1. a3 + b3 + c3 ≥ a2 . căn (bc) + b2 .căn (ac) + c2 .căn (ab)
2. (a2 + b2 + c2)(1/(a +b ) + 1/(b+c) +1/(a+c) ) ≥ (3/2)(a + b+c)
3. a4 + b4 +c4 ≥ (a + b+c)abc
cho a,b,c thỏa mãn a2+b2+c2=4;a3+b3+c3=8
tính a4+b4+c4
Phân tích thành nhân tử :
a. (a + b)(a2 - b2) + (b - c)(b2 - c2) + (c + a)(c2 - a2)
b. a3 (b - c) + b3(c - a) + c3 (a - b)
c. a3 (c - b2) + b3 (a -c3) + c3 (b - a2) + abc(abc - 1)
d.a ( b + c )2 ( b - c ) + b ( c + a )2 (c - a ) + c ( a + b )2 (a - b )
e. a ( b + c )3 + b ( c - a )3 + c ( a - b )3
f. a2 b2 ( a - b ) + b2 c2 ( b - c ) + c2 a2( c - a )
g. a ( b2 + c2) + b ( c2 + a2 ) + c ( a2 + b2) - 2abc - a3 - b3 - c3
h. a4 ( b - c ) + b4 ( c - a ) + c4 ( a - b )
Phân tích thành nhân tử :
a. (a + b)(a2 - b2) + (b - c)(b2 - c2) + (c + a)(c2 - a2)
b. a3 (b - c) + b3(c - a) + c3 (a - b)
c. a3 (c - b2) + b3 (a -c3) + c3 (b - a2) + abc(abc - 1)
d.a ( b + c )2 ( b - c ) + b ( c + a )2 (c - a ) + c ( a + b )2 (a - b )
e. a ( b + c )3 + b ( c - a )3 + c ( a - b )3
f. a2 b2 ( a - b ) + b2 c2 ( b - c ) + c2 a2( c - a )
g. a ( b2 + c2) + b ( c2 + a2 ) + c ( a2 + b2) - 2abc - a3 - b3 - c3
h. a4 ( b - c ) + b4 ( c - a ) + c4 ( a - b )
a, a( b + c)2(b - c) + b( c + a)2( c - a) + c( a + b)2( a - b)
b, a( b - c )3 + b( c - a)3 + c( a - b)3
c, a2b2( a - b) + b2c2( b - c) + c2a2( c - a)
d, a( b2 + c2) + b( c2 + a2) + c( a2 + b2) - 2abc - a3 - b3 - c3
e, a4( b - c) + b4( c - a) + c4( a - b)
Cho a + b + c = 5 ; ab + bc + ca = 17 4 ; abc = 1. Tính 1) a2 + b2 + c2
2) a2b2 + b2c2 + c2a2
3) a3 + b3 + c3
4) a4 + b4 + c4
Nhanh lên mọi người mik còn phải gửi bài cho giáo viên mình nữa
1: Ta có: \(a^2+b^2+c^2\)
\(=\left(a+b+c\right)^2-2\cdot\left(ab+bc+ca\right)\)
\(=5^2-2\cdot174=-323\)
(1) (a+b+c)2=a2+b2+c2+2ab+2bc+2ac(a+b+c)2=a2+b2+c2+2ab+2bc+2ac
(2) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) (a−b−c)2=a2+b2+c2−2ab−2ac+2bc(a−b−c)2=a2+b2+c2−2ab−2ac+2bc
(4) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) a3−b3=(a−b)3+3ab(a−b)a3−b3=(a−b)3+3ab(a−b)
(6) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)
(8) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2(a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2
(10) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33
(12)ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Chứng minh giùm mik hằng đẳng thức kia vs
cho a,b,c là 3 số dương thỏa mãn: a+b+c=2019. Tìm GTNN : a3/a2+b2+ab + b3/b2+c2+bc + c3/c2+a2+ca
Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)
Ta có: \(\dfrac{a^3}{a^2+b^2+ab}=a-\dfrac{ab\left(a+b\right)}{a^2+b^2+ab}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^3b^3}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)
Tương tự: \(\dfrac{b^3}{b^2+c^2+bc}\ge\dfrac{2b-c}{3}\) ; \(\dfrac{c^3}{c^2+a^2+ca}\ge\dfrac{2c-a}{3}\)
Cộng vế:
\(P\ge\dfrac{a+b+c}{3}=673\)
Dấu "=" xảy ra khi \(a=b=c=673\)
cho (a+b+c)2=a2+b2+c2 và a,b,c ≠0. Chứng minh 1/a3+1/b3+1/c3=3/abc
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
=>\(2\left(ab+bc+ac\right)=0\)
=>ab+bc+ac=0
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)
=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)
=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)
=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)
=>0=0(đúng)
Cho a,b,c>0 và a+b+c=3. Tìm GTNN của
a) M= a2/a+1 + b2/b+1 + c2/b+1
b) N= 1/a + 4/b+1 + 9/c+2
c) P= a2/a+b + b2/b+c + c2/c+a
d)Q= a4 + b4 + c4 + a2 + b2 + c2 +2020
a) Áp dụng Cauchy Schwars ta có:
\(M=\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\ge\frac{\left(a+b+c\right)^2}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
b) \(N=\frac{1}{a}+\frac{4}{b+1}+\frac{9}{c+2}\ge\frac{\left(1+2+3\right)^2}{a+b+c+3}=\frac{36}{6}=6\)
Dấu "=" xảy ra khi: x=y=1
c) \(P=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{9}{2.3}=\frac{3}{2}\)
Dấu "=" xảy ra khi: x=y=1