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\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)
\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)
\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)
\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)
1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)
2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)
\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)
\(=25\left(a-b\right)^2=25\cdot100=2500\)
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
b)\(2a^2-3+5a\)
\(=\left(2a^2+6a\right)-\left(a+3\right)\)
\(=\left(a+3\right)\left(2a-1\right)\)
d)\(2a^2-5-3a\)
\(=\left(2a^2+2a\right)-\left(5a+5\right)\)
\(=\left(a+1\right)\left(2a-5\right)\)
a) \(a^2-3-2a\)
\(=a^2-2a+1-4\)
\(=\left(a^2-2a+1\right)-2^2\)
\(=\left(a-1\right)^2-2^2\)
\(=\left(a-1-2\right)\left(a-1+2\right)\)
\(=\left(a-3\right)\left(a+1\right)\)
c) \(4a+a^2+3\)
\(=a^2+4a+4-1\)
\(=\left(a^2+4a+4\right)-1^2\)
\(=\left(a+2\right)^2-1^2\)
\(=\left(a+2-1\right)\left(a+2+1\right)\)
\(=\left(a+1\right)\left(a+3\right)\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
a)-4a^3b.2a^3b^2-4a^3b(-3ab^4)
-8a^6b^3+12a^4b^3
b)2x.x^2-2x.3x+2x.4-3.x^2-3.(-3x)-3.4
2x^3-6x^2+8x-3x^2+9x-12
2x^3-9x^2+17x-12
\(\left(-2a+3\right)\left(-2+3a\right)=4a-6a^2-6+9a=6a^2+13a-6\)
`(-2a+3) (-2+3a)`
`= -2a (-2+3a)+3 (-2+3a)`
`= 4a - 6a^2 -6 + 9a`
`= -6a^2 + 13a - 6`