\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)

b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)

Khi x=-1/2 thì B=2/5

c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)

hay \(x\in\left\{3;1\right\}\)

a, đk : x khác -2 ; 2 

\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)

b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)

Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)

Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)

c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)

a. \(A=\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{4x+2}{x^2-1}\)

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

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

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

\(A=\dfrac{4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\)

b) Ta có: \(A=\dfrac{4}{x-1}=\dfrac{4}{2015}\) (ĐK: \(x\ne\pm1\) )

\(\Leftrightarrow8060=4\left(x-1\right)\)

\(\Leftrightarrow8060=4x-4\)

\(\Leftrightarrow8064=4x\)

\(\Leftrightarrow x=\dfrac{8064}{4}=2016\left(tm\right)\)

c) Ta có: \(\dfrac{4}{x-1}\left(x\ne1\right)\)

Để \(\dfrac{4}{x-1}\) nhận giá trị nguyên thì \(4:\left(x-1\right)\Leftrightarrow x-1\in\text{Ư}\left(4\right)=\left\{1;4;2\right\}\)

Vậy với x ∈ {2; 5; 3; 0; -1; -3} thì biểu thức \(\dfrac{4}{x-1}\) nhận giá trị nguyên

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

\(\dfrac{4}{-\dfrac{1}{2}-1}=-3\)

Vậy biểu thức A có giá trị -3 tại \(x=-\dfrac{1}{2}\)

bạn ktra lại đề ở chỗ 2/3/-x 

\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)

a) ĐKXĐ: \(\hept{\begin{cases}x+3\ne0\\3-x\ne0\\x^2-9\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne3\\x\ne3;x\ne-3\end{cases}}}\)

Vậy ĐKXĐ: x khác -3; x khác 3 ( b vào tcn của mìnk để thấy chi tiết)

Rút gọn:

\(A=\frac{5}{x+3}-\frac{2}{3-x}-\frac{3x^2-2x-9}{x^2-9}\)

\(\Leftrightarrow A=\frac{5}{x+3}+\frac{2}{x-3}-\frac{3x^2-2x-9}{\left(x-3\right)\left(x+3\right)}\) MTC: (x-3)(x+3)

\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-\left(3x^2-2x-9\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{9x-3x^2}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{3x\left(3-x\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3x}{x+3}\)

Vậy A=-3x/x+3 với x khác 3 và x khác -3

b) |x-2|=1

Bỏ dấu gt tuyệt đối ta có 2 TH: (đối chiếu đkxđ)

* x-2=1=> x=1+2=>x=3 (o t/m)

*x-2=-1=>x=-1+2=>x=1 (tm)

Thay x=1 vào phân thức A rút gọn ta có:

\(A=\frac{-3x}{x+3}=\frac{-3.1}{1+3}=\frac{-3}{4}\)

Vậy A=-3/4 khi x=1

c) Để A có gt nguyên => A thuộc Z

=> \(A=\frac{-3x}{x+3}\in Z\)

Ta có:  -3x chia hết x+3

=> -3(x-3)-9 chia hết x+3

=> -9 chia hết cho x+3

=>  x+3 thược Ư(-9)={1;-1;9;-9;3;-3)

Lập bảng thay vào hoặc o cần cx được 

Vậy...


 

a) \(H=\left(\frac{x}{x+2}-\frac{x^3-8}{x^3+8}.\frac{x^2-2x+4}{x^2-4}\right).\frac{x+3}{x+2}\)

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

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

\(=\frac{-4}{\left(x+2\right)^2}.\frac{x+3}{x+2}=\frac{-4x-12}{\left(x+2\right)^3}\)

a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-3}{\sqrt{x}-3}\)