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(x + 2y - 3)2 - 4(x + 2y - 3) + 4
= (x + 2y - 3)2 - 2. 2. (x + 2y - 3) + 22 (hằng đẳng thức số 2, bình phương của một hiệu)
= ( x + 2y - 3 - 2)2
= ( x + 2y - 5)2
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
\(=x^4+x^3+x^2+3x^3+3x^2+3x+x^2+x+1+x^2\)
\(=x^4+4x^3+6x^2+4x+1\)
\(=\left(x+1\right)^4\)
a) 2x.(4x - 1)
câu b), c) mik ko biết
ko mong b cho mik
nhưng vẫn hi vọng b hoặc ai đó sẽ làm vậy
b) \(4x^4+1=4x^4+4x^2+1-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
a) \(8x^2-2x-1=8x^2-4x+2x-1=4x.\left(2x-1\right)+\left(2x-1\right)=\left(2x-1\right)\left(4x+1\right)\)
b) \(4x^4+1=\left(2x^2\right)^2+4x^2+1-4x^2=\left(2x^2+1\right)^2-4x^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
c) \(\left(x^2-2x\right)\left(x^2-2x-1\right)-6=x^4-2x^3-x^2-2x^3+4x^2+2x-6\)
\(=x^4-4x^3+3x^2+2x-6=\left(x^4-3x^3\right)-\left(x^3-3x^2\right)+\left(2x-6\right)\)
\(=x^3.\left(x-3\right)-x^2.\left(x-3\right)+2.\left(x-3\right)=\left(x-3\right).\left(x^3-x^2+2\right)\)
\(=\left(x-3\right)\left[\left(x^3+x^2\right)+\left(-2x^2-2x\right)+\left(2x+2\right)\right]\)
\(=\left(x-3\right)\left[x^2\left(x+1\right)-2x.\left(x+1\right)+2.\left(x+1\right)\right]=\left(x-3\right)\left(x+1\right)\left(x^2-2x+2\right)\)
a, 8x^2-2x-1 = 8x2-4x+2x-1 = 4x ( 2x -1) + (2x-1) = (4x+1)(2x-1)
b) 4x4+1 = (2x2)2 + 4x2+ 1 - 4x2 = (2x2+1)2-(2x)2 = (2x2+1-2x)(2x2+1+2x)
a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
Có \(\left(x^2+4x+4\right)^3-y^6=\left(x+2\right)^6-y^6\)
=\(\left[\left(x+2\right)^3+y^3\right]\left[\left(x+2\right)^3-y^3\right]\)=\(\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\left(x+2-y\right)\left(x^2+4x+4+xy+2y+y^2\right)\)