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a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)
\(3y^2\left(a-3x\right)-a\left(a-3x\right)=\left(3y^2-a\right)\left(a-3x\right)\)
b: \(x^4+324=x^4+36x^2+324-36x^2\)
\(=\left(x^2+18\right)^2-36x^2\)
\(=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
c: \(64a^4+b^8\)
\(=64a^4+b^8+16a^2b^4-16a^2b^4\)
\(=\left(8a^2+b^4\right)^2-16a^2b^4\)
\(=\left(8a^2-4ab^2+b^4\right)\left(8a^2+4ab^2+b^4\right)\)
g: \(a^6-b^6=\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(=\left(a-b\right)\cdot\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
a) \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)
\(=\left[3\left(a+b\right)+2\left(a-2b\right)\right]\left[3\left(a+b\right)-2\left(a-2b\right)\right]\)
\(=\left(3a+3b+2a-4b\right)\left(3a+3b-2a+4b\right)\)
\(=\left(5a-b\right)\left(a+7b\right)\)
b) \(\left(2a-b\right)^2-4\left(a-b\right)^2\)
\(=\left[\left(2a-b\right)-2\left(a-b\right)\right]\left[\left(2a-b\right)+2\left(a-b\right)\right]\)
\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)
\(=b\left(4a-3b\right)\)
c) \(125-\left(x+2\right)^3\)
\(=\left(5-x-2\right)\left[25+5\left(x+2\right)+\left(x+2\right)^2\right]\)
\(=\left(3-x\right)\left(25+5x+10+x^2+4x+4\right)\)
\(=\left(3-x\right)\left(x^2+9x+39\right)\)
d) \(\left(x+3\right)^3-8=\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+4\right]\)
\(=\left(x+1\right)\left(x^2+8x+19\right)\)
e) \(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2=\left(x^6-y^2\right)\left(x^6+y^2\right)\) 9 khai triển tiếp hđt 6,7)
\(1,\)
\(a,25+10a^2+a^4\)
\(=5^2+2.5.a^2+\left(a^2\right)^2\)
\(=\left(5+a^2\right)^2\)
\(b,\left(x^2+4x+4\right)-25y^2\)
\(=\left(x^2+2x.2+2^2\right)-\left(5y\right)^2\)
\(=\left(x+2\right)^2-\left(5y\right)^2\)
\(=\left(x+2-5y\right)\left(x+2+5y\right)\)
\(c,4b^2-\left(a^2-6a+9\right)\)
\(=\left(2b\right)^2-\left(a^2-2a.3+3^2\right)\)
\(=\left(2b\right)^2-\left(a-3\right)^2\)
\(=\left(2b-a+3\right)\left(2b+a-3\right)\)
Chúc bn học giỏi nhoa!!!
Dễ mà :
Ta có : 25 + 10a2 + a4
= 52 + 2.a2.5 + (a2)2
= (5 + a2)2
(áp dụng a2 + 2ab + b2 = (a + b)2 )
Bài 1:
a) \(x^2-y^2+10x+25\)
\(=\left(x^2+10x+25\right)-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+y+5\right)\left(x-y+5\right)\)
b) \(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^4+4y^4\)
\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
e) \(7x^2-10xy+3y^2\)
\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)
\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)
f) Sửa đề \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)
\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)
\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
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. (a - b + c - d).(a - b + c - d)
= (a - b + c - d)2
Câu 1 vậy là gọn nhé
2.
a) x2 - 10xy + 25y2
= x2 - 2x5y + (5y)2
= (x - 5y)2
b) 16a4 + 8a2b3 + b6
= (4a2)2 + 2.4a2.b3 + (b3)2
= (4a2 + b3)2
c) a4 - 1
= (a2)2 - 1
= (a2 - 1)(a2 + 1)
= (a - 1)(a + 1)(a2 + 1)
d) 16a4 - 81b4
= (4a2)2 - (9b2)2
= (4a2 - 9b2)(4a2 + 9b2)
= [(2a)2 - (3b)2](4a2 + 9b2)
= (2a - 3b)(2a + 3b)(4a2 + 9b2)
e) (a4 - 2a2b + b2) - b4
= [(a2)2 - 2a2b + b2] - (b2)2
= (a2 - b)2 - (b2)2
= (a2 - b - b2)(a2 - b + b2)
= [(a - b)(a + b) - b](a2 - b + b2)
f) 81x4 - (b2 - 2b + 1)
= (9x2)2 - (b - 1)2
= (9x2 - b + 1)(9x2 + b - 1)