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Câu hỏi của Access_123 - Toán lớp 8 - Học toán với OnlineMath
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)+b\left[\left(c^3-b^3\right)-\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(b^2+ac+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(b^2+ac+c^2-a^2-ab-b^2\right)\)
Đặt a+b-c=x;c+a-b=y;b+c-a=z
=>x+y+z=a+b-c+a+b-c+b+c-a=a+b+c
Ta có hăng đẳng thức:(x+y+z)3-x3-y3-z3=3(x+y)(y+z)(x+z)
=>(a+b+c)3-(a+b-c)3-(c+a-b)3-(b+c-a)3
=(x+y+z)3-x3-y3-z3
=3(x+y)(y+z)(z+x)
=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)
=3.2a.2c.2b
=24abc
\(A=a\left[\left(b-c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]+4abc\)
\(=a\left(b-c+a\right)\left(b-c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)+4abc\)
\(=\left(a+b-c\right)\left(ab-ac-a^2-bc+ab-b^2\right)+c\left(a^2-2ab+b^2-c^2+4ab\right)\)
\(=\left(a+b-c\right)\left[-c\left(a+b\right)-\left(a-b\right)^2\right]+c\left[\left(a+b\right)^2-c^2\right]\)
\(=\left(a+b-c\right)\left(-ca-cb-a^2+2ab-b^2+ac+cb+c^2\right)\)
\(=\left(a+b-c\right)\left(c^2-\left(a-b\right)^2\right)\)
\(=\left(a+b-c\right)\left(c+a-b\right)\left(a+b-c\right)\)
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=\left(a\left(b-c\right)\left(b^2+bc+c^2\right)\right)+\left(b\left(c-a\right)\left(c^2+ac+a^2\right)\right)+\left(c\left(a-b\right)\left(a^2+ab+b^2\right)\right)\)
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(\text{Phân tích đa thức thành nhân tử :}\)
\(-\left(ac^3-b^4c+a^3bc+\left(-a\right)b^3+b^3-a^3\right)\)
\(\text{Chúc bạn học tốt !}\)
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(\text{Phân tích đa thức thành nhân tử :}\)
\(-\left(ac^3-b^4c+a^3bc+\left(-a\right)b^3+b^3-a^3\right)\)
mình và bạn be be be là 1
= (a + b)3 + c3 + 3(a + b)c.(a + b + c)] - [(a + b)3 - c3 - 3(a+ b)c.(a + b - c)] - [c3 + (a - b)3 + 3c(a - b).(c + a - b)] - [c3 - (a - b)3 - 3c(a - b)(c - a + b)]
= 2.c3 + 3(a + b)c(a + b + c) + 3(a + b)c(a + b - c) - 2c3 - 3c(a - b)(c + a - b) + 3c(a - b)(c - a + b)
= 3(a+ b)2c + 3c2(a+ b) + 3(a+ b)2c - 3c2(a+ b) - 3c2(a - b) - 3c(a - b)2 + 3c2(a - b) - 3c(a - b)2
= 3(a + b)2c - 3c(a - b)2 = 3c.[(a + b)2 - (a - b)2] = 3c(a + b + a - b)(a + b- a + b) = 3c.2a.2b = 12abc
Ta có (a+b+c)3-(a+b-c)3-(c+a-b)3-(c-a+b)3
= (a + b)3 + c3 + 3(a + b)c.(a + b + c)] - [(a + b)3 - c3 - 3(a+ b)c.(a + b - c)] - [c3 + (a - b)3 + 3c(a - b).(c + a - b)] - [c3 - (a - b)3 - 3c(a - b)(c - a + b)]
= 2.c3 + 3(a + b)c(a + b + c) + 3(a + b)c(a + b - c) - 2c3 - 3c(a - b)(c + a - b) + 3c(a - b)(c - a + b)
= 3(a+ b)2c + 3c2(a+ b) + 3(a+ b)2c - 3c2(a+ b) - 3c2(a - b) - 3c(a - b)2 + 3c2(a - b) - 3c(a - b)2
= 3(a + b)2c - 3c(a - b)2
= 3c.[(a + b)2 - (a - b)2]
= 3c(a + b + a - b)(a + b- a + b)
= 3c.2a.2b
= 12abc
hok giỏi