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b) Xét VP ta có :
\(\left(a+b+c\right)\cdot\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=a^3+ab^2+ac^2-ab^2-abc-ca^2+ba^2+b^3+bc^2-ab^2-bc^2-abc+ca^2+cb^2+c^3-abc-bc^2-c^2a\)
\(=a^3+b^3+c^3-abc-abc-abc\)
\(=a^3+b^3+c^3-3abc\)
\(=VT\)
Vậy đẳng thức đã được Cm
Câu hỏi của nguyen van quyen - Toán lớp 8 - Học toán với OnlineMath
a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)
B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
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 \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Bài 1:
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac+a^2+ab+ac+a^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\)
\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)
\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Bài 2:
Từ câu 1b ta đã chứng minh được:
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Thay a + b + c = 0 vào ta được
\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
a)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[a^2+b^2+c^2-ab-bc-ca\right]\)
\(=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
b/
\(a+b+c=0\Rightarrow c=-\left(a+b\right)\Rightarrow c^2=\left(a+b\right)^2\)
\(\Leftrightarrow c^2=a^2+b^2+2ab\)\(\Leftrightarrow a^2+b^2+ab=c^2-ab\)
\(2x^4=\left(a^2+b^2+ab\right)^2+\left(c^2-ab\right)^2\)
\(=a^4+b^4+a^2b^2+2a^2b^2+2a^3b+2ab^3+c^4-2abc^2+a^2b^2\)
\(=a^4+b^4+c^4+\left(4a^2b^2+2a^3b+2ab^3-2abc^2\right)\)
\(=a^4+b^4+c^4+2ab\left(2ab+a^2+b^2-c^2\right)\)
\(=a^4+b^4+c^4+0\)
\(=a^4+b^4+c^4\)
Có: \(a^3+b^3+c^3-3abc\)
\(=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\left(đpcm\right)\)