Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐK:\(\begin{cases} x + 2≠0\\ x - 2≠0 \end{cases}\)⇔\(\begin{cases} x ≠ -2\\ x≠ 2 \end{cases}\)
Vậy biểu thức P xác định khi x≠ -2 và x≠ 2
b) P= \(\dfrac{3}{x+2}\)-\(\dfrac{2}{2-x}\)-\(\dfrac{8}{x^2-4}\)
P=\(\dfrac{3}{x+2}\)+\(\dfrac{2}{x-2}\)-\(\dfrac{8}{(x-2)(x+2)}\)
P= \(\dfrac{3(x-2)}{(x-2)(x+2)}\)+\(\dfrac{2(x+2)}{(x-2)(x+2)}\)-\(\dfrac{8}{(x-2)(x+2)}\)
P= \(\dfrac{3x-6+2x+4-8}{(x-2)(x+2)}\)
P=\(\dfrac{5x-10}{(x-2)(x+2)}\)
P=\(\dfrac{5(x-2)}{(x-2)(x+2)}\)
P=\(\dfrac{5}{x+2}\)
Vậy P=\(\dfrac{5}{x+2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: a2-1 ≠0 ⇔ (a-1)(a+1)≠0 ⇔\(\left[{}\begin{matrix}a-1\ne0\\a+1\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)
b) A=\(\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\) , a≠1, -1
=\(\dfrac{2a^2}{\left(a-1\right)\left(a+1\right)}-\dfrac{a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}+\dfrac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
=\(\dfrac{2a^2-a\left(a-1\right)+a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
=\(\dfrac{2a^2-a^2+a+a^2+a}{\left(a-1\right)\left(a+1\right)}\)
=\(\dfrac{2a^2+2a}{\left(a-1\right)\left(a+1\right)}\) =\(\dfrac{2a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\) =\(\dfrac{2a}{a-1}\)
vậy A =\(\dfrac{2a}{a-1}\) với a≠1,-1.
c) Có:A= \(\dfrac{2a}{a-1}\) = \(\dfrac{2a-2+2}{a-1}=\dfrac{2\left(a-1\right)+2}{a-1}=2+\dfrac{2}{a-1}\)
Để a∈Z thì a-1 ∈ Z ⇒ (a-1) ∈ Ư(2) =\(\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
a-1 | 1 | -1 | 2 | -2 |
a | 2 | 0 | 3 | -1 |
Thử lại | TM | TM | TM | ko TM(vì a≠-1 |
Vậy để biểu thức A có giá trị nguyên thì a∈\(\left\{2;0;3\right\}\)
a) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)
b) Ta có: \(A=\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\)
\(=\dfrac{2a^2}{\left(a+1\right)\left(a-1\right)}-\dfrac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}+\dfrac{a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a^2-a^2+a+a^2+a}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a^2+2a}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(=\dfrac{2a}{a-1}\)
c) Để A nguyên thì \(2a⋮a-1\)
\(\Leftrightarrow2a-2+2⋮a-1\)
mà \(2a-2⋮a-1\)
nên \(2⋮a-1\)
\(\Leftrightarrow a-1\inƯ\left(2\right)\)
\(\Leftrightarrow a-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow a\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(a\in\left\{0;2;3\right\}\)
Vậy: Để A nguyên thì \(a\in\left\{0;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: x<>4; x<>-4
b: \(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x-1}{x+4}\)
c: Để A nguyên thì x+4-5 chia hết cho x+4
=>\(x+4\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-3;-5;1;-9\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ: \(a\ne1;a\ne-1\)
Ta có:
\(P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}=\frac{2a^2}{\left(a-1\right)\left(a+1\right)}\) \(+\frac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\frac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(\Rightarrow P=\frac{2a^2+a^2-a-a^2-a}{\left(a-1\right)\left(a+1\right)}=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(\Rightarrow P=\frac{2a}{a+1}\)
b. Để P có giá trị nguyên \(\Rightarrow2a⋮a+1\Rightarrow2\left(a+1\right)-2a⋮a+1\Rightarrow2a+2-2a⋮a+1\)
\(\Rightarrow2⋮a+1\) vì \(a\in Z\Rightarrow a+1\in\left\{-2;-1;1;2\right\}\Rightarrow a\in\left\{-3;-2;0;1\right\}\)
Vậy \(a\in\left\{-3;-2;0;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)
a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)
= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)
= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)
Có vài bước mình làm tắc á nha :>
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\left(2x-1\right)\left(4x^2+2x+1\right)-7\left(x^3+1\right)\)
\(=\left(2x\right)^3-1^3-7x^3-7\)
\(=8x^3-1-7x^3-7=x^3-8\)
b: Thay x=-1/2 vào A, ta được:
\(A=\left(-\dfrac{1}{2}\right)^3-8=-\dfrac{1}{8}-8=-\dfrac{65}{8}\)
Biểu thức đâu bạn ? :)))
Sau khi ib với Đinh Lan Anh thì \(P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)
\(a,ĐKXĐ:\hept{\begin{cases}a+1\ne0\\a-1\ne0\end{cases}\Leftrightarrow a\ne\pm1}\)
\(b,P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}\)
\(=\frac{2a^2+a\left(a-1\right)-a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{2a^2+a^2-a-a^2-q}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{2a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{2a}{a+1}\)
\(c,P=\frac{2a}{a+1}=\frac{2a+2}{a+1}-\frac{2}{a+1}=2-\frac{2}{a+1}\)
Để \(P\inℤ\)thì \(2-\frac{2}{a+1}\inℤ\)
\(\Leftrightarrow\frac{2}{a+1}\inℤ\)
Mà \(a\inℤ\Rightarrow a+1\inℤ\)
Ta có bảng
Kết hợp ĐKXĐ \(a\ne\pm1\)ta được \(a\in\left\{-3;-2;0\right\}\)
Vậy //////