\(3x^n\left(6x^{n-3}+1\right)-2x^n\left(9x^{n-3}-1\right)\)

\(=18x^{n-3}n^n+3n^n-18x^{2n-3}+2x^n\)

a, lm kiểu j ??? => \(3x-3a+yz^2-ya\)

b, \(x^3-2x^2+x-xy^2=x\left(x^2-2x+1-y^2\right)\)

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

c, \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

3x - 3a + yx - ya

= ( 3x - 3a ) + ( yx - ya )

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

= ( x - a )( 3 + y )

x³ - 2x² + x - xy²

= x( x² - 2x + 1 - y² )

= x[ ( x² - 2x + 1 ) - y² ]

= x[ ( x - 1 )² - y² ]

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

x² - 5x + 4 

= x² - x - 4x + 4

= ( x² - x ) - ( 4x - 4 )

= x( x - 1 ) - 4( x - 1 )

= ( x - 1 )( x - 4 )

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

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

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

\(=x^4-1.\)

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

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

                                                        =x^4 -1

k đúng cho mik nha!

( 2x - 3 )² - ( x + 5 )² = 0

<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0

<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0

<=> ( x - 8 )( 3x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

7x² - 28 = 0

<=> 7( x² - 4 ) = 0

<=> x² - 4 = 0

<=> x² = 4

<=> x² = (±2)²

<=> x = ±2

a) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left[\left(2x-3\right)-\left(x+5\right)\right].\left[\left(2x-3\right)+\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(2x-3-x-5\right).\left(2x-3+x+5\right)=0\)

\(\Leftrightarrow\left(x-8\right).\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+8\\3x=0-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\3x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

Vậy tập hợp nghiệm của phương trình là:  \(S=\left\{8;-\frac{2}{3}\right\}.\)

b) \(7x^2-28=0\)

\(\Leftrightarrow7x^2=0+28\)

\(\Leftrightarrow7x^2=28\)

\(\Leftrightarrow x^2=28:7\)

\(\Leftrightarrow x^2=4\)

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

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

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

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

Vậy tập hợp nghiệm của phương trình là:  \(S=\left\{2;-2\right\}.\)

\(A=\left(x-2\right)\left(x^4+2x^3+4x^2+8x+16\right)\)

\(=x^5+2x^4+4x^3+8x^2+16x-2x^4-4x^3-8x^2-16x-32\)

\(=x^5-32\) 

a) x³ - 1 + 5x² - 5 + 3x - 3

= x³ + 5x² + 3x - 9

= x³ + 6x² - x² + 9x - 6x - 9

= ( x³ + 6x² + 9x ) - ( x² + 6x + 9 )

= x( x² + 6x + 9 ) - ( x² + 6x + 9 ) 

= ( x² + 6x + 9 )( x - 1 )

= ( x + 3 )²( x - 1 )

b) a⁵ + a⁴ + a³ + a² + a + 1

= ( a⁵ + a⁴ + a³ ) + ( a² + a + 1 )

= a³( a² + a + 1 ) + 1( a² + a + 1 )

= ( a² + a + 1 )( a³ + 1 )

= ( a² + a + 1 )( a + 1 )( a² - a + 1 )

c) x³ - 3x² + 3x - 1 - y³

= ( x³ - 3x² + 3x - 1 ) - y³

= ( x - 1 )³ - y³

= ( x - 1 - y )[ ( x - 1 )² + ( x - 1 )y + y² ]

= ( x - 1 - y )( x² - 2x + 1 + xy - y + y² ) 

d) 5x³ - 3x²y - 45xy² + 27y³

= ( 5x³ - 45xy² ) - ( 3x²y - 27y³ )

= 5x( x² - 9y² ) - 3y( x² - 9y² )

= ( 5x - 3y )( x² - 9y² )

= ( 5x - 3y )[ x² - ( 3y )² ]

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

Sửa đề : x³ + y³ - xy( x + y ) = ( x + y )( x - y )²

x³ + y³ - xy( x + y )

= x³ + y³ - x²y - xy²

= x³ + 3x²y + 3xy² + y³ - 4x²y - 4xy²

= ( x³ + 3x²y + 3xy² + y³ ) - 4xy( x + y )

= ( x + y )³ - 4xy( x + y )

= ( x + y )[ ( x + y )² - 4xy ]

= ( x + y )( x² + 2xy + y² - 4xy )

= ( x + y )( x² - 2xy + y² )

= ( x + y )( x - y )²

=> đpcm

x³ + y³ = (x + y)(x² - xy + y²) = (x + y)(x² - 2xy + y² + xy) = (x + y)[(x - y)² + xy] (đpcm)

            Bài làm :

Ta có :

x³ + y³ 

= (x + y)(x² - xy + y²)

= (x + y)(x² - 2xy + y² + xy)

= (x + y)[(x - y)² + xy]

=> Điều phải chứng minh

Bị tự tin quá khả năng nhẩm mồm, sai em xin lỗi ...

a, Ta có \(P\left(x\right)=8x^3+2x^2-3x-3x^3+10-x-2x^2-3\)

\(=5x^3-4x-7\)

\(Q\left(x\right)=9x^3-4x^2+2x-3+2x+3x^2+4x^3-2\)

\(=13x^3-x^2+4x-5\)

b, Ta có : \(P\left(-\frac{1}{2}\right)=5.\left(-\frac{1}{2}\right)^3-4.\left(-\frac{1}{2}\right)-7=-\frac{45}{8}\)

c , \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\) 

  \(5x^3-4x-7+13x^3-x^2+4x-5=18x^3-x^2-12\)

\(N\left(x\right)=P\left(x\right)-Q\left(x\right)\)

\(5x^3-4x-7-13x^3+x^2-4x+5=-8x^3-8x-2+x^2\)

d, Đặt \(5x^3-4x-7=0\)( vô nghiệm )