Các đơn thức là :

\(\left(1-\dfrac{1}{\sqrt[]{3}}\right)x^2;x^2.\dfrac{7}{2}\)

Chọn D

Bài 1.

\(\left(x^2-x-6\right)\left(x^2-5\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

Mà \(x\in Q\)

\(\Rightarrow x=\left\{-2;3\right\}\)

Pt\(\Leftrightarrow\)\(\left[{}\begin{matrix}x^2-x-6=0\\x^2-5=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}3\\-2\\-\sqrt{5}\\\sqrt{5}\end{matrix}\right.\)

   Đáp án A

a/ \(=\lim\limits_{x\rightarrow-\infty}x^2\left(1+\dfrac{x}{x^2}-\dfrac{1}{x^2}\right)=+\infty\)

b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x}+\lim\limits_{x\rightarrow+\infty}2.\dfrac{x^2-x^2+x}{\sqrt{x^2-x}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}+\dfrac{x}{x}}+2\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{x}{x^2}}+\dfrac{x}{x}}=\dfrac{1}{2}+\dfrac{2}{2}=\dfrac{3}{2}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\dfrac{x^2+2x-x^2}{\sqrt{x^2+2x}+x}+2.\dfrac{x^2-x^2-x}{\sqrt{x^2+x}+x}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{2x^2}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{2x}{x^2}+\dfrac{x}{x^2}}}+2\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{x^2}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}=0\)

a: 3x+4⋮x-3

=>3x-9+13⋮x-3

=>13⋮x-3

=>x-3∈{1;-1;13;-13}
=>x∈{4;2;16;-10}

b: \(x^2+3x-13\vdots x+3\)

=>x(x+3)-13⋮x+3

=>-13⋮x+3

=>x+3∈{1;-1;13;-13}

=>x∈{-2;-4;10;-16}

c: \(x^2+3\vdots x-1\)

=>\(x^2-x+x-1+4\vdots x-1\)

=>4⋮x-1

=>x-1∈{1;-1;2;-2;4;-4}

=>x∈{2;0;3;-1;5;-3}

d: \(x^2+2x-11\vdots x+2\)

=>x(x+2)-11⋮x+2

=>-11⋮x+2

=>x+2∈{1;-1;11;-11}

=>x∈{-1;-3;9;-13}

a) Ta có:  x 2 = 2 2  nên x = 2.

b) Ta có: x 2 = 5 2 nên x = 5.

c) Ta có:  3 x 5 = 3  nên  x 5 = 1 . Do đó x = 1.

d) Ta có:  6 x 3 = 48  nên  x 3 = 8 . Do đó x = 2.

e) Ta có:  x - 1 2 = 2 2  nên  x - 1 = 2 . Do đó x = 3.

f) Ta có:  x + 1 2 = 5 2  nên x +1 = 5. Do đó x = 4.

g) Ta có:  x - 1 3 = 3 3  nên  x - 1 = 3 . Do đó x = 4.

h) Ta có:  x + 1 3 = 4 3 nên x +1 = 4. Do đó x = 3