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\(1,\)
\(x^2+x-12\)
\(=x^2-3x+4x-12\)
\(=x\left(x-3\right)+4\left(x-3\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
\(2,\)
\(x^2-9x+20\)
\(=x^2-4x-5x+20\)
\(=x\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-5\right)\left(x-4\right)\)
\(3,\)
\(x^2+x-20\)
\(=x^2-4x+5x-20\)
\(=x\left(x-4\right)+5\left(x-4\right)\)
\(=\left(x+5\right)\left(x-4\right)\)
\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+1+2\left(x+y\right)\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2+2xy+1+2x+2y-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy+1+2x+2y\right)\)
\(=\left(x+y-1\right)\left[\left(x^2+1+2x\right)\left(y^2-xy+2y\right)\right]\)
\(=\left(x+y-1\right)\left(x+1\right)^2\left(y-x+2\right)y\)
Phần a? phải là \(4a^2-4a+1\)chứ
a) \(4a^2-4a+1=\left(2a\right)^2+2.2a+1\)
\(=\left(2a+1\right)^2\)
b) \(9x^2-25y^2=\left(3x\right)^2-\left(5y\right)^2\)
\(=\left(3x-5y\right)\left(3x+5y\right)\)
c) \(1-2x+a^2=\left(1-a\right)^2\)
d) \(\left(2x+1\right)-2.\left(2x+1\right)\left(3x-y\right)+\left(3x-y\right)^2\)
\(=\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)
nếu có sai thì bn thông cảm
1.
b) nó là hằng đẳng thức rồi bn nhá
c) \(1-2a+a^2\)= \(1^2-2a1+a^2\)=\(\left(1-a\right)^2\)
d)\(\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)=\(\left(2x+1-3x+y\right)^2\)=\(\left(1-x+y\right)^2\)
2.
a)\(\left(\frac{1}{2}x\right)^2-\left(3y\right)^2\)=\(\left(\frac{x}{2}-3y\right)\left(\frac{x}{2}+3y\right)\)
b) Ko khai triển đc
c) \(4x^2+2xy+\frac{1}{4}y^2\)
1,
\(\left(\frac{2}{3}x+y\right)^2=\left(\frac{2}{3}x\right)^2+2.\frac{2}{3}x.y+\left(y\right)^2=\frac{4}{9}x^2+\frac{4}{3}xy+y^2\)
\(\left(3a+\frac{1}{2}b\right)^2=\left(3a\right)^2+2.3a.\frac{1}{2}b+\left(\frac{1}{2}b\right)^2=9a^2+3ab+\frac{1}{4}b^2\)
2,
\(25a^2+4b^2+20ab=\left(5a\right)^2+\left(2b\right)^2+2.5a.2b=\left(5a+2b\right)^2\)
\(x^2+2x+1=\left(x\right)^2+2.x.1+\left(1\right)^2=\left(x+1\right)^2\)
\(9x^2+6x+1=\left(3x\right)^2+2.3x.1+\left(1\right)^2=\left(3x+1\right)^2\)
\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
\(\left(x-2\right)^3-1=\left(x-2\right)\left[\left(x-3\right)^2+x-2\right]=\left(x-2\right)\left(x^2+5x+7\right)\)
\(\left(x+3y\right)^2-9y^2=x\left(x+6y\right)\)
\(\left(x+3\right)^2-\left(x-1\right)^2=4\left(2x+4\right)=8\left(x+2\right)\)
a) \(\left(x-2\right)^3-1=\left(x-2\right)^3-1^3=\left(x-2-1\right)\left[\left(x-2\right)^2+\left(x-2\right)\cdot1+1^2\right]\)\(=\left(x-3\right)\left(x^2-4x+4+x-2+1\right)\)
\(=\left(x-3\right)\left(x^2-3x+3\right)\)
b) \(\left(x+3y\right)^2-9y^2\)
\(=\left(x+3y\right)^2-\left(3y\right)^2\)
\(=\left(x+3y+3y\right)\left(x+3y-3y\right)\)
\(=x\left(x+6y\right)\)
c) \(\left(x+3\right)^2-\left(x-1\right)^2\)
\(=\left(x+3-x+1\right)\left(x+3+x-1\right)\)
\(=4\left(2x+2\right)\)
\(=8\left(x+1\right)\)
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
(x+1)3-(x-1)3=(x+1-x+1)[(x+1)2+(x+1)(x-1)+(x-1)2 ]=2(3x2+1)