(a+b+c)3-a^3-b^3-c^3=3*(a+b)*(b+c)*(c+a)
Chúng minh hẳng dang thuc
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chung minh dang thuc sau;(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)
cho a,b,c,d la cac so thuc thoa ma dang thuc a+b+c+d=0.chung minh rang:
\(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
Ta có \(a+b+c+d=0\Leftrightarrow a+c=-\left(b+d\right)\Leftrightarrow\left(a+c\right)^3=\left[-\left(b+d\right)\right]^3\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-3b^2d-3bd^2-d^3\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2c-3ac^2-3b^2d-3bd^2\Leftrightarrow a^3+b^3+c^3+d^3=-3ac\left(a+c\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\Leftrightarrow a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)Vậy \(a+b+c+d=0\) thì \(a^3+b^3+c^3+d^3=3\left(b+d\right)\left(ac-bd\right)\)
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tim cac so nguyen a,b,c thoa man dong thoi 3 dang thuc
a*b*c+a=1333;a*b*c+b=1335;a*b*c+c=1341
Giup mink cau nay nha
1,chung minh dang thuc
-(-a+b+c)+(b-c-1)=(b-c+6)-(7-a+b)+c
2,Tim x thuoc Z
(x+1)+(x+3)+(x+5)+.....+(x+99)=0
(x-3)+(x-2)+(x-1)+...+10+11=0
\(-\left(-a+b+c\right)+\left(b-c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)
\(a-b-c+b-c-1=b-c+6-7+a-b+c\)
\(a-2c-1=a-1\)
\(-2c\ne0\)hay đẳng thức ko xảy ra
viet cac bieu thuc sau duoi dang tong cua 3 binh phuong
a)(a+b+c)\(^2\)+a\(^2\)+b\(^2\)+c\(^2\)
b)2(a-b)(c-b)+2(b-c)(c-a)+2(b-c)(a-c)
ai giai giup minh minh tich cho
tim cac so nguyen a;b;c thoa man dong thoi 3 dang thuc :a.b.c+a=2011;a.b.c+b=2013;a.b.c+c=2015
xac dinh cac so a b c de co dang thuc x^3-ax^2+bx-c=(x-a)(x-b)(x-c)
Gợi ý thôi.
\(x^3-ax^2+bx-c=\left(x-a\right)\left(x-b\right)\left(x-c\right)\)
\(\Rightarrow x^3-ax^2+bx-c\)có ba nghiệm \(x=a,x=b,x=c\)
Theo định lí Vi-et:\(\hept{\begin{cases}a+b+c=a\\ab+bc+ca=b\\abc=c\end{cases}\Leftrightarrow}\hept{\begin{cases}b=-c\\ab+bc+ca=b\\c\left(ab-1\right)=0\end{cases}}\)
okeee cam on ban
Cho ba so thuc khong am a,b,c thoa man a+b+c=3. Chung minh a^3+b^3+c^3+ab+ac+bc>=6
\(a^3+a^3+1\ge3a^2\Rightarrow a^3+\frac{1}{2}\ge\frac{3}{2}a^2\)
\(\Rightarrow VT+\frac{3}{2}\ge\frac{3}{2}a^2+\frac{3}{2}b^2+\frac{3}{2}c^2+ab+bc+ca\)
\(\Rightarrow VT+\frac{3}{2}\ge a^2+b^2+c^2+\frac{1}{2}\left(a+b+c\right)^2\)
\(\Rightarrow VT+\frac{3}{2}\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{2}\left(a+b+c\right)^2=\frac{15}{2}\)
\(\Rightarrow VT\ge\frac{15}{2}-\frac{3}{2}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Sau khi đưa BĐT về dạng thuần nhất ta có:
\(VT-VP=\frac{1}{18} \sum\limits_{cyc} (7a+7b+c)(a-b)^2 \geq 0\)
cho a,b,c la cac so thuc duong chung minh \(a^3/(b+3 c)^3+b^3/(c+3 a)^3 + c^3/(a+3b)^3 >+3/64 \)