-6x³ + x² + 5x - 2 

= -6x³ - 6x² + 7x² + 7x - 2x - 2

= -6x²( x + 1 ) + 7x( x + 1 ) - 2( x + 1 )

= ( x + 1 )( -6x² + 7x - 2 )

= ( x + 1 )( -6x² + 3x + 4x - 2 )

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

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

\(-6x^3+x^2+5x-2=-6x^3-6x^2+7x^2+7x-2x-2\)

\(=\left(-6x^2+7x-2\right)\left(x+1\right)=\left(x+1\right)\left(3x-2\right)\left(1-2x\right)\)

x( 2x - 1 ) + 1/3( 1 - 2x ) = 0

<=> x( 2x - 1 ) - 1/3( 2x - 1 ) = 0

<=> ( 2x - 1 )( x - 1/3 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x-\frac{1}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{3}\end{cases}}\)

Ta có: x(2x-1)+1/3(1-2x)=0
=> x(2x-1)-1/3(2x-1)=0
=>(x-1/3)(2x-1)=0
=> \(\orbr{\begin{cases}x-\frac{1}{3}=0\\2x-1=0\end{cases}}\)

=> \(\hept{\begin{cases}x=\frac{1}{3}\\x=\frac{1}{2}\end{cases}}\)

Vậy ........

( 5a - 3b + 8c )( 5a - 3b - 8c ) 

= [ ( 5a - 3b ) + 8c ][ ( 5a - 3b ) - 8c ]

= ( 5a - 3b )² - ( 8c )²

= 25a² - 30ab + 9b² - 64c²

= 25a² - 30ab + 9b² - 16.4c²

= 25a² - 30ab + 9b² - 16( a² - b² ) < vì a² - b² = 4c² >

= 25a² - 30ab + 9b² - 16a² + 16b²

= 9a² - 30ab + 25b²

= ( 3a - 5b )²

=> đpcm

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)

\(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)

\(=\left(5a-3b\right)^2-\left(8c\right)^2\)

\(=25a^2-30ab+9b^2-64c^2\)

\(=25a^2-30ab+9b^2-16.4c^2\)

\(=25a^2-30ab+9b^2-16.\left(a^2-b^2\right)\)

\(=25a^2-30ab+9b^2-16a^2+16b^2\)

\(=9a^2-30ab+25b^2\)

\(=\left(3a-5b\right)^2\left(đpcm\right)\)

Xét \(\Delta BAD\)và \(\Delta ABC\)có:

\(\widehat{A}=\widehat{B}\)

\(AD=BC\)

\(AB\)chung

\(\Rightarrow\Delta BAD=\Delta ABC\left(c.g.c\right)\)

\(\Rightarrow AC=BD\)(2 cạnh t.ư)

=>tứ giác ABCD là HTC

Cách 1 : Kẻ thêm đường phụ AC 

Và đường phụ BD 

Xét tam giác ADC và tam giác ABC ta có : 

AC chung 

AD = BC (gt)

^A = ^B (gt) 

=> tam giác ADC = tam giác ABC 

=> AB = DC ( 2 cạnh tương ứng bằng nhau ) 

hay 2 góc kề cạnh đáy bằng nhau => ABCD là hình thang 

Cách 2 : Ta có : AD = BC gt 

=> 2 cạnh bên bằng nhau Vậy ABCD là hình thang :)) 

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

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

\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)

\(=-7x-13\)

2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)

\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)

3. \(2x^3-x^2+2x-1=0\)

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

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

Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)

\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

Bài 1.

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

= ( x² - 4 )( x + 3 ) - ( x³ + 3x² + 3x + 1 )

= x³ + 3x² - 4x - 12 - x³ - 3x² - 3x - 1

= -7x - 13

Bài 2.

64 - x² - y² + 2xy

= 64 - ( x² - 2xy + y² )

= 8² - ( x - y )²

= ( 8 -  x + y )( 8 + x - y )

Bài 3.

2x³ - x² + 2x - 1 = 0

<=> ( 2x³ - x² ) + ( 2x - 1 ) = 0

<=> x²( 2x - 1 ) + 1( 2x - 1 ) = 0

<=> ( 2x - 1 )( x² + 1 ) = 0

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

a) ( x² - 1 )( x - 101 ) + 101x( x + 1 ) = 101

<=> x³ - 101x² - x + 101 + 101x² + 101x - 101 = 0

<=> x³ + 100x = 0

<=> x( x² + 100 ) = 0

<=> \(\orbr{\begin{cases}x=0\\x^2+100=0\end{cases}}\Leftrightarrow x=0\)( vì x² + 100 ≥ 100 > 0 ∀ x )

b) x⁴ - 3x²( 2x - 3 ) = 0

<=> x⁴ - 6x³ + 9x² = 0

<=> x²( x² - 6x + 9 ) = 0

<=> x²( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x^2=0\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

a,\(\left(x^2-1\right)\left(x-101\right)+101x\left(x+1\right)=101\)

\(\Leftrightarrow x^3-101x^2-x+101+101x^2+101x=101\)

\(\Leftrightarrow x^3+100x=101-101\)

\(\Leftrightarrow x^3+101x=0\)

\(\Leftrightarrow x\left(x^2+101\right)=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x^2+101\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-101\end{cases}\Rightarrow}x=0}\)