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a)<=>5x^3-5x^2+10x^2y-10xy-5y^2-5y^2x
<=>x(5.x^2-5x+10xy)-5y^2-5x.y^2
<=>x[x.(5x-5+10y)]-y^2(5+5x)
<=>x[x.(5(x-1)+10y)]-y^2(5(x+1)
hok tốt các câu hỏi còn lại tương tự
1.\(4a^2b^2-\left(a^2+b^2-1\right)^2\)
\(=\left(2ab\right)^2-\left(a^2+b^2-1\right)^2\)
\(=\left(2ab+a^2+b^2-1\right)\left(2ab-a^2-b^2+1\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[1-\left(a-b\right)^2\right]\)
2.\(\left(xy+4\right)^2-\left(2x+2y\right)^2\)
\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)
\(=\left(x-2\right)\left(y-2\right)\left(x+2\right)\left(y+2\right)\)
1. x3 - 4x2 - 8x +8
= (x3 + 8) - (4x2 + 8x)
= (x + 2)(x2 - 3x + 4) - 4x(x + 2)
= (x + 2)(x2 - 3x + 4 - 4x)
= (x + 2)([x - 2]2 - 3x)
2. 1 + 6x - 6x2 - x3
= 6x(1 - x) + 1 - x3
= 6x(1 - x) + (1 - x)(1 + x + x2)
= (1 - x)(6x + 1 + x + x2)
= (1 - x)(7x + 1 + x2)
\(A=a^4+a^3+a^3b+a^2b\)
\(A=a\left(a^3+a^2\right)+b\left(a^3+a^2\right)\)
\(A=\left(a+b\right)\left(a^3+a^2\right)\)
\(A=a^2\left(a+1\right)\left(a+b\right)\)
Bài 1:
\(A=x^2+6x+5=x^2+5x+x+5=x\left(x+5\right)+\left(x+5\right)=\left(x+1\right)\left(x+5\right)\)
Đặt \(a=x^2-x+2\) ta có:
\(B=\left(a-1\right).a-12=a^2-a-12=a^2+3a-4a-12=a\left(a+3\right)-4\left(a+3\right)=\left(a+3\right)\left(a-4\right)\)
Thay a = x2 - x + 2 vào ta được:
\(\left(x^2-x+2-4\right)\left(x^2-x+2+3\right)=\left(x^2-x-2\right)\left(x^2-x+5\right)=\left(x+1\right)\left(x-2\right)\left(x^2-x+5\right)\)
1) x3 - 1 + x - x2
= (x3 + x) + (-x2 - 1)
= x(x2 + 1) + x2 + 1
= (x2 + 1) (x + 1)
chả bít đúng ko!!!!! 4534645756756876876356546576778983244343543545654
a) x4-4x3+4x2
= x2(x2-4x+4)
= x2(x-2)2
b) 2ab2-a2b-b3
= -b(-2ab+a2+b2)
=-b(a-b)2
e) x3+3x2-3x-1
= x3-x2+4x2-4x+x-1
=(x3-x2)+(4x2-4x)+(x-1)
=x2(x-1)+4x(x-1)+(x-1)
(x-1)(x2+4x+1)
f) x3-3x2-3x+1
=x3+x2-4x2-4x+x+1
=(x3+x2)-(4x2+4x)+(x+1)
=x2(x+1)-4x(x+1)+(x+1)
=(x+1)(x2-4x+1)
g)x3-4x2+4x-1
=x3-x2-3x2+3x+x-1
=(x3-x2)-(3x2-3x)+(x-1)
=x2(x-1)-3x(x-1)+(x-1)
=(x-1)(x2-3x+1)
a ) \(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a^2-2ab+b^2\right)\right]\)
\(=\left(a+b-c\right)\left(a+b+c\right)\left[c^2-\left(a-b\right)^2\right]\)
\(=\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(c+a-b\right)\)
b ) \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)