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a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
b) \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2-x+9\right)\)
d) \(a^6-a^4+2a^3+2a^2=a^2\left(a^4-a^2+2a+2\right)=a^2\left[a^2\left(a-1\right)\left(a+1\right)+2\left(a+1\right)\right]=a^2\left(a+1\right)\left(a^3-a^2+2\right)=a^2\left(a+1\right)\left[a^3+a^2-2a^2+2\right]=a^2\left(a+1\right)\left[a^2\left(a+1\right)-2\left(a-1\right)\left(a+1\right)\right]=a^2\left(a+1\right)^2\left(a^2-2a+2\right)\)
a) Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
b) Ta có: \(x^3+2x^2+2x+1\)
\(=\left(x^3+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
\(\left(1+2x\right).\left(1-2x\right)-x.\left(x+2\right).\left(x-2\right)\))
\(=1-\left(2x\right)^2-x.x^2-2^2\)
\(=1-4x^2-x^3-4\)
Ko bt có đúng ko nữa
\(x^4+2x^2+1=\left(x^2+1\right)^2\) (Nhớ k cho mình với nhé!)
x(x+2)(x^2+2x+2)+1 = (x^2+2x)(x^2+2x+1)+1
Đặt x^2+2x+1=y ta được:
(y-)(y+1)+1=y^2-1+1=y^2
= (x^2+2x+1)^2
= ( x + 1 )^4
Đặt \(A=\left(x-y+4\right)^2-\left(3x+3y-1\right)^2\)
Ta có:
\(\left(x-y+4\right)^2=x^2-xy+4x-yx+y^2-4y+4x-4y+16\)
\(=x^2+y^2-2xy+8x-8y+16\)
\(\left(3x+3y-1\right)^2=9x^2+9xy-3x+9xy+9y^2-3y-3x-3y+1\)
\(=9x^2+9y^2-6x-6y+18xy+1\)
Mình làm đến đây bạn trừ 2 kết quả cho nhau rồi sẽ ra
Ta có : x3 + 2x2 + 2x + 1
= x3 + x2 + (x2 + 2x + 1)
= x2(x + 1) + (x + 1)2
= (x + 1) ( x2 + x + 1)
a)\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2\)
\(=a^2b^2+a^2+b^2+1\)
\(=a^2\left(b^2+1\right)+\left(b^2+1\right)\)
\(\left(a^2+1\right)\left(b^2+1\right)\)
b)\(x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
\(=\left(x^2+x+1\right)\left(x+1\right)\)