Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4.\)
\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4.\)
\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4.\)
\(=\left(x+5ax+4a^2+a^2\right)^2.\)
\(=\left(x+5ax+5a^2\right)^2.\)
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)
\(=\)\(\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\)
\(=\)\(\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\)\(\left[\left(x^2+5ax+5a^2\right)-a^2\right].\left[\left(x^2+5ax+5a^2\right)-a^2\right]+a^4\)
\(=\)\(\left(x^2+5ax+5a^2\right)^2-a^4+a^4\)
\(=\)\(\left(x^2+5ax+5a^2\right)^2\)
Chúc bạn học tốt ~
(x + a)(x + 2a)(x + 3a)(x + 4a) + a4
= (x + a)(x + 4a)(x + 2a)(x + 3a) + a4
= (x2 + 4ax + ax + 4a2)(x2 + 3ax + 2ax + 6a2) + a4
= (x2 + 5ax + 4a2)(x2 + 5ax + 6a2) + a4
Đặt x2 + 5ax + 4a2 = t
= t(t + 2a2) + a4
= (t + a2)2
= (x2 + 5ax + 4a2 + a2)2
= (x2 + 5ax + 5a2)2
1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)
2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)
3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)
\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)
\(=25\left(a-b\right)^2=25\cdot100=2500\)
c) Ta có: \(6x^4-11x^2+3\)
\(=6x^4-2x^2-9x^2+3\)
\(=\left(6x^4-2x^2\right)-\left(9x^2-3\right)\)
\(=2x^2\left(3x^2-1\right)-3\left(3x^2-1\right)\)
\(=\left(3x^2-1\right)\left(2x^2-3\right)\)
d) Ta có: \(\left(x^2+x\right)+3\left(x^2+x\right)+2\)
\(=4\left(x^2+x\right)+2\)
\(=2\left[2\left(x^2+x\right)+1\right]\)
a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)
\(=\left(x-y\right)\left(a+b+c\right)\)
b: \(a^m-a^{m+2}\)
\(=a^m-a^m\cdot a^2\)
\(=a^m\left(1-a^2\right)\)
\(=a^m\left(1-a\right)\left(1+a\right)\)
1.
\(3x^2-16x+5\\ =3x^2-x-15x+5\\ =x\left(3x-1\right)-5\left(3x-1\right)\\ =\left(x-5\right)\left(3x-1\right)\)
2.
\(3x^3-14x^2+4x+3\\ =\left(3x^3+x^2\right)-\left(15x^2+5x\right)+\left(9x+3\right)\\ =x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\\ =\left(x^2-5x+3\right)\left(3x+1\right)\)
3. \(x^8+x^7+1\\ =\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\\ =x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x^3-1\right)+x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x\left(x^3+1\right)\left(x+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)[x^2\left(x^3+1\right)\left(x-1\right)+x\left(x^3+1\right)\left(x-1\right)+1]\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+x^5-x^4+x^2-x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)4.
\(64x^4+y^4\\ =\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\\ =\left(8x^2+y^2\right)^2-16x^2y^2\\ =\left(8x^2+y^2-4xy\right)\left(8x^2+y+4xy\right)\)
5.
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\\ =\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\\ =\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+4a^2+2a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)+2a^2\left(x^2+5ax+4a^2\right)+a^4\\ =\left(x^2+5ax+5a^2\right)^2\)
Ta có:
(x+a)(x+2a)(x+3a)(x+4a) + a4
=(x+a)(x+4a)(x+3a)(x+2a) +a4
=(x2+5ax+4a2)(x2+5ax+6a2) + a4
Đặt x2+5ax+5a2=y
=>(x2+5ax+4a2)(x2+5ax+6a2) + a4=(y-a2)(y+a2)+a4
=y2-a4+a4
=y2
=(x2+5ax+5a2)2
k mik nha