a) 6xy - 54xz² = 6x( y - 9z² )

b) x⁴ + 2x³ - 4x² - 8x

= ( x⁴ + 2x³ ) - ( 4x² + 8x )

= x³( x + 2 ) - 4x( x + 2 )

= ( x + 2 )( x³ - 4x )

= ( x + 2 )x( x² - 4 )

= ( x + 2 )x( x - 2 )( x + 2 )

= ( x + 2 )²x( x - 2 )

c) 3x² + 5x - 2

= 3x² - x + 6x - 2

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

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

d) 5x² + 6xy + y²

= 5x² + 5xy + xy + y²

= 5x( x + y ) + y( x + y )

= ( x + y )( 5x + y )

e) -14x² + 39x - 10

= -14x² + 4x + 35x - 10

= -2x( 7x - 2 ) + 5( 7x - 2 )

= ( 7x - 2 )( 5 - 2x )

f) x³ + 2x² - 25x - 50

= ( x³ + 2x² ) - ( 25x + 50 )

= x²( x + 2 ) - 25( x + 2 )

= ( x + 2 )( x² - 25 )

= ( x + 2 )( x - 5 )( x + 5 )

ta có tam giác ADH vuông tại H
=> AH^2+HD^2=AD^2
=>HD^2=AD^2-AH^2
            =5^2-4^2
            =9
=>HD=3 cm
kẻ BK vuông góc với CD
=>ABKH là hình chữ nhật 
=>AH=BK=4cm 
tam giác BKC vuông tại K
=>BK^2+KC^2=BC^2
=>KC^2=BC^2-BK^2
            =80-16
           =64
=>KC=8 (cm)
lại có DH+HK+KC=20
=>HK=20-3-8=9 (cm)
=>AB+HK=9 cm
ta có chu vi hình thang ABCD là AB+BC+CD+DA=9+√80+20+5=34+√80(cm)
 

Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)

    \(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)

    \(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)

    \(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)

    \(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)

Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)

    \(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)

    \(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)

    \(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)

    \(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)

    \(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)

A = a( b + 2 ) + b( 2 + b )

= a( b + 2 ) + b( b + 2 )

= ( a + b )( b + 2 )

Với a = 2 ; b = 3

A = ( 2 + 3 )( 3 + 2 ) = 5.5 = 25

B = b² + b + c( b + 1 )

= b( b + 1 ) + c( b + 1 )

= ( b + c )( b + 1 )

Với b = 1 ; c = 2

B = ( 1 + 2 )( 1 + 1 ) = 6

C = xy( x - y ) - 2x + 2y

= xy( x - y ) - 2( x - y )

= ( x - y )( xy - 2 )

Với xy = 8 ; x - y = 5

C = 5.( 8 - 2 ) = 30

D = x( x + y ) - xy( x + y )

= ( x + y )( x - xy )

= ( x + y )x( 1 - y )

Với x = 1 ; y = -5

D = ( 1 - 5 ).1.[ 1 - ( -5 ) ] = -24

Bài làm:

a) Ta có: \(x\left(x-3\right)-2x+6=0\)

\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

b) \(\frac{x}{3}+\frac{x^2}{2}=0\)

\(\Leftrightarrow\frac{3x^2+2x}{6}=0\)

\(\Leftrightarrow x\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{2}\end{cases}}\)

c) \(x-2=\left(x-2\right)^2\)

\(\Leftrightarrow x^2-4x+4+2-x=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

d) \(\left(x^2+3\right)\left(x+1\right)+x=-1\)

\(\Leftrightarrow\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+4\right)\left(x+1\right)=0\)

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

a. \(x\left(x-3\right)-2x+6=0\)

\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

b. \(\frac{x}{3}+\frac{x^2}{2}=0\)

\(\Leftrightarrow\frac{2x+3x^2}{6}=0\)

\(\Leftrightarrow x\left(2+3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)

c. \(x-2=\left(x-2\right)^2\)

\(\Leftrightarrow x-2-x^2+4x-4=0\)

\(\Leftrightarrow-\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

d. \(\left(x^2+3\right)\left(x+1\right)+x=-1\)

\(\Leftrightarrow x^3+x^2+3x+3+x+1=0\)

\(\Leftrightarrow x^3+x^2+4x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=-4\left(vo-ly\right)\end{cases}}\)

<=> x = - 1

a) x( x + 2018 ) - 2x - 4036 = 0 

<=> x( x + 2018 ) - 2( x + 2018 ) = 0

<=> ( x + 2018 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x+2018=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2018\\x=2\end{cases}}\)

b) x + 5 = 2( x + 5 )²

<=> x + 5 = 2( x² + 10x + 25 )

<=> x + 5 = 2x² + 20x + 50

<=> 2x² + 20x + 50 - x - 5 = 0

<=> 2x² + 19x + 45 = 0

<=> 2x² + 10x + 9x + 45 = 0

<=> 2x( x + 5 ) + 9( x + 5 ) = 0

<=> ( x + 5 )( 2x + 9 ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{9}{2}\end{cases}}\)

c) ( x² + 1 )( 2x - 1 ) + 2x = 1

<=> 2x³ - x² + 4x - 1 - 1 = 0

<=> 2x³ - x² + 4x - 2 = 0

<=> x²( 2x - 1 ) + 2( 2x - 1 ) = 0

<=> ( 2x - 1 )( x² + 2 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}\Leftrightarrow}x=\frac{1}{2}\)( vì x² + 2 ≥ 2 > 0 ∀ x )

d) \(\frac{x}{3}-\frac{x^2}{4}=0\)

\(\Leftrightarrow\frac{4x}{12}-\frac{3x^2}{12}=0\)

\(\Leftrightarrow\frac{4x-3x^2}{12}=0\)

\(\Leftrightarrow4x-3x^2=0\)

\(\Leftrightarrow x\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

Bài này đề sửa thành: \(H=a+4b+1\) mk ms lm được ạ

Ta có: \(a=111...1\) (2020 chữ số 1)

\(a=111...1\cdot100...0+111...1\)

\(a=b.\left(9b+1\right)+b\)

Thay vào:

\(H=a+4b+1=b\left(9b+1\right)+b+4b+1=9b^2+6b+1=\left(3b+1\right)^2\)

=> đpcm