để C nguyên thì 2 chia hết cho n-1 nên n-1 thuộc Ư(2)= 2;-2;1;-1

nên n-1 = 2;-2;1;-1 <=> n = -1;0;2;3. (1)

để D nguyên thì n+4 chia hết cho n+1.

ta có : n+4 = (n+1) + 3 .

vì n + 1 chia hết cho n + 1 nên 3 chia hết cho n + 1 hay n + 1 thuộc Ư(3) = -3;3;-1;1.

nên n + 1 = -3;3;-1;1 <=> n = -4;-2;0;2 (2)

từ (1) và (2) thì n = 0;2 thì C và D nguyên.

\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)

Não không "sấm sét" nên không nghĩ được cách nào hay cả :)

\(x^2+\left(2x\right)^2+\left(3x\right)^2+\left(4x\right)^2+\left(5x\right)^2=220\)

\(\Leftrightarrow x^2+4x^2+9x^2+16x^2+25x^2=220\)

\(\Leftrightarrow x^2\cdot\left(1+4+9+16+25\right)=220\)

\(\Leftrightarrow x^2\cdot55=220\)

\(\Leftrightarrow x^2=4\)

Mà \(x>0\Rightarrow x=\sqrt{4}=2\)

Vậy x = 2.

\(x^2+\left(2x\right)^2+\left(3x\right)^2+\left(4x\right)^2+\left(5x\right)^2=220\\ \Leftrightarrow x^2+4x^2+16x^2+25x^2=220\\ \Leftrightarrow x^2\left(1+4+9+16+25\right)=220\)

\(\Leftrightarrow55x^2=220\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=2\)

vậy x = 2

a) ĐKXĐ: \(x\notin\left\{0;-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

Ta có: \(A=\left(\dfrac{1}{2x-1}+\dfrac{3}{1-4x^2}-\dfrac{2}{2x+1}\right):\left(\dfrac{x^2}{2x^2+x}\right)\)

\(=\left(\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{3}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\right):\left(\dfrac{x^2}{x\left(2x+1\right)}\right)\)

\(=\dfrac{2x+1-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}:\dfrac{x}{2x+1}\)

\(=\dfrac{-2x}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x}\)

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

b) Ta có: \(\left|2x-1\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)

Thay \(x=\dfrac{3}{2}\) vào biểu thức \(A=\dfrac{-2}{2x-1}\), ta được:

\(A=-2:\left(2\cdot\dfrac{3}{2}-1\right)=-2:\left(3-1\right)=-2:2=-1\)

Vậy: Khi \(\left|2x-1\right|=2\) thì A=-1

c) Để \(A=\dfrac{1}{3}\) thì \(\dfrac{-2}{2x-1}=\dfrac{1}{3}\)

\(\Leftrightarrow2x-1=-6\)

\(\Leftrightarrow2x=-5\)

hay \(x=-\dfrac{5}{2}\)(thỏa ĐK)

Vậy: Để \(A=\dfrac{1}{3}\) thì \(x=-\dfrac{5}{2}\)

mình tự bình loạn các bạn ạ

Ta có: \(Ư\left(36\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm9;\pm12;\pm36\right\}\)

Mà \(x\ge6\Rightarrow x\in\left\{6;9;12;36\right\}\)

Vậy \(x=6;9;12;36\)

Ta có Ư(36)={1;2;3;4;6;9;12;18;36}

=> x thuộc {1;2;3;4;6;9;12;18;36}

Nhưng x >= 6 => x thuộc {6;9;12;18;36}

yOz=180

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)