\(5x^3-5x^2y-10x^2+10xy=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2\right)\left(x-y\right)\)

Ta có : \(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+5x^2+10x-6x-12=0\)

\(\Leftrightarrow x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^3+5x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^3-x^2+x^2-x+6x-6\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+6\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\)\(x+2=0\)

hoặc   \(x-1=0\)

hoặc   \(x^2+x+6=0\)

\(\Leftrightarrow\) \(x=-2\)(tm)

hoặc    \(x=1\)(tm)

hoặc   \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)

a: 3x-2=2x-3

=>x=-1

b: 2x+3=5x+9

=>-3x=6

=>x=-2

c: 5-2x=7

=>2x=-2

=>x=-2

d: 10x+3-5x=4x+12

=>5x+3=4x+12

=>x=9

e: 11x+42-2x=100-9x-22

=>9x+42=78-9x

=>18x=36

=>x=2

f: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

viết bằng công thức ở chỗ \(\sum\) đi bạn

Bạn bảo cái gì cơ

5x² - 10xy + 5y² - 20z²

=5(x² - 2xy + y² - 4z²)

= 5[ (x² - 2xy + y²) - (2z)²]

= 5[(x-y)² - (2z)²]

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

5x²-10xy+5y²-20z²

=5(x²-2xy+y²-4z²)

=5[(x-y)²-(2z)²]

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

\(2x^2-5x-7\)

\(=2x^2-2x+7x-7\)

\(=2x.\left(x-1\right)+7.\left(x-1\right)\)

\(=\left(2x+7\right).\left(x-1\right)\)

sr  :>

\(2x^2-5x-7\)

\(=2x^2+2x-7x-7\)

\(=2x.\left(x+1\right)-7.\left(x+1\right)\)

\(=\left(2x-7\right).\left(x+1\right)\)

2x² - 5x - 7

= 2x² + 7x - 2x - 7

= ( 2x² - 2x ) + ( 7x - 7 )

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

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

ta có :

\(x^2-5x+5y-y^2=\left(x^2-y^2\right)-5\left(x-y\right)\)

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

x²-5x+5y-y²

=(x²-y²)-(5x-5y)

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

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

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