Chứng minh
\(\left(a+b\right)^3=\left(a-b\right)\left(a^2+ab+b^2\right)-3ab\left(a-b\right)\)
Những câu hỏi liên quan
Chứng minh rằng:
a)\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
b)\(\left(a-b\right)^3+3ab\left(a-b\right)=a^3-b^3\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
Đọc tiếp
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
left(a+b+cright)^3-a^3-b^3-c^3left[left(a+bright)+cright]^3-a^3-b^3-c^3left(a+bright)^3+3left(a+bright)^2c+3left(a+bright)c^2+c^3-a^3-b^3-c^3a^3+3a^2b+3ab^2+b^3+3cleft(a^2+2ab+b^2right)+3ac^2+3bc^2-a^3-b^33a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^23left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abcright)3left[left(a^2b+ab^2right)+left(a^2c+abcright)+left(ac^2+bc^2right)+left(b^2c+abcright)right]3left[ableft(a+bright)+acleft(a+bright)+c^2left(a+bright)+bcleft(a+bright)right]3left(a+bright)left(ab+ac+c^2+bcrigh...
Đọc tiếp
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)
\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)
\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)
Châu ơi!đăng làm j z
Chứng minh rằng:
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)
=\(a^3+b^3+\left(a^3-b^3\right)\)
=\(a^3+b^3+a^3-b^3\)
=\(2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
=\(\left(a+b\right)\left(a^2-2ab+b^2-ab\right)\)
=\(\left(a+b\right)\left[\left(a^2-2ab+b^2\right)-ab\right]\)
=\(\left(a+b\right)\left[\left(a-b\right)^2-ab\right]\)
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a. \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=a^3+b^3+a^3-b^3=2a^3\)
b. \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
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Chứng minh rằng :
a) \(\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)=2a^3\)
b) \(a^3+b^3=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)
c) \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(a,\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a^2+ab+b^2\right)\)\(=\left(a^3+b^3\right)+\left(a^3-b^3\right)=2a^3\Rightarrowđpcm\)
\(b,\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left(a^2-2ab+b^2+ab\right)=\left(a+b\right)\left(a^2-ab+b^2\right)\)\(=\left(a^3+b^3\right)\Rightarrowđpcm\)
\(c,\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\Rightarrowđpcm\)
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a) (a+b)(a2-ab+b2)+(a-b)(a2+ab+b2)
= a3+b3+a3-b3 = 2a3
b) a3+b3
= (a+b)(a2-ab+b2)
= (a+b)(a2- 2ab+b2)+ab
= (a+b)(a2-b2)+ab
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a. Biến đổi vế trái:
(a+b)(a2−ab+b2)+(a−b)(a2+ab+b2)=a3+b3+a3−b3=2a3
=>VT bằng VP (đpcm)
b. Biến đổi vế phải:
(a+b)[(a−b)2+ab]=(a+b)[a2−2ab+b2+ab]
=(a+b)(a2−ab+b2)=a3+b3
=>VP bằng VT (đpcm)
c. Biến đổi vế phải:
(ac+bd)2+(ad−bc)2=a2c2+2abcd+b2d2+a2d2−2abcd+b2c2
=a2c2+b2d2+a2d2+b2c2=c(a2+b2)+d2(a2+b2)=(a2+b2)(c2+d2)
=>VP bằng VT (đpcm)
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Chứng minh
\(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b) \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Biến đổi VT ta có :
+) \(a^3+b^3+c^3=ab+bc+ca\)
\(\Leftrightarrow3a^3+3b^3+3c^3=3ab+3bc+3ca\)
\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=0\)
\(\Rightarrow a=b=c\)
< => VT = VP
=> đpcm
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\(VP=\left(a+b\right)^3-3ab\left(a+b\right)=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3=VT\)
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Phân tích thành nhân tử:
\(a)ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\\ b)a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\\ c)a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(a-b+1\right)\)
#)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) \(a^2\left(b-c\right)-b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
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chứng minh đẳng thức: \(a^3-b^3=\left(a-b^3\right)+\left(a-b\right)^3+3ab\left(a-b\right)\)
Cho ba số thực a, b, c. Chứng minh rằng:\(\left(a^2-bc\right)^3+\left(b^2-ca\right)^3+\left(c^2-ab\right)^3\ge3\left(a^2-bc\right)\left(b^2-ca\right)\left(c^2-ab\right)\)
Đặt \(\left\{{}\begin{matrix}a^2-bc=x\\b^2-ca=y\\c^2-ab=z\end{matrix}\right.\)
\(\Rightarrow x+y+z\ge0\)
\(\)Đẳng thức cần c/m trở thành: \(x^3+y^3+z^3\ge3xyz\left(1\right)\)
Áp dụng Bất đẳng thức AM-GM cho 3 số x,y,z, ta có:
\(x^3+y^3+z^3\ge3\sqrt[3]{x^3.y^3.z^3}=3xyz\)
=> Đẳng thức (1) luôn đúng với mọi x
Dấu = xảy ra khi: x=y=z hay \(a^2-bc=b^2-ca=c^2-ab\)
và \(a^2+b^2+c^2-\left(ab+bc+ca\right)=0\)\(\Rightarrow a=b=c\)
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