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a) \(5x-10x^2\) = \(5x\left(1-2x\right)\)
b) Mạn phép sửa đề:
\(\dfrac{1}{2}x\left(x^2-4\right)+4\left(x+2\right)\) = \(\left(x+2\right)\left[\dfrac{1}{2}x\left(x-2\right)+4\right]\)
= \(\left(x+2\right)\left(\dfrac{1}{2}x^2-x+4\right)\)
c) \(x^4-y^6=\left(x^2-y^3\right)\left(x^2+y^3\right)\)
e) \(x^3-4x^2+4x-1=x^3-x^2-3x^2+3x+x-1\)
= \(x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-3x+1\right)\)
g) \(x^4+6x^3-12x^2-8x\)
= \(x\left(x^3-2x^2+8x^2-16x+4x-8\right)\)
= \(x\left[x^2\left(x-2\right)+8x\left(x-2\right)+4\left(x-2\right)\right]\)
= \(x\left(x-2\right)\left(x^2+8x+4\right)\)
h) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (*)
Đặt \(x^2+4x+8=a\) => (*) trở thành:
\(a^2+3ax+2x^2\) = \(a^2+ãx+2ax+x^2\)
= \(a\left(a+x\right)+2x\left(a+x\right)\)
= \(\left(a+x\right)\left(a+2x\right)\) (1)
Thay \(a=x^2+4x+8\) vào (1) ta được:
\(\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
=\(\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)\)
= \(\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)
= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
P/s: Còn câu f đang suy nghĩ!
a)\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
Đặt: \(a-b=x;\)\(b-c=y;\)\(c-a=z\)
thì: \(x+y+z=0\)
Dễ dàng chứng minh đc:
\(x+y+z=0\)
thì \(x^3+y^3+z^3=3xyz\)
đến đây bạn thay trở lại nhé
a) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3b-a^3c+b^3\left(c-a\right)+c^3a-c^3b\)
\(=\left(a^3b-c^3b\right)+\left(c^3a-a^3c\right)+b^3\left(c-a\right)\)
\(=-b\left(c^3-a^3\right)+ca\left(c^2-a^2\right)+b^3\left(c-a\right)\)
\(=-b\left(c-a\right)\left(c^2-ac+a^2\right)+ca\left(c+a\right)\left(c-a\right)+b^3\left(c-a\right)\)
\(=\left(c-a\right)\left(-c^2b+abc-a^2b\right)+\left(c-a\right)\left(c^2a+ca^2\right)+b^3\left(c-a\right)\)
\(=\left(c-a\right)\left(-c^2b+abc-a^2b+c^2a+ca^2+b^3\right)\)
a) a3 (b-c) + b3 (c-a) +c3 (a-b)
<=> a3b – a3c +b3c – b3a + c3a – c3b
<=> b(a3 – c3) +c(a3 – b3) + a(b3 - c3)
(Tự áp dụng hằng đẳng thức)
b)
A) \(\left(x^2y-4\right)^2+4\left(x^2+y\right)^2\)
\(\Leftrightarrow x^4y^2-8x^2y+16+4\left(x^4+2x^2y+y^2\right)\)
\(\Leftrightarrow x^4y-8x^2y+16+4x^4+8x^2y+4y^2\)
\(\Leftrightarrow\left(x^4y^2+4x^4\right)+\left(16+4y^2\right)\)
\(\Leftrightarrow x^4\left(y^2+4\right)+4\left(y^2+4\right)\)
\(\Leftrightarrow\left(y^2+4\right)\left(x^4+4\right)\)
nếu thấy đúng thì tkkkk