a) 5x³ - 40 = 5( x³ - 8 ) = 5( x - 2 )( x² + 2x + 4 )

b) x²z + 4xyz + 4y²z = z( x² + 4xy + 4y² ) = z( x + 2y )²

c) 4x² - y² - 6x + 3y = ( 4x² - y² ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )

d) x² + 2x - 4y² + 1 = ( x² + 2x + 1 ) - 4y² = ( x + 1 )² - ( 2y )² = ( x - 2y + 1 )( x + 2y + 1 )

e) 3x² - 3y² - 12x + 12y = 3( x² - y² - 4x + 4y ) = 3[ ( x² - y² ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )

f) x³ + 5x² + 4x + 20 = x²( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x² + 4 )

g) x³ - x² - 25x + 25 = x²( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x² - 25 ) = ( x - 1 )( x - 5 )( x + 5 )

a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)

b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)

c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)

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

d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)

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

e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)

\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)

\(=3\left(x-y\right)\left(x+y+4\right)\)

f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)

\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)

g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)

\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)

\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)

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

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

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

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

\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x^2-4x+5=0\end{cases}}\Leftrightarrow x=-3\)< vì x² - 4x + 5 = ( x² - 4x + 4 ) + 1 = ( x - 2 )² + 1 ≥ 1 > 0 ∀ x

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

\(\Leftrightarrow x^3-3x^2+5x+3x^2-9x+15=x^2+3x\)

\(\Leftrightarrow x^3-4x+15=x^2+3x\)

\(\Leftrightarrow x^3-x^2-7x+15=0\)

\(\Leftrightarrow\left(x^2-4x+5\right)\left(x+3\right)=0\Leftrightarrow x=-3\)

a)B=3x³ -2y³-6x²y²+xy

   B=(3x³-6x²y²)+(xy-2y³)

   B=3x²(x-2y²)+y(x-2y²)

    B=(x-2y²)(3x²+y)
tại x=\(\frac{2}{3}\)và y=\(\frac{1}{2}\)ta có B=(x-2y²)(3x²+y)=(\(\frac{2}{3}\)-2*^{\(\frac{1}{2}\)}^2 )(3*\(\frac{2}{3}\)^2+\(\frac{1}{2}\))=\(\frac{1}{6}\)*\(\frac{11}{6}\)=\(\frac{11}{36}\)

b)C= 2x+xy²-x²y-2y

   C=(2x-2y)+(xy²-x²y)

   C=2(x-y)-xy(x-y)

   C=(2-xy)(x-y)

tại x=\(-\frac{1}{2}\)và y=\(-\frac{1}{3}\)ta có C=(2-xy)(x-y)=(2-\(-\frac{1}{2}\)*\(-\frac{1}{3}\))(\(-\frac{1}{2}\)+\(\frac{1}{3}\))=\(\frac{-11}{36}\)

Ta có: \(44\equiv2\left(mod7\right)\Rightarrow44^{2005}\equiv2^{2005}\left(mod7\right)\) (*)

Lại có: \(2^3\equiv1\left(mod7\right)\Rightarrow\left(2^3\right)^{668}\equiv1\left(mod7\right)\Rightarrow\left(2^3\right)^{668}.2\equiv2\left(mod7\right)\)

            \(\Leftrightarrow2^{2005}\equiv2\left(mod7\right)\)(**)

Từ (*) và (**) suy ra \(44^{2005}\equiv2\left(mod7\right)\)

Vậy \(44^{2005}\)chia 7 dư 2

bạn có thể giúp mình trả lời 2 câu b và c đk ko