Lời giải:
BĐT cần chứng minh tương đương với:

$18a^2+3b^2+7c^2+18-16ac+6bc-12a\geq 0$

$\Leftrightarrow (16a^2-16ac+4c^2)+3(b^2+2bc+c^2)+2(a^2-6a+9)\geq 0$

$\Leftrightarrow (4a-2c)^2+3(b+c)^2+2(a-3)^2\geq 0$

(luôn đúng với mọi $a,b,c$ thực)

Do đó ta có đpcm.

a: \(\frac{x}{y}=\frac13\)

=>y=3x

\(\frac{14x+5y}{3x-11y}=\frac{11x+5\cdot3x}{3x-11\cdot3x}=\frac{26x}{3x-33x}=\frac{26x}{-30x}=\frac{-26}{30}=-\frac{13}{15}\)

b: \(\frac{a}{b}=\frac12\)

=>b=2a

\(\frac{11a^4-3ab^3+15a^3b+7b^4}{3a^2b^2+ab^3-6a^3b-2b^4}\)

=\(\frac{11a^4-3a\cdot\left(2a\right)^3+15a^3\cdot2a+7\left(2a\right)^4}{3a^2\cdot\left(2a\right)^2+a\cdot\left(2a\right)^3-6a^3\cdot2a-2\cdot\left(2a\right)^4}\)

\(=\frac{11a^4-3a\cdot8a^3+30a^4+7\cdot16a^4}{3a^2\cdot4a^2+a\cdot8a^3-6a^3\cdot2a-2\cdot16a^4}\)

\(=\frac{11a^4-24a^4+30a^4+112a^4}{12a^4+8a^4-12a^4-32a^4}=\frac{11-24+30+112}{12+8-12-32}=\frac{129}{-24}=\frac{-43}{8}\)

mọi người giải giúp em bài này với 

a³ - 3a²+ 5a – 17 = 0   ,   b³ - 3b² + 5b + 11 = 0   .   Tính a+b

a) dk: \(\left\{{}\begin{matrix}a,d\ne0\\5a\ne3b\\5c\ne3d\end{matrix}\right.\) \(VT=\dfrac{5a+3b}{5a-3b}=\dfrac{5.\dfrac{a}{b}+3}{5\dfrac{a}{b}-3}=\dfrac{5.\dfrac{c}{d}+3}{5\dfrac{c}{d}-3}=\dfrac{\dfrac{5c+3d}{d}}{\dfrac{5c-3d}{d}}=\dfrac{5c+3d}{d}.\dfrac{d}{5c-3d}=\dfrac{5c+3d}{5c-3d}=VP\)

b)

\(\left\{{}\begin{matrix}b,d\ne0\\11a^2\ne8b^2\\11c^2\ne8d^2\end{matrix}\right.\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\right)\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7.\dfrac{a^2}{b^2}+3\dfrac{a}{b}}{11\dfrac{.a^2}{b^2}-8}=\dfrac{7.\dfrac{c^2}{d^2}+3\dfrac{c}{d}}{11\dfrac{.c^2}{d^2}-8}=\dfrac{7c^2+3cd}{11c^2-8d^2}=VP\)