Đây nhé! Tích giúp c nha

1: Ta có: \(a^2+b^2+c^2\)

\(=\left(a+b+c\right)^2-2\cdot\left(ab+bc+ca\right)\)

\(=5^2-2\cdot174=-323\)

ké ý (b) ạ!!!

a: \(\left(a+b\right)\left(a^2-b^2\right)+\left(b-c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)

\(=a^3-ab^2+a^2b-b^3+b^3-bc^2-b^2c+c^3+\left(c+a\right)\left(c^2-a^2\right)\)

\(=a^3+c^3-ab^2-b^2c+a^2b-bc^2+\left(c+a\right)\left(c+a\right)\left(c-a\right)\)

\(=\left(c+a\right)\left(c^2-ac+a^2\right)-b^2\left(c+a\right)-b\left(c-a\right)\left(c+a\right)+\left(c+a\right)^2\cdot\left(c-a\right)\)

=(c+a)\(\left(c^2-ac+a^2-b^2-bc+ba+c^2-a^2\right)\)

=(c+a)\(\left(2c^2-2a^2-b^2-ac-bc+ba\right)\)

b: \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)+b^3\left(c-b+b-a\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)

\(=\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\)

=(b-c)(a-b)\(\left(a^2+ab+b^2-b^2+bc-c^2\right)\)

=(b-c)(a-b)\(\left(a^2+ab+bc-c^2\right)\)

=(b-c)(a-b)\(\left\lbrack\left(a-c\right)\left(a+c\right)+b\left(a+c\right)\right\rbrack\)

=(b-c)(a-b)(a+c)(a-c+b)

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

a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)

b) Ta có:

\(a^3+b^3+c^3-3abc\)

\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)

Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi

a) Ta có: (a + b + c)³ - a³ - b³ - c³ = [ (a + b + c)³ - a³ ] - ( b³ + c³)

= (a + b + c - a) ( a² + b² + c² + 2ab + 2bc + 2ac + a² + ab + ac + a²) - (b + c) ( b² - bc + c³)

= (b + c) ( 3a² + b² + c² + 3ab + 2bc + 3ac) - (b + c) ( b² - bc + c³)

= ( b + c) ( 3a² + b² + c² + 3ab + 2bc + 3ac - b² + bc - c³)

= ( b + c) ( 3a² + 3ab + 3bc + 3ac)

= 3 (b + c) [a (a + b) + c (a + b)]

= 3 (b + c) (a + b) (a + c) (đpcm)