Ta có:

a³b³ + 2b³c³ + 3a³c³

=a³b³ -b³c³ + 3b³c³ + 3a³c³

= b³ ( a³ - c³ ) +3c³ (b³ + a³ )

= b³ (-b³ - 2c³ )  +3c³ ( -c³)

= -b⁶ - 2 b³ c³  - 3 c⁶ \(\le\)0

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

Ta có :

$a^3{b}^3+2b^3{c}^3+3a^3{c}^3=b^3\left(a^3+2c^3\right)+3a^3{c}^3$

Từ $a^3+b^3+c^3=0\Rightarrow a^3+2c^3=c^3-b^3$, thì:

$b^3(c^3-b^3)+3a^3{c}^3=-b^6+c^3(b^3+3a^3)$

Và từ $a^3+b^3+c^3=0\Rightarrow b^3+3a^3=2a^3-c^3$

Suy ra $-b^6+c^3(2a^3-c^3)=-(b^3-c^3{)}^2\le 0$

bạn ơi viết khó nhìn quá

Đặt ab = x, bc = y, ca = z     (x, y, z ≠ 0 thỏa mãn x^3 + y^3 + z^3 = 3xyz)

⇔ (x+y)^3 − 3xy(x + y) + z^3 = 3xyz <=> (x+y)^3 − 3xy(x + y) + z^3 = 3xyz

⇔ (x + y)^3 + z^3 − 3xy(x + y+ z) = 0 ⇔ (x + y)^3 + z^3 − 3xy(x + y + z) = 0

⇔ (x + y + z)[(x + y)^2 − z (x + y) + z^2] − 3xy(x + y + z) = 0 ⇔ (x + y + z)[(x + y)^2 − z(x + y) + z2] − 3xy(x + y + z) = 0

⇔ (x + y + z)(x^2 + y^2 + z^2 − xy − yz − xz) = 0 ⇔ (x + y + z)(x^2 + y^2 + z^2 − xy − yz − xz) = 0

<=> x + y + z = 0   (1)        và           x^2 + y^2 + z^2 − xy − yz − xz = 0   (2)

Với (1): ⇔ ab + bc + ac = 0 ⇔ ab + bc + ac = 0

P = (1 + a/b)(1 + b/c)(1 + c/a) = (a + b)(b + c)(c + a)/abc=(ab + bc + ac)(a + b + c) − abc/abc = 0 − abc/abc = −1

Với (2) ⇔ (x − y)^2 + (y − z)^2 + (z − x)^2/2 = 0

⇔ (x − y)^2 + (y − z)^2 + (z − x)^2 = 0 

Ta thấy (x − y)^2; (y − z)^2; (z − x)^2 ≥ 0 ∀x, y, z nên để tổng của chúng bằng 0 thì:

(x − y)^2 = (y − z)^2 = (z − x)^2 = 0 ⇒ x = y = z

⇔ ab = bc = ac ⇔ a=b=c (do a, b, c ≠ 0)

⇒ A = (1 + 1)(1 + 1)(1 + 1) = 8 

Vậy...........

đề: \(a^3b^3+2b^3c^3+3c^3a^3\le0\) đúng ko?

\(L=a^3b^3+2b^3c^3+3c^3a^3\)

\(L=a^3b^3+2b^3c^3+2c^3a^3+c^3a^3\)

\(L=a^3\left(b^3+c^3\right)+2c^3\left(a^3+b^3\right)\)

Vì \(a^3+b^3+c^3=0\) nên: \(\left\{{}\begin{matrix}b^3+c^3=-a^3\\a^3+b^3=-c^3\end{matrix}\right.\)

nên: \(L=-a^6-2c^6\le0\)

Dấu "=" khi \(a^3=b^3=c^3=0\Leftrightarrow a=b=c=0\)