
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


Câu hỏi của giang ho dai ca - Toán lớp 8 - Học toán với OnlineMath
https://olm.vn/hoi-dap/detail/9389242563.html
y đúc như bài trên

Ta có : \(x^8+14x^4+1\)
\(=x^8+2.x^4.7+1\)
\(=x^8+2.x^4.7+49-48\)
\(=\left(x^4+7\right)^2-48\)
\(=\left(x^4+7-\sqrt{48}\right)\left(x^4+7+\sqrt{48}\right)\)
a/\(=\left(x^4+1\right)^2+12x^4=\left(x^4+1\right)^2+4x^2\left(x^4+1\right)+4x^4-4x^2\left(x^4+1\right)+8x^4\)
\(=\left(x^4+1+2x^2\right)^2-4x^2\left(x^4+1-2x^2\right)=\left(x^4+2x^2+1\right)-\left(2x^3-2x\right)^2\)
\(=\left(x^4+2x^3+2x^2-2x+1\right)\left(x^4-2x^3+2x^2+2x+1\right)\)
b/\(=\left(x^4+1\right)^2+96x^4=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)
\(=\left(x^4+1+8x^2\right)^2-16x^2\left(x^4+1-2x^2\right)=\left(x^4+8x^2+1\right)-\left(4x^3-4x\right)^2\)
\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)

a,x8 +x4 +1=x6 .x2 +x3 .x+1=x6 .x2-x2 +x3 .x-x+1+x+x2=x2.(x6-1)+x.(x3-1)+1+x+x2=x2.(x3-1).(x3+1)+x.(x-1).(x2+x+1)+1+x+x2

\(A=x^8-2x^4-8\)
\(A=x^8-2x^4+4x^4-8\)
\(A=x^4\left(x^4-2\right)-4\left(x^4-2\right)\)
\(A=\left(x^4-4\right)\left(x^4-2\right)\)
\(a=\left(x^2-2\right)\left(x^2+2\right)\left(x^4-2\right)\)
\(A=\left(x^4\right)^2-4x^4+2x^4-8\)
\(=x^4\left(x^4-4\right)+2\left(x^4-4\right)\)
\(=\left(x^4+2\right)\left(x^4-4\right)\)
\(=\left(x^4+2\right)\left(x^2+2\right)\left(x^2-2\right)\)
\(=\left(x^4+2\right)\left(x^2+2\right)\left(x-2\right)\left(x+2\right)\)
Bạn cũng có thể đặt \(t=x^4\)để bài toán dễ làm hơn

a ) \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
b ) \(4x^8+1\)
\(=4x^8+1+4x^2-4x^2\)
\(=\left(2x^2+1\right)^2-\left(2x\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)\)

\(x^8+x^4+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(x^5-x^4-1\)
\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)
\(=\left(x^5-x^4+x^3\right)-\left(x^3-x^2+x\right)-\left(x^2-x+1\right)\)
\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)

mk chỉ lm đc câu b thôi !
mk k viết đề đâu nha !:
=(x4-8x2+16)+(5x2-20x+20)+(3x2+12x+12)+15
=(x2-4)2+5(x-2)2+3(x-2)2+15
=[(x-2)2+3][(x-2)2+5]
=(x2-4x+7)(x2+4x+9)
đúng 100 %

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/ \(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2+4x^4+1-4x^4=\left(2x^4+1\right)^2-4x^2=\left(2x^4-2x+1\right)\left(2x^4+2x+1\right)\)

\(a,4x^4+4x^3-x^2-x=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)