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\(a^3+4a^2-7a-10\)
\(=a^3+3a^2+a^2-10a+3a-10\)
\(=\left(a^3+a^2\right)+\left(3a^2+3a\right)-\left(10a+10\right)\)
\(=a^2\left(a+1\right)+3a\left(a+1\right)-10\left(a+1\right)\)
\(=\left(a+1\right)\left(a^2+3a-10\right)\)
\(=\left(a+1\right)\left[\left(a^2+5a-2a-10\right)\right]\)
\(=\left(a+1\right)\left[a\left(a+5\right)-2\left(a+5\right)\right]\)
\(=\left(a+1\right)\left(a+5\right)\left(a-2\right)\)
b. \(\left(a^2+a\right)+a\left(a^2+a\right)-12\)
<=>\(\left(x^3+3x^2-4\right)+\left(3x^2+9x-12\right)\)
<=>\(x\left(x^2+3x-4\right)+3\left(x^2+3x-4\right)\)
<=>\(\left(x^2+3x-4\right)\left(x+3\right)\)
<=>\(\left(x+3\right)\left(x^2+4x\right)-\left(x-4\right)\)
đóngmở ngoặc nhé mk ngại ghi lại
<=>(x+3)(x(x+4)-(x+4))
<=>(x+3)(x-1)(x+4)
kết pn fb mk nhé longtrangv@gmail.com
c) \(x^3-x^2-4x^2+8x-4\)
= \(x^3-x^2-4x^2+4x+4x-4\)
= \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-4x+4\right)\)
= \(\left(x-1\right)\left(x-2\right)^2\)
Bài làm:
1) Ta có: \(2x^2+5xy+2y^2\)
\(=\left(2x^2+4xy\right)+\left(xy+2y^2\right)\)
\(=2x\left(x+2y\right)+y\left(x+2y\right)\)
\(=\left(2x+y\right)\left(x+2y\right)\)
2) Ta có: \(2x^2+2xy-4y^2\)
\(=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)\)
\(=2x\left(x-y\right)+4y\left(x-y\right)\)
\(=2\left(x+2y\right)\left(x-y\right)\)
\(1)2x^2+5xy+2y^2=2x^2+4xy+xy+2y^2=\left(2x^2+4xy\right)+\left(xy+2y^2\right)=2x\left(x+2y\right)+y\left(x+2y\right)=\left(2x+y\right)\left(x+2y\right)\)\(2)2x^2+2xy-4y^2=2x^2+4xy-2xy-4y^2=\left(2x^2-2xy\right)+\left(4xy-4y^2\right)=2x\left(x-y\right)+4y\left(x-y\right)=\left(2x+4y\right)\left(x-y\right)\)
1) \(=5\left(x+y\right)-\left(x-y\right)\left(x+y\right)=\left(x+y\right)\left(5-x+y\right)\)
2) \(=3\left(x^2-4x+4\right)=3\left(x-2\right)^2\)
3) \(=\left(x^2-1\right)\left(x^2+1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
4) \(=\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\)
5) \(=3\left(a^2-10a+25-b^2\right)=3\left(\left(a-5\right)^2-b^2\right)=3\left(a-5-b\right)\left(a-5+b\right)\)
6) \(=a\left(x-y\right)\left(x+y\right)+b\left(x+y\right)=\left(x+y\right)\left(ax-ay+b\right)\)
a) \(6x^2+6\)
\(=6\left(x^2+1\right)\)
b) \(2x^2-18\)
\(=2\left(x^2-9\right)\)
\(=2\left(x-3\right)\left(x+3\right)\)
c) \(3x^2-3xy+4x-4y\)
\(=\left(3x^2-3xy\right)+\left(4x-4y\right)\)
\(=3x\left(x-y\right)+4\left(x-y\right)\)
\(=\left(3x-4\right)\left(x-y\right)\)
a) \(\left(x^3-9x^2+27x-27\right)\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^3\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^2\)
c) \(\frac{x^2-4}{2x}:\frac{3x-6}{6}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{2x}.\frac{6}{3\left(x-2\right)}\)
\(=\frac{\left(x+2\right)}{x}\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
Đề thiếu rồi bạn