Bài làm:

Ta có: \(3x^2+3x-6\)

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

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

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

\(3x^2+3x-6\)

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

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

\(=3\left[x\left(x+2\right)-\left(x+2\right)\right]\)

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

a) \(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)\)

\(=27\left(1-x^3\right)+81\left(x^2-x\right)\)

\(=27-27x^3+81x^2-81x\)

b) \(y\left[x^2+x\left(x-y\right)+\left(x-y\right)^2\right]+\left(x-y\right)^3\)

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

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

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

a, \(27\left(1-x\right)\left(x^2+x+1\right)+81x\left(x-1\right)=27-27x^3+81x^2-81x\)

b, \(y\left[x^2+x\left(x-y\right)+\left(x-y\right)^2\right]+\left(x-y\right)^3\)

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

a + b + c = 0

=> a + b = - c

=> ( a + b )³ = - c³

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(-c\right)\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\) ( đpcm )

Ta dễ có được

\(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)

Suy ra điều cần chwgns minh

a) \(pt< =>x^3-3.x^2.3+3.x.9-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=4\)

\(< =>x^3-27-x^3+27-9x^2+27x+9x^2+18x+9=4\)

\(< =>45x=4-9=-5< =>x=-\frac{5}{45}=-\frac{1}{9}\)

b) \(pt< =>x\left(x^2-25\right)-\left(x^3+8\right)=17\)

\(< =>x^3-25x-x^3-8=17< =>25x=-8-17=-25< =>x=-1\)

a) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 9( x + 1 )² = 4

<=> x³ - 9x² + 27x - 27 - ( x³ - 27 ) + 9( x² + 2x + 1 ) = 4

<=> x³ - 9x² + 27x - 27 - x³ + 27 + 9x² + 18x + 9 = 4

<=> 45x + 9 = 4

<=> 45x = -5

<=> x = -5/45 = -1/9

b) x( x - 5 )( x + 5 ) - ( x + 2 )( x² - 2x + 4 ) = 17

<=> x( x² - 25 ) - ( x³ + 2³ ) = 17

<=> x³ - 25x - x³ - 8 = 17

<=> -25x - 8 = 17

<=> -25x = 25

<=> x = -1

a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 ) = 16

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 16

<=> x² + 5x + 6 - x² - 3x + 10 = 16

<=> 2x + 16 = 16

<=> 2x = 0

<=> x = 0

b) 3x( 2x - 4 ) - 2x( 3x + 5 ) = 44

<=> 6x² - 12x - 6x² - 10x = 44

<=> -22x = 44

<=> x = -2

c) 2( 5x - 8 - 3 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 5x - 11 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 20x² - 69x + 55 ) = 12x - 16

<=> 40x² - 138x + 110 = 12x - 16

<=> 40x² - 138x + 110 - 12x + 16 = 0

<=> 40x² - 150 + 126 = 0 ( chưa học nghiệm vô tỉ nên để vô nghiệm nha :) )

=> Vô nghiệm

a. (x+2) (x+3)-(x-2) (x+5)= 16

x²+5x+6-x²-3x+10=16

2x+16=16

2x=0

x=0

b,3x (2x-4)-2x (3x+5)= 44

6x²-12x-6x²-10x=44

-22x=44

x=-2

Ý c bạn tự lm,tương tự nhưa,b

a) 5x( x - 1 ) = x - 1

<=> 5x² - 5x = x - 1

<=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0

<=> 5x² - 5x - x + 1 = 0

<=> 5x( x - 1 ) - 1( x - 1 ) = 0

<=> ( x - 1 )( 5x - 1 ) = 0

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

b) 2( x + 5 ) - x² - 5x = 0

<=> 2x + 10 - x² - 5x = 0

<=> -x² - 3x + 10 = 0

<=> -x² - 5x + 2x + 10 = 0

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

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

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

c) x² - 2x - 3 = 0

<=> x² + x - 3x - 3 = 0

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

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

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

d) 2x² + 5x - 3 = 0

<=> 2x² - x + 6x - 3 = 0

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

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

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

a) 5x ( x - 1 ) = x - 1 <=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0 <=> 5x² - x - ( 5x - 1 ) = 0 

<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0

<=> x = 1 hoặc x = 1/5

b) 2 ( x + 5 ) - x² - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0

<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5

c) x² - 2x - 3 = 0 <=> x² + x - 3x - 3 = 0 

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

<=> x = 3 hoặc x = -1

d) 2x²  + 5x - 3 = 0

Ta có : delta = 5² - 4.2.3 = 25 - 24 = 1

Khi đó : x = -1 hoặc x = 3/2