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a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
b: \(C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
Để C=0 thì x-1=0
hay x=1
c: Để C>0 thì x-1>0
hay x>1
Vậy: \(\left\{{}\begin{matrix}x\in Z\backslash\left\{1\right\}\\x\notin\left\{2;-2\right\}\end{matrix}\right.\)
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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=\(\dfrac{5}{x+2}\)
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a) Ta có: A = \(\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+4x}{4-x^2}\)
A = \(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
A = \(\frac{x^2+3x+2+x^2-3x+2-x^2-4x}{\left(x-2\right)\left(x+2\right)}\)
A = \(\frac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
A = \(\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)
b) Với x = 4 => A = \(\frac{4-2}{4+2}=\frac{2}{8}=\frac{1}{4}\)
c) ĐKXĐ: \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\\4-x^2\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne2\\x\ne-2\\x\ne\pm2\end{cases}}\) <=> \(x\ne\pm2\)
Ta có: A = \(\frac{x-2}{x+2}=\frac{\left(x+2\right)-4}{x+2}=1-\frac{4}{x+2}\)
Để A nhận giá trị nguyên dương <=> \(1-\frac{4}{x+2}\) nguyên dương
<=> \(-\frac{4}{x+2}\) nguyên dương <=> -4 \(⋮\)x + 2
<=> x + 2 \(\in\)Ư(-4) = {1; -1; 2; -2; 4; -4}
Lập bảng:
x + 2 | 1 | -1 | 2 | -2 | 4 | -4 |
x | -1(tm) | -3(tm) | 0(tm) | -4(tm) | 2(ktm) | -6(tm) |
Vậy ....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
=> C nguyên dương khi và chỉ khi x -1 >0 => x > 1 như vậy với x nguyên dương lớn hơn 1 thì C nguyên dương
\(C=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}=\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{x^2-4}=\frac{x^3-x^2-2x-2x+4}{x^2-4}\)
\(C=\frac{x\left(x^2-4\right)-\left(x^2-4\right)}{x^2-4}=\frac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}=x-1\)
\(\Rightarrow C\in Z^+\)với \(x>1\)
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a,ĐKXĐ: \(x^2-4\ne0\) \(\Leftrightarrow x\ne\pm2\)
b,Rút gọn:
\(C=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^3-4x\right)-\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x^2-4\right)-\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}\)
\(=x-1\)
Để C = 0 thì x - 1 = 0
=> x = 1
Vậy : Để C = 0 thì x = 1
c,Để C nhận giá trị dương thì C > 0
Hay: x - 1 > 0
<=> x > 1
Vậy: Để C dương thì x > 1
=.= hok tốt!!
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BÀI 1:
a) \(ĐKXĐ:\) \(x-3\)\(\ne\)\(0\)
\(\Leftrightarrow\)\(x\)\(\ne\)\(3\)
b) \(A=\frac{x^3-3x^2+4x-1}{x-3}\)
\(=\frac{\left(x^3-3x^2\right)+\left(4x-12\right)+11}{x-3}\)
\(=\frac{x^2\left(x-3\right)+4\left(x-3\right)+11}{x-3}\)
\(=x^2+4+\frac{11}{x-3}\)
Để \(A\)có giá trị nguyên thì \(\frac{11}{x-3}\)có giá trị nguyên
hay \(x-3\)\(\notinƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
Ta lập bảng sau
\(x-3\) \(-11\) \(-1\) \(1\) \(11\)
\(x\) \(-8\) \(2\) \(4\) \(14\)
Vậy....
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a: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{4}{x-2}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)
\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)
\(=\dfrac{6\left(x+1\right)\cdot x\left(x+2\right)}{3\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{x-2}\)
Đáp án x ∈ 1 ; 3 ; 5 .