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1.
= (x^3 + 125 ) -(x^2 +5x)
=(x +5) (x^2 -5x +25) -x(x+5)
=(x+5)(x^2 -5x +25 -x)
=(x+5)(x^2 -6x +25)
2.
= (x^3 -27) + (2x^2 -6x)
=(x-3) (x^2 +3x +9) +2x (x-3)
=(x-3) (x^2 +3x +9 +2x)
=(x-3) (x^2 +5x +9)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(b,x^3-2x^2-4xy^2+x\)
\(=x\left(x^2-2x-4y^2+1\right)\)
\(=x\left[\left(x^2-2x+1\right)-4y^2\right]\)
\(=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)
\(=x\left(x-1-2y\right)\left(x-1+2y\right)\)
\(=x\left(x-2y-1\right)\left(x+2y-1\right)\)
\(---\)
\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)
Đặt \(y=x^2+7x+10\), thay vào (1) ta được:
\(y\left(y+2\right)-8\)
\(=y^2+2y+1-9\)
\(=\left(y+1\right)^2-3^2\)
\(=\left(y+1-3\right)\left(y+1+3\right)\)
\(=\left(y-2\right)\left(y+4\right)\)
\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)
\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
#Ayumu
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)
b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)
\(=\left(2x+y\right)\left(4x^2-8xy+y^2\right)\)
c) \(\left(3x+2\right)^2-2\left(x-1\right)\left(3x+2\right)+\left(x-1\right)^2\)
\(=[\left(3x+2\right)-\left(x-1\right)]^2=\left(3x+2-x+1\right)^2=\left(2x+3\right)^2=\left(2x+3\right)\left(2x+3\right)\)
a) \(x^2-xz-9y^2+3yz\)
\(=\left(x^2-9y^2\right)-\left(xz-3yz\right)\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
c) \(x^3+2x^2-6x-27\)
\(=\left(x^3-27\right)+\left(2x^2-6x\right)\)
\(=\left(x-3\right)\left(x^2-3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-3x+9+2x\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c/ Ta có:
\(x^2-3xy+x-3y\)
\(=x^2+x-3xy-3y\)
\(=x\left(x+1\right)-3y\left(x+1\right)\)
\(=\left(x+1\right)\left(x-3y\right)\)
d/ Ta có:
\(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^2-3xy+x-3y\)
\(=x\left(x-3y\right)+\left(x-3y\right)\)
\(=\left(x+1\right)\left(x-3y\right)\)
\(x^3-x^2-5x+125\) k có nghiệm
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
p/s: chúc bạn học tốt
x3 - x2 - 5x + 125
=( x3 + 125 ) - ( x2 + 5x )
=(x+5)( x2- 5x +25 ) - x( x + 5 )
=(x+5)( x2- 5x + 25 - x )
=( x + 5 )( x2-4x + 25 )
b) x3 – x2 – 5x + 125
= (x3 + 125) - (x2 + 5x)
= (x + 5)(x2 - 5x + 25) - x(x + 5)
= (x + 5)(x2 - 5x + 25 - x)
= (x + 5)(x2 - 6x + 25)