Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)=x^2y-xy^2+y^2z-yz^2+z^2z-zx^2=x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(z-y\right)\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)-z^2\left(y-z\right)=\left(y-z\right)\left(x-z\right)\left(x+z\right)-y^2\left(x-z\right)=\left(x-z\right)\left(xy-yz-zx-z^2-y^2\right)\)
t cx k bt có đúng hay k đâu nha, nhớ xem kĩ lại
a) 16(4x+5)2 - 25(2x+2)2
\(=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2\)
\(=\left[4\left(4x+5\right)+5\left(2x+2\right)\right]\left[4\left(4x+5\right)-5\left(2x+2\right)\right]\)
\(=\left(16x+20+10x+10\right)\left(16x+20-10x-10\right)\)
\(=\left(26x+30\right)\left(6x+10\right)\)
\(b,\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-2y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-3y+5\right)\)
\(c,\left(x+1\right)^4-\left(x-1\right)^4\)
\(=\left(x+1\right)^{2^2}-\left(x-1\right)^{2^2}\)
\(=\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\)
\(=\left(x^2+2x+1+x^2-2x+1\right)\left[\left(x+1+x-1\right)\left(x+1-x+1\right)\right]\)
\(=\left(2x^2+2\right)2x.2\)
\(=4x.2\left(x^2+1\right)\)
\(=8x\left(x^2+1\right)\)
1) \(25x^4-10x^2y+y^2\)
\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)
\(\Leftrightarrow\left(5x^2+y\right)^2\)
2) \(x^4+2x^3-4x-4\)
\(\Leftrightarrow\left(x^4-4\right)+\left(2x^3-4x\right)\Leftrightarrow\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2+2x\right)\)
3) \(x^4+x^2+1\)
\(\Leftrightarrow x^4+x^2-x+x+1\)
\(\Leftrightarrow\left(x^4-x\right)+\left(x^2+x+1\right)\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)
\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)
5) \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)
\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)
\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\)
\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)
\(a,\left(a^3-b^3\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
\(b,\left(x^2+1\right)^2-4x^2\)
\(=x^4+2x^2+1-4x^2\)
\(=x^4-2x^2+1\)
\(\left(x^2-1\right)^2\)
\(c\left(y^3+8\right)+\left(y^2-4\right)\)
\(=\left(y+2\right)\left(y^2-8y+4\right)+\left(y-2\right)\left(y+2\right)\)
\(=\left(y+2\right)\left(y^2-8y+4+y-2\right)\)
\(=\left(y+2\right)\left(y^2-7y+2\right)\)
a) ( a3 - b3) + ( a - b)2
= (a-b) (a2 + ab + b2 ) + (a-b)2
= (a-b) (a2 + ab + b2 +a -b )
hok tốt
a) 2x + 2y - x2 - xy
= 2(x + y) + x(x + y)
= (x + y) (x + 2)
mk ko bít phân tích đúng ko đúng thì t i c k nhé!! 245433463463564564574675687687856856846865855476457
a)\(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
b)\(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x+3\right)-\left(2x-5\right)\right]\)
\(=\left(x+3\right)\left(8-x\right)\)
c)\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x-2\right)^2\)
\(=\left(3x+2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]+\left(3x-2\right)\left[\left(3x-2\right)-\left(3x+2\right)\right]\)
\(=4\left(3x+2\right)-4\left(3x-2\right)\)
\(=4\left(3x+2-3x+2\right)\)
=4.4=16
(x^2-x+2)^2+(x-2)^2
= [(x^2-x+2)+(x-2)]^2-2[(x^2-x+2)*(x-2)] (áp dụng (a^2+b^2)=(a+b)^2-2ab
=(x^2)^2- 2((x^3-3x^2+4x-4)
=x^4-2x^3+6x^2-8x+8
giờ phân tích đa thức
x^4-2x^3+6x^2+8x-8
=(x^4-2x^3+2x^2)+(4x^2-8x+8) (cái này làm bài tập nhiêu nhìn ra nhanh)
=[x^2(x^2-2x+2)]+4(x^2-2x+2) dẹp luôn
=(x^2-2x+2)(x^2+4)
\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=\left[\left(x-2\right)\left(x+1\right)\right]^2+\left(x-2\right)^2\)
\(=\left(x-2\right)^2\left(x+1\right)^2+\left(x-2\right)^2\)
\(=\left(x-2\right)^2\left(x^2+2x+1\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)^2\left(x^2+2x+2\right)\)
Phân tích thành nhân tử:
(4x + 3y)2 + (6xy - 2)2
=\((16x^2+24xy+9y^2)+(36x^2y^2-24xy+4)\)
=\(16x^2+24xy+9y^2+36x^2y^2-24xy+4\)
=\(16x^2+9y^2+36x^2y^2+4\)
=\((4x)^2+(3y)^2+(6xy)^2+2^2\)
MÌNH CHỈ LÀM ĐC TỚI ĐÂY
a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz
=y.(4\(x^3\) + \(\dfrac{1}{2}\)z)
b, (a2 + b2 - 5)2 - 2.(ab + 2)2
= [a2 + b2 - 5 - \(\sqrt{2}\)(ab + 2) ].[ a2 + b2 - 5 + \(\sqrt{2}\)(ab +2)]
a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)
b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)
\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)
\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)
\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)
\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)