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Đặt x2 + 4x + 8 = A. Ta sẽ được:
A2 + 3xA + 2x2
= A2 - xA - 2xA + 2x2
= A(A-x) - 2x(A-x)
= (A-x)(A-2x)
= (x2+3x+8)(x2+2x+8)
e) \(\left(x-y\right)^2+4\left(x-y\right)-12\)
\(=\left(x-y\right)\left[\left(x-y\right)+4\right]-12\)
\(=\left(x-y\right)\left(x-y+4\right)-12\)
f) \(x^2-2xy+y^2+3x-3y-10\)
\(=\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\)
\(=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(=\left(x-y\right)\left[\left(x-y\right)+3\right]-10\)
\(=\left(x-y\right)\left(x-y+3\right)-10\)
g) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left[\left(x^2+4x+8\right)+3x\right]+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+3x\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+7x+8\right)+2x^2\)
a,\(x^2+4x-25y^2+4=\left(x^2+4x+4\right)-\left(5y\right)^2=\left(x+2\right)^2-\left(5y\right)^2\)
\(=\left(x+2-5y\right)\left(x+2+5y\right)\)
b, \(x^3-4x^2-8x+8=\left(x^3+8\right)-\left(4x^2+8x\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)-4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4-4x\right)=\left(x+2\right)\left(x^2-2x+4\right)\)
c,\(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(4x+1\right)\)
Mk làm sai câu b nha , phảo làm như sau
\(x^3-4x^2-8x+8=\left(x^3+8\right)-\left(4x^2+8x\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=\left(x+2\right)\left(x^2-6x+4\right)\)
a, \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\text{[}x\left(x+1\right)+2\left(x+1\right)\text{]}\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
b, \(2x^3+3x^2+3x+2\)
\(=2x^3+2x^2+x^2+x+2x+2\)
\(=2x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+x+2\right)\)
c, \(x^3-4x^2-8x+8\)
\(=x^3+2x^2-6x^2-12x+4x+8\)
\(=x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-6x+4\right)\)
b. x4 - x2 - 2x - 1
=x4-(x2+2x+1)
=x4-(x+1)2
=(x2-x-1)(x2+x+1)
d. ( x2 + 3x + 1 ) ( x2 + 3x - 3 ) - 5
Đặt x2+3x=y
=> (y+1)(y-3)-5=y2-2y-8=(y-1)2-9
=(y-4)(y+2)
=(x2+3x-4)(x2+3x+2)=(x-1)(x+4)(x+1)(x+2)
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right),\left(x+4\right)\)
ko biết
thế mà xl cc ak