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Câu 1:
a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
b: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
a) \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)\left(ac+bc+c^2\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
b) \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)
a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
Lời giải:
Ta có:
$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc$
$=(a+b)^3+c^3-3ab(a+b+c)$
$=(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)$
$=(a+b+c)[(a+b)^2-c(a+b)+c^2-3ab]=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
$=\frac{1}{2}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ac)$
$=\frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
$=\frac{1}{2}(a+b+c).6abc=3abc(a+b+c)$
$\Rightarrow a^3+b^3+c^3=3abc(a+b+c+1)$ (đpcm)
Câu 1:
a) a(a+2b)3 - b(2a+b)3 = a( a3 + 6a2b + 12ab2 + 8b2) - b
= a( a3 + 6a2b + 12ab2 + 8b3) - b( 8a3 + 12a2b + 6ab2 + b3)
= a4 + 6a3b + 12a2b2 + 8ab3 - 8a3b -12a2b2 - 6ab3 - b4
= a4 - 2a3b + 2ab3 - b4
= (a - b )(a + b)(a2 +b2) - 2ab(a - b)(a + b)
= (a - b )(a + b)(a2 +b2 -2ab)
= (a - b )3(a + b)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)[\left(a+b+c\right)^2-3ab-3ac-3bc]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right).2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right)[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)]\)
\(=\frac{1}{2}\left(a+b+c\right)[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2]\)
Sửa đề : CM \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Ta có : \(VT=a^3+b^3+c^3-3abc\)
\(=\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3a^2b-3b^2a-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VP\)
\(\left(đpcm\right)\)
a. \(2\left(a^2+b^2\right)=\left(a-b\right)^2\)
\(\Leftrightarrow2a^2+2b^2=a^2+b^2-2ab\)
\(\Leftrightarrow a^2+b^2=-2ab\)
\(\Leftrightarrow a^2+2ab+b^2=0\)
\(\Leftrightarrow\left(a+b\right)^2=0\)
\(\Leftrightarrow a+b=0\Leftrightarrow a=-b\) (đpcm)
b. \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Vì \(\left(a-1\right)^2;\left(b-1\right)^2;\left(c-1\right)^2\ge0\)
\(\Rightarrow\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)
\(\Leftrightarrow a-1=b-1=c-1=0\Leftrightarrow a=b=c=1\)
c. \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tương tự câu b ta có a = b = c
Xét vế trái:
\(2\left(a^3+b^3+c^3-3abc\right)\)
\(=2\left[\left(a^3+b^3\right)+c^3-3abc\right]\)
\(=2\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\right]\)
\(=2\left\{\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]\right\}\)
\(=2\left\{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\right\}\)
\(=2\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc-c^2-3ab\right)\)
\(=2\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)\)
\(=\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\right]\)
\(=\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\left(đpcm\right)\)
Chúc bạn học tốt!
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b\right)-3abc\)\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrow2\left(a^3+b^3+c^3-3abc\right)=\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\)\(\Rightarrow2\left(a^3+b^3+c^3-3abc\right)=\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\left(đpcm\right)\)