Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) \(x^3-3x^2+2\)
\(=x^3-2x^2-x^2+2\)
\(=x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x^2-x-2\right)\left(x-2\right)\)
c) \(x^4y^4+64\)
\(=x^4y^4+16x^2+64-16x^2\)
\(=\left(x^2y^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2y^2-4x+8\right)\left(x^2y^2+4x+8\right)\)
d) \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+1\)
\(=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^6-x^4-x+x^3-1\right]\)
a) \(3x^2-9x+30=3\left(x^2-3x+10\right)\)
b) \(3x^2-5x-2=3x^2-6x+x-2\)
\(=3x\left(x-2\right)+\left(x-2\right)=\left(3x+1\right)\left(x-2\right)\)
c) \(x^4+4y^4\)
\(=x^4+4y^4+2x^2y^2+2x^2y^2-4x^2y^2+4xy^3-4xy^3+2x^3y-2x^3y\)
\(=\left(4y^4-4xy^3+2x^2y^2\right)+\left(4xy^3-4x^2y^2+2x^3y\right)\)
\(+\left(2x^2y^2-2x^3y+x^4\right)\)
\(=2y^2\left(2y^2-2xy+x^2\right)+2xy\left(2y^2-2xy+x^2\right)\)
\(+x^2\left(2y^2-2xy+x^2\right)\)
\(=\left(2y^2+2xy+x^2\right)\left(2y^2-2xy+x^2\right)\)
d) \(x^5+x+1\)
\(=x^5+x+1+x^4-x^4+x^3-x^3+x^2-x^2\)
\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
a )\(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
b )\(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+2x+2\right)\left(x^2-2\right)\)
c ) \(x^2\left(1-x^2\right)-4-4x^2=x^2-x^4-4-4x^2\)
\(=x^2-\left(x^2+2\right)^2=\left(x-x^2-2\right)\left(x^2+x+2\right)\)
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
a) \(45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(a,45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)
\(=x^2\left(x-5\right)+9\left(5-x\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)\)
\(=\left(x-5\right)\left(x^2-3^2\right)\)
\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)
\(c,2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\left(2\right)\)
(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\left(3\right)\)
Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)
1.
x4y4+4=[(x2y2)2+2.x2y2.2+22]-4x2y2
=(x2y2+2)2-(2xy)2
bạn tính nốt đi, câu 2, 4, 6 tương tự
câu 4 khá dài bạn lấy số đấy chia cho (x+1) ra nháp rồi tính ngược lại sẽ ra
1: \(=x^4y^4+4+4x^2y^2-4x^2y^2\)
\(=\left(x^2y^2+2\right)^2-4x^2y^2\)
\(=\left(x^2y^2+2xy+2\right)\left(x^2y^2-2xy+2\right)\)
2: \(=x^4y^4+16x^2y^2+64-16x^2y^2\)
\(=\left(x^2y^2+8\right)^2-16x^2y^2\)
\(=\left(x^2y^2+8-4xy\right)\left(x^2y^2+8+4xy\right)\)
3: \(=x^4+4x^2+4-x^2\)
\(=\left(x^2+2\right)^2-x^2\)
\(=\left(x^2+x+2\right)\left(x^2-x+2\right)\)
4: \(=4x^4y^4+1+4x^2y^2-4x^2y^2\)
\(=\left(2x^2y^2+1\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2y^2+1-2xy\right)\left(2x^2y^2+1+2xy\right)\)
6: \(=x^4+4y^4+4x^2y^2-4x^2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2+2xy\right)\left(x^2+2y^2-2xy\right)\)
1) \(x^2+6x+8\)
\(=x^2+2x+4x+8\)
\(=x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+4\right)\left(x+2\right)\)
2) \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
3) \(2x^2+5x+3\)
\(=2x^2+2x+3x+3\)
\(=2x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+3\right)\)
4) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
1, \(=\left(2y\right)^2-\left(x^2-2x+1\right)=\left(2y\right)^2-\left(x-1\right)^2=\left(2y-x+1\right)\left(2y+x-1\right)\)
2, \(=2\left(x^2-y^2\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1+4\right)=2\left(x+1\right)\left(x+3\right)\)
3, \(=\left(x^2+6x+9\right)-\left(2y\right)^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
4, \(=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
\(4y^2-x^2+2x-1\)
\(=4y^2-\left(x^2-2x+1\right)\)
\(=\left(2y\right)^2-\left(x-1\right)^2\)
\(=\left(2y-x+1\right)\left(2y+x-1\right)\)
hk tốt
^^
Biết câu nào làm câu đấy thoy nha :))
3. \(x^4y^4+4\)
\(=\left(x^2y^2\right)^2+2\cdot x^2y^2\cdot2+2^2-2\cdot x^2y^2\cdot2\)
\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)
4. \(x^4+4y^4\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot2y^2+\left(2y^2\right)^2-2\cdot x^2\cdot2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)
2. \(x^4+x^2+1\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot1+1^2-2x^2\)
\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)
\(=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)\)