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f) \(x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
g) \(x^4+64=\left(x^2+4x+8\right)\left(x^2-4x+8\right)\)
\(x^2-6x+5\)
\(=\left(x^2-2.3x+3^2\right)-4\)
\(=\left(x-3\right)^2-2^2\)
\(=\left(x-3-2\right)\left(x-3+2\right)\)
\(=\left(x-5\right)\left(x-1\right)\)
Ta có : \(x^3-64\)
\(=x^3-4^3\)
Áp dung hằng dẳng thức : \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\Rightarrow x^3-4^3=\left(x-4\right)\left(x^2+4x+4^2\right)\)
\(x^3\)\(-64\)\(=x^3\)\(-4^3\)=\(\left(x-4\right)\)\(\left(x^2+4x+16\right)\)
= ( x 4) 2 + 82 - 16x4 + 16x4 = ( x4 + 4) - ( 4x ) 2 = ( x4 +4 - 4x)( x4 +4 + 4x )
=
( x4 ) 2 + 16x4 + 16x4 = ( x4 + ) 2 = ( x4 + 4 - 4x ) + ( x4 + 4 + 4x )
Đáp số : ....
1) = \(x^2-1=\left(x-1\right)\left(x+1\right)\)
2) \(=\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
3)
\(=x^4-x+x^2+x+1=x\left(x^3-1\right)+x^2+x+1=x\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(=x^5-x^2+x^2+x+1=x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
1.\(x^2-2021+2020=x^2-1=\left(x+1\right)\left(x-1\right)\)
2. \(x^4+64=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
3. \(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2+x+1\right)\)
4. \(x^5+x+1=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Bạn tham khảo link này nhé :
https://olm.vn/hoi-dap/detail/11579055142.html
~Study well~
#SJ
\(a,\)\(x^{16}-1\)
\(=\left(x^8+1\right)\left(x^8-1\right)\)
\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^4-1\right)\)
\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x^2-1\right)\)
\(=\left(x^8+1\right)\left(x^4+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\)
hdt thức nha bạn
~~~~~~~~~~~~~
^_^
\(x^4+64\)
\(=\left(x^2\right)^2+8^2+2x^2.8-2x^2.8\)
\(=\left(x^2+8\right)^2-\left(4x^2\right)\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)