\(\left(-\frac{1}{3}ab^3-2a^3b\right)^3\)

\(=\left(-\frac{1}{3}ab^3\right)-3\left(-\frac{1}{3}ab^3\right)^2.2a^3b+3\left(-\frac{1}{3}ab^3\right)\left(2a^3b\right)^2-\left(2a^3b\right)^3\)

\(=-\frac{1}{27}a^3b^9+\frac{2}{3}a^5b^7-4a^7b^5-8a^9b^3\)

\(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=\left(2x-y\right)((2x)^2+2xy+y^2-\left(2x+y\right)((2x)^2-2xy+y^2\)

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

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

\(=-2y^3\)

à..........cái đó thì mình chưa tính ra được

(2x+y)²=(2x)²+2.2x.y+y2

=4x²+4xy+y²

\(H=\frac{-1}{3}x^2-5x+1\)

\(=\frac{-1}{3}\left(x^2+15x-3\right)\)

\(=\frac{-1}{3}\left(x^2+2x.\frac{15}{2}+\frac{225}{4}-\frac{237}{4}\right)\)

\(=\frac{-1}{3}\left(x+\frac{15}{2}\right)^2+\frac{79}{4}\le\frac{79}{4}\forall x\)

Dấu '' = '' xảy ra khi: \(\left(x+\frac{15}{2}\right)^2=0\Rightarrow x+\frac{15}{2}=0\Rightarrow x=\frac{-15}{2}\)

Vậy \(MaxH=\frac{79}{4}\) khi \(x=\frac{-15}{2}\)

\(L=-3x^2+6x-y^2+6y-12\)

\(=\left(-3x^2+6x-3\right)+\left(-y^2+6y-9\right)\)

\(=-3\left(x^2-2x+1\right)-\left(y^2-6y+9\right)\)

\(=-3\left(x-1\right)^2-\left(y-3\right)^2\le0\forall x;y\)

Dấu '' = '' xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\\\left(y-3\right)^2=0\Rightarrow y-3=0\Rightarrow y=3\end{cases}}\)

Vậy \(MaxL=0\) khi \(\hept{\begin{cases}x=1\\y=3\end{cases}}\)

???

?????????????////////////////////////////////////

mik có cách giải rồi

xy+yz+xz>=\(3\sqrt[3]{xyz^2}\) mà xyz=1

suy ra xy+yz+xz>=3 mà xy+yz+xz=3

suy ra 3=3 suy ra dấu "=" xảy ra suy ra x=y=z

A=3x/3x³=1/x²

b, x.y².z³=64 mà x=y=z

nên x⁶=64 suy ra x=2

thay x=2 vào A=3x/3x³=1/x²

A=1/4

à..........cái đó thì mình chưa tính ra được

a) ĐKXĐ : \(\hept{\begin{cases}a\ne0\\a\ne-1\\a\ne1\end{cases}}\)

Khi đó P = \(\left[\frac{2}{3a}-\frac{2}{a+1}\left(\frac{a+1}{3a}-a-1\right)\right]:\frac{a-1}{a}\)

\(=\left[\frac{2}{3a}-\frac{2}{a+1}.\frac{a+1}{3a}+\frac{2}{a+1}.\left(a+1\right)\right]:\frac{a-1}{a}\)

\(=\left(\frac{2}{3a}-\frac{2}{3a}+2\right):\frac{a-1}{a}=2:\frac{a-1}{a}=\frac{2a}{a-1}\)

b) Ta có P = \(\frac{2a}{a-1}=\frac{2a-2+2}{a-1}=2+\frac{2}{a-1}\)

\(P\inℤ\Leftrightarrow2⋮a-1\Leftrightarrow a-1\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)

<=> \(a\in\left\{2;3;0;-1\right\}\)

c) Để P \(\le1\)

<=> \(\frac{2a}{a-1}\le1\)

<=> \(\frac{a+1}{a-1}\le0\)

Xét 2 trường hợp 

TH1 : \(\hept{\begin{cases}a+1\ge0\\a-1\le0\end{cases}}\Leftrightarrow-1\le a\le1\)

Kết hợp điều kiện => -1 < a < 1 (a \(\ne0\))

TH2 : \(\hept{\begin{cases}a+1\le0\\a-1\ge0\end{cases}}\Leftrightarrow a\in\varnothing\)

Vậy - 1 < a < 1 (a \(\ne0\))