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a) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2+bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)
b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24
(a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ba+b^2
(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2
(a+3)^3=(a+b)^2*(a+b)
=(a^2+2ab+b^2)(a+b)
=a^3+a^2b+2a^2b+2ab^2+b^2a+b^3
=a^3+3a^2b+3ab^2+b^3
a) \(\dfrac{a^2+a+1}{a^2-a+1}=\dfrac{\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Thấy tử và mẫu của phân số đều lớn hơn 0 => \(\dfrac{a^2+a+1}{a^2-a+1}>0\)
b)\(a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2a+1\right)+\left(c^2-2a+1\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (luôn đúng với mọi a,b,c)
Dấu = xra khi a=b=c=1
b)
\(a^2-2a+1+b^2-2b+1+c^2-2c+1\ge0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) ( Luôn đúng)
Dấu "=" xảy ra khi a=b=c=1
a) \(\left(A+B\right)^2=\left(A+B\right)\left(A+B\right)=A^2+AB+AB+B^2=A^2+2AB+B^2\)
b) \(\left(A+B\right)^3=\left(A+B\right)^2\left(A+B\right)=\left(A^2+2AB+B^2\right)\left(A+B\right)\)( NHÂN ra nốt hộ mk nha ) :D !
c)\(\left(A+B\right)\left(A-B\right)=A^2+AB-AB-B^2=A^2-B^2\)
ý d tương tự nha :D !
1) a3+b3+c3-3abc = (a+b)3-3ab(a+b)+c3-3abc
= (a+b+c)(a2+2ab+b2-ab-ac+c2) -3ab(a+b+c)
= (a+b+c)( a2+b2+c2-ab-bc-ca)
Áp dụng bất đẳng thức Cosi, ta có:
\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)Do đó, để chứng minh bất đẳng thức đã cho, ta chỉ cần chứng minh rằng:
\(\frac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\le\sqrt{3}\)
Áp dụng bất đẳng thức Côsi lần thứ hai ta nhận được:
\(VT=\frac{\sqrt{a}\sqrt{a\left(1+b+c\right)}+\sqrt{b}\sqrt{b\left(1+c+a\right)}+\sqrt{c}\sqrt{c\left(1+a+b\right)}}{a+b+c}\)
\(\le\frac{\sqrt{\left(a+b+c\right)\left[a\left(1+b+c\right)+b\left(1+c+a\right)+c\left(1+a+b\right)\right]}}{a+b+c}\)
\(=\sqrt{1+\frac{2\left(ab+bc+ca\right)}{a+b+c}}\)
\(\le\sqrt{1+\frac{2\left(a+b+c\right)}{3}}\)
\(\le\sqrt{1+\frac{2\sqrt{3\left(a^2+b^2+c^2\right)}}{3}}=\sqrt{3}\left(đpcm\right)\)
Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.
a) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(a-b\right)^2.\left(a+b\right)^2\)( đpcm )
b) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)
\(-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3-\left(a-c\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)
\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)( đpcm )
1) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2\)
\(=a^4+2a^2b^2+b^4-4a^2b^2\)
\(=a^4-2a^2b^2+b^4\)
\(=\left(a^2-b^2\right)^2\)
\(=\left[\left(a-b\right)\left(a+b\right)\right]^2\)
\(=\left(a-b\right)^2\left(a+b\right)^2\)
2) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=\left(a-b+b-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\)
\(=\left(a-c\right)\left(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2\right)+\left(c-a\right)^3\)
\(=-\left(c-a\right)\left(a^2+3b^2+c^2-3ab+ac-3bc\right)+\left(c-a\right)\left(c^2-2ca+a^2\right)\)
\(=\left(c-a\right)\left(c^2-2ca+a^2-a^2-3b^2-c^2+3ab-ac+3bc\right)\)
\(=\left(c-a\right)\left(3ab+3bc-3b^2-3ac\right)\)
\(=3\left(c-a\right)\left(ab-b^2-ac+bc\right)\)
\(=3\left(c-a\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)