M\)là trung điểm của \(BC\)nên \(MN\)là đường trung bình của hình thang \(BDEC\).

suy ra \(N\)là trung điểm của \(DE\Rightarrow DN=NE\)(1).

Mà tam giác \(MKH\)cân tại \(M\)nên \(MN\)là đường cao đồng thời cũng là đường trung tuyến suy ra \(KN=HN\)(2)

(1) (2) suy ra \(DN-KN=EN-HN\Leftrightarrow DK=HE\).

Ta có đpcm.

cho 5 nha

chia hết cho 5 nha k hộ mk

D.7 nha!

d .7 nha bn 

chúc bn hok tốt

Trả lời:

a, ( x + 1 )² + ( x - 2 ) ( x + 3 ) - 4x 

= x² + 2x + 1 + x² + 3x - 2x - 6 - 4x

= 2x² - x - 5

b, ( x - 2 )² + ( x + 1 )² + 2 ( x - 2 ) ( - 1 - x ) 

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

= 2x² - 2x - 3 - 2x - 2x² + 4x + 4x

= 4x - 3

a) \(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x\)

\(=\left(x^2+2x+1\right)+\left(x^2+x-6\right)-4x\)

\(=x^2+2x+1+x^2+x-6-4x\)

\(=2x^2-x-5\)

b) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)

\(=\left(x^2-4x+4\right)+\left(x^2+2x+1\right)+\left(2x-4\right)\left(-1-x\right)\)

\(=x^2-4x+4+x^2+2x+1+\left(-2x-2x^2+4+4x\right)\)

\(=x^2-4x+4+x^2+2x+1-2x-2x^2+4+4x\)

\(=9\)

\(KL//MN\Rightarrow\widehat{K}+\widehat{N}=180^o\)

mà \(8\widehat{K}=\widehat{N}\)

suy ra \(\widehat{K}+8\widehat{K}=180^o\Leftrightarrow\widehat{K}=20^o\Rightarrow\widehat{N}=8.20^o=160^o\)

a) (x + 3)² - 2(x + 3)(x - 2) + (x - 2)² 

= (x + 3 - x + 2)² = 5² = 25

b) (2x + 5)² + 2(2x + 5)(3x - 1) + (3x - 1)² 

= (2x  + 5 + 3x - 1)² = (5x  + 4)²

Trả lời:

a/ ( x + 3 )² - 2 ( x + 3 ) ( x - 2 ) + ( x - 2 )²

=  [ ( x + 3 ) - ( x - 2 ) ]² 

= ( x + 3 - x + 2 )²

= 5²

= 25

b/ ( 2x + 5 )² + 2 ( 2x + 5 ) ( 3x - 1 ) + ( 3x - 1 )²

= ( 2x + 5 + 3x - 1 )²

= ( 5x + 4 )²

a.\(xz+yz-5\left(x+y\right)\)

\(=z\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(z-5\right)\)

b.\(3x^2-3xy-5x+5y\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?

d.\(3x^2+6xy+3y^2-3z^2\)

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

\(=3\left[\left(x+y\right)^2-z^2\right]\)'

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

Trả lời:

a, xz + yz - 5 ( x + y )

= ( xz + yz ) - 5 ( x + y )

= z ( x + y ) - 5 ( x + y )

= ( x + y ) ( z - 5 )

b, 3x² - 3xy - 5x + 5y

= ( 3x² - 3xy ) - ( 5x - 5y )

= 3x ( x - y ) - 5 ( x - y )

= ( x - y ) ( 3x - 5 )

c, x² + 6x - y² - 3z²

= - ( 3x² - x² + y² - 6x )

d, 3x² + 6xy + 3y² - 3z²

= 3 ( x² + 2xy + y² - x² )

= 3 [ ( x² + 2xy + y² ) - z² ]

= 3 [ ( x + y )² - z² ]

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

a) \(6x-6y=6\left(x-y\right)\)

b)\(2xy+3x+6y+xz\)

\(=\left(2xy+xz\right)+\left(6y+3z\right)\)

\(=x\left(2y+z\right)+3\left(2y+z\right)\)

\(=\left(2y+z\right)\left(x+3\right)\)

c)\(x^2+6x+9-y^2\)

\(=\left(x^2+6x+9\right)-y^2\)

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

\(=\left(x-y+3\right)\left(x+y+3\right)\)

d) \(9x-x^3\)

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

\(=x\left(3-x\right)\left(3+x\right)\)

e)\(x^2-xy+x-y\)

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

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

\(=\left(x-y\right)\left(x+1\right)\)

a, 6x - 6y = 6( x-y )

b, 2xy + 3z + 6y + xz

  = ( 2xy + 6y ) + ( 3z + xz )

  = 2y( x + 3 ) + z ( 3 + x )

  = 2y( 3 + x ) + z ( 3 + x )

  = ( 3 + x ) ( 2y + z )

c, x² + 6x + 9 - y² = ( x² + 6x + 9 ) - y²

                             = ( x + 3 )² - y²

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

d , 9x - x³ = x ( 9 - x² )

                = x ( 3 - x ) ( 3 + x )

e, x² - xy + x - y =( x ² - xy ) + ( x - y )

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

                          = ( x - y ) ( x + 1 )

\(a,\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)\)

\(\sqrt{x}^2-6^2\)

\(x-36\)

\(b,\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\)

\(\left(2\sqrt{x}\right)^2-1\)

\(4x-1\)

\(\left(\sqrt{x}-6\right)\left(6+\sqrt{x}\right)\)

\(=\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)\)

\(=\left(\sqrt{x}\right)^2-6^2\)

\(=x-36\)

b.\(\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\)

\(=\left(2\sqrt{x}\right)^2-1^2\)

\(=4x-1\)