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a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
1,\(\left(a^2+b^2-c^2\right)^2\)\(-\left(a^2-b^2+c^2\right)^2\)
\(=\left(a^2+b^2-c^2\right)^2\)\(-\left(b^2-a^2-c^2\right)^2\)
\(=\left(a^2+b^2-c^2-b^2+a^2+c^2\right)\)\(\left(a^2+b^2-c^2+b^2-a^2-c^2\right)\)
\(=2a^2\left(2b^2-2c^2\right)\)
\(=4a^2b^2-4a^2c^2\)
\(=\left(2ab-2ac\right)\left(2ab+2ac\right)\)
2,\(\left(a+b+c\right)^2\)\(+\left(a+b-c\right)^2\)\(-2\left(a+b\right)^2\)
\(=\left(\left(a+b+c\right)^2-\left(a+b\right)^2\right)\)\(+\left(\left(a+b-c\right)^2-\left(a+b\right)^2\right)\)
\(=\left(a+b+c-a-b\right)\)\(\left(a+b+c+a+b\right)+\)\(\left(a+b-c-a-b\right)\)\(\left(a+b-c+a+b\right)\)
\(=c\left(2a+2b+c\right)\)\(-c\left(2a+2b-c\right)\)
\(=c\left(2a+2b+c-2a-2b+c\right)\)
\(=c.2c\)
\(=2c^2\)
a) \(x^4-x^2+3=\left[\left(x^2\right)^2-2\cdot x^2\cdot\frac{1}{2}+\frac{1}{4}\right]+\frac{11}{4}=\left(x^2-\frac{1}{2}\right)^2+\frac{11}{4}>0\)
=>đpcm
b) \(x^2-x+1=\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
=>đpcm
c) \(x^2+x+2=\left(x^2+2\cdot x+\frac{1}{2}+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=>đpcm
d) \(\left(x+3\right)\left(x-11\right)+20\)
\(=x^2-11x+3x-33+20\)
\(=x^2-8x-13\)
\(=\left(x^2-8x+16\right)-29=\left(x+4\right)^2-29\)
Xem lại đề
Từ đề bài,ta suy ra\(\Rightarrow p^2-2pa+a^2+p^2-2pb+p^2-2pc+c^2\)
\(=a^2+b^2+c^2+3q^2-2p\left(a+b+c\right)=a^2+b^2+c^2+3q^2-4p^2\)(vì a+b+c=2p)
\(=a^2+b^2+c^2-p^2\)(ĐIỀU PHẢI CHỨNG MINH)