Sửa đề: Tìm n để A nguyên

Để A nguyên thì n-3+5 chia hết cho n-3

=>n-3 thuộc {1;-1;5;-5}

=>n thuộc {4;2;8;-2}

\(\Leftrightarrow x^3-3x^2+3x-1-2x+3x^2-2+6x=-3\)

\(\Leftrightarrow x^3+7x-5=0\)

27:(x-3/2)^3=(x-3/2):3

Ta có: \(\dfrac{27}{\left(x-\dfrac{3}{2}\right)^3}=\dfrac{\left(x-\dfrac{3}{2}\right)}{3}\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^3.\left(x-\dfrac{3}{2}\right)\)=27.3

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4\)=81

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^4=3^4\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}+\dfrac{3}{2}\\x=\dfrac{-8}{2}+\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)

Vậy x∈\(\left\{\dfrac{11}{2};\dfrac{-5}{2}\right\}\)

\(\Leftrightarrow2x^3-3x^2+x+a=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow2\left(-27\right)-3\cdot9-3+a=0\\ \Leftrightarrow-54-27-3+a=0\\ \Leftrightarrow-84+a=0\\ \Leftrightarrow a=84\)

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)

Vì \(x\le3\Rightarrow\dfrac{2}{\sqrt{x}}\ge\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow-\dfrac{2}{\sqrt{x}}\le-\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{3}}\le1-\dfrac{2}{\sqrt{3}}\)

\(\Rightarrow\)\(P\le\dfrac{3-2\sqrt{3}}{3}\)

Dấu = xra khi x=3

Vậy \(P_{max}=\dfrac{3-2\sqrt{3}}{3}\)

a = 12
7a6b = 7161

ta có 7+a+6+b =13+a+b chia hết cho 3=>a+b ko chia hết cho 3 mà a-b=4(a,b là số có 1 chữ số)

=>a=8;b=4

ta có 7+a+6+b =13+a+b chia hết cho 3=>a+b ko chia hết cho 3 mà a-b=4(a,b là số có 1 chữ số)

=>a=9;b=5

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)