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1 tháng 7 2017
\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)=\left(c-a\right)\left(c-b\right)\left(b-a\right)\)
7 tháng 8 2016
\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=a^3-ab^2+a^2b-b^3+b^3-bc^2+b^2c-c^3+c^3-a^2c+ac^2-a^3\)
\(=-ab^2+a^2b-bc^2+b^2c-a^2c+ac^2\)
\(=\left(a^2b-ab^2\right)+\left(ac^2-bc^2\right)-\left(a^2c-b^2c\right)\)
\(=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)
\(=\left(a-b\right)\left[\left(ab-ac\right)+\left(c^2-bc\right)\right]\)
\(=\left(a-b\right)\left[a\left(b-c\right)+c\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
18 tháng 8 2019
\(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left[\left(a-b\right)+\left(c-a\right)\right]+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)+c^2a^2\left(c-a\right)-b^2c^2\left(c-a\right)\)
\(=\left(a-b\right)b^2\left(a-c\right)\left(a+c\right)+\left(c-a\right)c^2\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(ab^2+cb^2-c^2a-c^2b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+ac+bc\right)\)
DD
Đoàn Đức Hà
Giáo viên
27 tháng 5 2021
Bài 1:
\(a^2\left(b-2c\right)+b^2\left(c-a\right)+2c^2\left(a-b\right)+abc\)
\(=2c^2\left(a-b\right)+a^2b-ab^2+b^2c-a^2c+abc-a^2c\)
\(=2c^2\left(a-b\right)+ab\left(a-b\right)-c\left(a+b\right)\left(a-b\right)-ac\left(a-b\right)\)
\(=\left(a-b\right)\left(2c^2+ab-ac-cb-ac\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-2c\right)\)
DD
Đoàn Đức Hà
Giáo viên
27 tháng 5 2021
Bài 2:
\(x^2+3x+1=0\Leftrightarrow x+\frac{1}{x}=-3\)(vì \(x=0\)không là nghiệm)
Ta có:
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right).x.\frac{1}{x}=-3^3-3.\left(-3\right)=-18\)
\(x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2=\left[\left(x+\frac{1}{x}\right)^2-2\right]^2-2=47\)
\(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)=x^7+\frac{1}{x^7}+x+\frac{1}{x}\)
\(\Leftrightarrow x^7+\frac{1}{x^7}=\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)=-18.47-\left(-3\right)=-843\)
CM
1
23 tháng 10 2019
\(\left(x^2+4x+6\right)\left(x^2+6x+6\right)-3x^2\left(1\right)\)
Đặt \(x^2+5x+6=t\)Thay vào (1) ta được:
\(\left(t-x\right)\left(t+x\right)-3x^2\)
\(=t^2-x^2-3x^2\)
\(=t^2-4x^2\)
\(=\left(t-2x\right)\left(t+2x\right)\)Thay \(t=x^2+5x+6\)ta được:
\(\left(x^2+5x+6-2x\right)\left(x^2+5x+6+2x\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+x+6x+6\right)\)
\(=\left(x^2+3x+6\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x^2+3x+6\right)\left(x+1\right)\left(x+6\right)\)
T
31 tháng 7 2019
#)Giải :
a)\(ab\left(b-a\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=a\left(a-b\right)+b^2c-bc^2+ac^2-a^2c\)
\(=ab\left(a-b\right)-\left(a-b\right)\left(a+b\right)c+c^2\left(a-b\right)\)
\(=\left(ab-ac-bc+c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
b) \(a^2\left(b-c\right)-b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)