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a)x(x2+2xy+y2-4)
=x[(x+y)2-22 ]
=x(x+y-2)(x+y+2)
b)x4+4=x4+4x2+4-4x2=(x2+2)2-4x2
=(x2+2-2x)(x2+2+2x)
\(x^3+2x^2y+xy^2-4x=x\)\(\left(x^2+2xy+y^2-4\right)\)
\(=x\left[\left(x+y\right)^2-4\right]\)
\(=x\left(x+y+2\right)\left(x+y-2\right)\)
\(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
a) \(x^4+4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) \(4x^8+1\)
\(=\left(2x^4\right)^2+2\cdot2x^4\cdot1+1^2-2\cdot2x^4\cdot1\)
\(=\left(2x^4+1\right)-\left(2x^2\right)^2\)
\(=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)
a/ \(x^4+4=\left(x^2\right)^2+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-4x+2\right)\left(x^2+4x+2\right)\)
b/ \(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2+4x^4+1-4x^4=\left(2x^4+1\right)^2-4x^2=\left(2x^4-2x+1\right)\left(2x^4+2x+1\right)\)
a) \(x^4+4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) \(4x^8+1\)
\(=\left(2x^4\right)^2+2\cdot2x^4\cdot1+1^2-2\cdot2x^4\cdot1\)
\(=\left(2x^4+1\right)^2-\left(2x^2\right)^2\)
\(=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)
\(x^4+4\)
\(=\left(x^2\right)^2+2.x^2.2+2^2-2.x^2.2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\)
a) \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) \(4x^8+1=4x^8+4x^4+1-4x^4=\left(2x^4+1\right)^2-4x^4=\left(2x^4-2x^2+1\right)\left(2x^4+2x^2+1\right)\)
d) \(x^2+14x+48=\left(x+7\right)^2-1=\left(x+7+1\right)\left(x+7-1\right)=\left(x+8\right)\left(x+6\right)\)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)
a, 4x4+1
=(2x)2+1
=(2x+1)(2x-1)
b,c tách làm bình phương rồi làm tương tự
\(\left(a\right)x^8+98x^4+1\)
\(\text{ Phân tích thành nhân tử}\)
\(\left(x^4-4x^3+8x^2+4x+1\right)\left(x^4+4x^3+8x^2+\left(-4\right)x+1\right)\)
\(\left(b\right)4x^4-32x^2+1\)
\(\text{ Phân tích thành nhân tử}\)
\(-\left(28x^2-1\right)\)
cái này phân tích thành nhân tử:
vì máy tính nên ko viết đc mũ
(x mũ 4-4xmũ 3+8x mũ 2+4x+1)vì vậy biểu thức ko thể rút gọn
a) \(\Rightarrow\left(x^2\right)^2+\left(2^2\right)^2+2.2x^2-2.2x^2\Rightarrow\left(x^2+2\right)^2-\left(2x\right)^2\Rightarrow\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
b) \(\Rightarrow\left(2x^4\right)^2+2.2.x^4.1+1-2.2.x^4.1\Rightarrow\left(2x^4+1\right)^2-\left(2x^2\right)^2\Rightarrow\left(2x^4+1-2x^2\right)\left(2x^4+1+2x^2\right)\)
CHÚC BẠN học tốt
T I C K cho mình nha cảm ơn
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)