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1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
3)(9a)2-(5a-3b)2
= (9a-5a+3b)(9a+5a-3b)
= (4a+3b)(14a-3b)
1) 9(a+b)2-4(a-2b)2
=[3(a+b)]2-[2(a-2b)]2=....
2) 8x3+27y3=(2x)3+(3y)3=...
3) (a+b)3-c3=(a+b-c).[(a+b)2+(a+b)c+c2]
4) x3+3x2+3x+1=(x+1)3
4/ a/ Ta có \(x^2-2xy+y^2+a^2=\left(x-y\right)^2+a^2\)
Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\a^2\ge0\end{cases}}\)=> \(\left(x-y\right)^2+a^2\ge0\)
=> \(x^2-2xy+y^2+a^2\ge0\)
Vậy \(x^2-2xy+y^2\)chỉ nhận những giá trị không âm.
b/ Ta có \(x^2+2xy+2y^2+2y+1=\left(x^2+2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x+y\right)^2+\left(y+1\right)^2\)
Mà \(\hept{\begin{cases}\left(x+y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)=> \(\left(x+y\right)^2+\left(y+1\right)^2\ge0\)
=> \(x^2+2xy+2y^2+2y+1\ge0\)
Vậy \(x^2+2xy+2y^2+2y+1\)chỉ nhận những giá trị không âm.
c/ Ta có \(9b^2-6b+4c^2+1=\left(3b-1\right)^2+4c^2\)
Mà \(\hept{\begin{cases}\left(3b-1\right)^2\ge0\\4c^2\ge0\end{cases}}\)=> \(\left(3b-1\right)^2+4c^2\ge0\)
=> \(9b^2-6b+4c^2+1\ge0\)
Vậy \(9b^2-6b+4c^2+1\)chỉ nhận những giá trị không âm.
d/ Ta có \(x^2+y^2+2x+6y+10=\left(x+1\right)^2+\left(y+3\right)^2\)
Mà \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)=> \(\left(x+1\right)^2+\left(y+3\right)^2\ge0\)
=> \(x^2+y^2+2x+6y+10\ge0\)
Vậy \(x^2+y^2+2x+6y+10\)chỉ nhận những giá trị không âm.
1/
a/ \(x^4-y^4=\left(x^2-y^2\right)\)
b/ \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left[a^2+2ab+b^2-\left(a^2-b^2\right)+\left(a^2-2ab+b^2\right)\right]\)
\(=2b\left(a^2+b^2\right)\)
c/ \(\left(a^2+2ab+b^2\right)+\left(a+b\right)\)
= \(\left(a+b\right)^2+\left(a+b\right)\)
= \(\left(a+b\right)\left(a+b+1\right)\)
1. \(x^3+9x^2+27x+27=\left(x+3\right)^3\)
2.\(8x^6-27y^3=\left(2x\right)^3-\left(3y\right)^3=\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(8x^6+6xy+27y^3\right)\)
3.\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)4.câu cuối là \(8b^3\)bạn nhé !!
tik mik nhé
1)\(144a^2-81=\left(12a\right)^2-9^2=\left(12a-9\right)\left(12a+9\right)\)
2)\(a^4-4b^2=a^2-\left(2b\right)^2=\left(a-2b\right)\left(a+2b\right)\)
3)Ko hỉu
4)\(\left(a-5b\right)^2-16b^2=\left(a-5b\right)^2-\left(4b\right)^2=\left(a-9b\right)\left(a-b\right)\)
5)\(9\left(a+b\right)^2-4\left(a-b\right)^2=\left[3\left(a+b\right)+2\left(a-b\right)\right]\left[3\left(a+b\right)-2\left(a-b\right)\right]\)
\(=\left(5a+b\right)\left(a+5b\right)\)
6)\(x^2+12x+36=\left(x+6\right)^2\)
7)Đề sai
8)\(9x^4+24x^2+16=\left(3x^2+4\right)^2\)
9)Chịu