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Bài toán 1.2 :
\(3a\left(b^2-2c\right)-\left(a-4\right)\left(2c-b^2\right)\)
\(=3a\left(b^2-2c\right)+\left(a-4\right)\left(b^2-2c\right)\)
\(=\left(b^2-2c\right)\left(3a+a-4\right)\)
\(=\left(b^2-2c\right)\left(4a-4\right)\)
\(Q=\left(x+2z\right)\left(3x^2+5x^2y\right)-\left(7x^2-3x^2y\right)\left(2z+x\right)\)
\(Q=3x^3+5x^3y+6x^2z+10x^2yz-14x^2z+6x^2yz-7x^3+3x^3y\)
\(Q=-4x^3+8x^3y+16x^2yz-8x^2z\)
\(Q=-4x^3\left(1-2y\right)+8x^2z\left(2y-1\right)\)
\(Q=4x^3\left(2y-1\right)+8x^2z\left(2y-1\right)\)
\(Q=\left(2y-1\right)\left(4x^3+8x^2z\right)\)
a) \(C=ax-ay-bx+by+\left(y-x\right)^2\)
\(=a\left(x-y\right)-b\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(a-b+x-y\right)\)
b) \(M=\left(5x-10\right)\left(x^2-1\right)\left(3x-6\right)\left(x^2-2x+1\right)\)
\(=5\left(x-2\right)\left(x-1\right)\left(x+1\right)-3\left(x-2\right)\left(x-1\right)^2\)
\(=\left(x-2\right)\left(x-1\right)\left[5\left(x+1\right)-3\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(5x+5-3x+3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(2x+8\right)\)
\(=2\left(x-2\right)\left(x-1\right)\left(x+4\right)\)
f, 9x4-\(\frac{1}{4}\)= (3x2)2-\(\left(\frac{1}{2}\right)^2\)= (3x2-\(\frac{1}{2}\))(3x2+\(\frac{1}{2}\))
h, (x-y)2-(m+n)2=(x-y+m+n)(x-y-m-n)
i, 9(a-b)2-4(x-y)2= (3a-3b)2-(2x-2y)2=(3a-3b+2x-2y)(3a-3b-2x+2y)
x^4 - 2x^3 - 2x^2 - 2x - 3
= ( x^4 - 3x^3 ) + ( x^3 - 3x^2 )+ ( x^2 - 3x ) + ( x - 3)
= x^3 ( x - 3 ) + x^2 ( x - 3 ) + x ( x - 3 ) + ( x - 3 )
= ( x - 3 ) ( x^3 + x^2 + x + 1 )
= ( x - 3 ) [( x^3 + x^2 ) + ( x + 1 )]
= ( x - 3 ) [ x^2 ( x + 1 ) + ( x + 1)]
= ( x - 3 ) ( x + 1 ) ( x^2 + 1 )
\(x^3-2x+y^3-2y=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
theo mình đề câu c là 6x2
\(x^3+6x^2+9x-xz^2=x\left(x^2-6x+9-z^2\right)\)
\(=\left(x-3-z\right)\left(x-3+z\right)\)
\(x^2-11x+30=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)=\left(x-5\right)\left(x-6\right)\)
\(4x^2-3x-1=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+x-1=\left(4x+1\right)\left(x-1\right)\)
\(9x^2-7x-2=9x^2-9x+2x-2\)
\(=9x\left(x-1\right)+2\left(x-1\right)=\left(9x+2\right)\left(x-1\right)\)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-5=\left(x^2+x-1\right)^2-4\)
\(=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
còn lại lát mình làm tiếp
Bài 1:
a, \(x^3-2x-y^3-2y=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
b, \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
c, \(x^3+6x^2+9x-xz^2=x\left(x^2+6x+9-z^2\right)\)
\(=x\left[\left(x+3\right)^2-z^2\right]=x\left(x+3+z\right)\left(x+3-z\right)\)
Mỗi bài mình sẽ làm một câu mẫu ạ
Bài 1:
a) \(x^3-2x+y^3-2y\)
\(=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
Bài 2:
a) \(x^2-11x+30\)
\(=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-6\right)\left(x-5\right)\)
Bài 3:
a) \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Bài 2:
b: \(=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
c: \(=9x^2-9x+2x-2=\left(x-1\right)\left(9x+2\right)\)
e: Sửa đề: \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-2\)
\(=\left(x^2+3x\right)^2+3\left(x^2+3x\right)+2-2\)
\(=\left(x^2+3x\right)\left(x^2+3x+3\right)\)
\(=x\left(x+3\right)\left(x^2+3x+3\right)\)
