giúp  mk vs 

hu...hu..hu

a: Sửa đề: \(-x^3-12x^2-48x-64\)

\(=-\left(x+4\right)^3\)

\(=-\left(-6+4\right)^3=-\left(-2\right)^3=-\left(-8\right)=8\)

b: \(=8x^3-y^3-8x^3+27y^3=26y^3=26\cdot\left(-3\right)^3=-702\)

c: \(=-\left(4x^4-12x^2y+9y^2\right)\)

\(=-\left(2x^2-3y\right)^2\)

\(=-\left(2x^2-2x-11\right)^2\)

Áp dụng bđt Svác xơ, ta có:

\(A\ge\dfrac{\left(\sqrt{2x}+\sqrt{3y}+\sqrt{4z}\right)^2}{2\left(4x^2+9y^2+16z^2\right)}\)\(=\dfrac{2x+3y+4z+2\left(\sqrt{6xy}+\sqrt{12yz}+\sqrt{8xz}\right)}{2}\)\(\ge\dfrac{1+2\left(3\sqrt[3]{\sqrt{576x^2y^2z^2}}\right)}{2}\)(BĐT Cô-si)\(\ge\dfrac{1+6}{2}=\dfrac{7}{2}\)

Vậy A_min=\(\dfrac{7}{2}\Leftrightarrow\)\(\left\{{}\begin{matrix}\dfrac{2x}{9y^2+16z^2}=\dfrac{3y}{4x^2+16z^2}=\dfrac{4z}{4x^2+9y^2}\\\sqrt{6xy}=\sqrt{12yz}=\sqrt{8xz}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{3}{2}y=2z\)

Viết lại bài toán: Cho \(a^2+b^2+c^2=1\). Tìm max \(\sum\dfrac{a}{b^2+c^2}\)

với a=2x, b=3y, c=4z.

Áp dụng BĐT AM-GM:

\(a\left(b^2+c^2\right)=\dfrac{1}{\sqrt{2}}\sqrt{2a^2\left(1-a^2\right)\left(1-a^2\right)}\le\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{8}{27}}=\dfrac{2}{3\sqrt{3}}\)

Do đó \(VT\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)

Vậy \(A_{Min}=\dfrac{3\sqrt{3}}{2}\)

\(\frac{1+3y}{12}=\frac{1+6y}{16}\)

<=> (1 + 3y).16 = (1 + 6y).12

<=> 16 + 48y = 12 + 72y

<=> 16 - 12 = 72y - 48y

<=> 24y = 4

<=> y = 1/6

Thay y = 1/6 vào đề bài ta được:

\(\frac{1+3.\frac{1}{6}}{12}=\frac{1+9.\frac{1}{6}}{4x}\)

\(\Leftrightarrow\frac{\frac{3}{2}}{12}=\frac{\frac{5}{2}}{4x}\Leftrightarrow6x=30\Leftrightarrow x=5\)

Vậy x = 5; y = 1/6

\(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)

\(\Leftrightarrow12\left(1+6y\right)=16\left(1+3y\right)\)

\(\Leftrightarrow3\left(1+6y\right)=4\left(1+3y\right)\)

\(\Leftrightarrow3+18y=4+12y\)

\(\Leftrightarrow3-4=12y-18y\)

\(\Leftrightarrow-1=-6y\)

\(\Rightarrow y=\frac{1}{6}\)

\(\Rightarrow\frac{1+3y}{12}=\frac{1+3.\frac{1}{6}}{12}=\frac{1+9.\frac{1}{6}}{4x}\)

\(\Leftrightarrow\frac{\frac{3}{2}}{12}=\frac{\frac{5}{2}}{4x}\)

\(\Leftrightarrow\frac{1}{8}=\frac{\frac{5}{2}}{4x}\)

\(\Rightarrow4x=8.\frac{5}{2}=20\)

\(\Rightarrow x=20:4=5\)

Vậy \(x=5\)