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\(\frac{x^3-2x^2+x+2}{x-2}=\frac{x^2\left(x-2\right)+\left(x-2\right)+4}{x-2}=\frac{\left(x-2\right)\left(x^2+1\right)+4}{x-2}\)
\(=\frac{\left(x-2\right)\left(x^2+1\right)}{x-2}+\frac{4}{x-2}=x^2+1+\frac{4}{x-2}\)
\(x^2+1+\frac{4}{x-2}\) nguyên khi và chỉ khi 4 chia hết cho x-2
<=>\(x-2\inƯ\left(4\right)=\left\{-4;-1;1;4\right\}\)
<=>\(x\in\left\{-2;1;3;6\right\}\)
Vậy ..................
a: Thay x=5 vào B, ta được:
\(B=\dfrac{5-1}{5-3}=\dfrac{4}{2}=2\)
b: \(A=\dfrac{2x^2+6x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-1}{\left(x+3\right)\left(x-3\right)}\)
a) ĐKXĐ: \(x\ne-1;0;1.\)Ta có:
\(A=\left[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\)
\(=\left[\frac{2}{\left(x+1\right)^3}\cdot\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}\cdot\frac{x^2+1}{x^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\left[\frac{2}{x\left(x+1\right)^2}+\frac{x^2+1}{x^2\left(x+1\right)^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\left[\frac{2x}{x^2\left(x+1\right)^2}+\frac{x^2+1}{x^2\left(x+1\right)^2}\right]\cdot\frac{x^3}{x-1}\)
\(=\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\cdot\frac{x^3}{x-1}\)
\(=\frac{\left(x+1\right)^2\cdot x}{\left(x+1\right)^2\left(x-1\right)}=\frac{x}{x-1}.\)
Vậy \(A=\frac{x}{x-1}\)với \(x\ne-1;0;1.\)
b) A < 1 \(\Leftrightarrow\frac{x}{x-1}< 1\Leftrightarrow\frac{x}{x-1}-1< 0\Leftrightarrow\frac{x}{x-1}-\frac{x-1}{x-1}< 0\)\(\Leftrightarrow\frac{1}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)(do 1 > 0)\(\Leftrightarrow x< 1.\)
Kết hợp ĐKXĐ, A < 1 khi \(x< 1\)và \(x\ne-1;0.\)
c) \(A\inℤ\Leftrightarrow\frac{x}{x-1}\inℤ.\)Mà \(x\inℤ\)\(\Rightarrow x⋮\left(x-1\right)\Rightarrow\left(x-1+1\right)⋮\left(x-1\right)\Rightarrow1⋮\left(x-1\right)\Rightarrow\left(x-1\right)\inƯ\left(1\right)=\left\{1;-1\right\}.\)Ta lập bảng sau:
| \(x-1\) | 1 | -1 |
| \(x\) | 2 | 0 |
| Kết luận | x thoả mãn ĐKXĐ | x không thoả mãn ĐKXĐ |
Vậy để A nguyên thì x = 2.
Để A là số nguyên thì \(x^2\left(x-2\right)+x-2+4⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{3;1;4;0;6;-2\right\}\)
\(A=\frac{x^3-x^2+2}{x-1}=x^2+\frac{2}{x-1}\inℤ\Leftrightarrow\frac{2}{x-1}\inℤ\)
mà \(x\inℤ\)nên \(x-1\inƯ\left(2\right)=\left\{-2,-1,1,2\right\}\)
\(\Leftrightarrow x\in\left\{-1,0,2,3\right\}\).
ta có x^2 -4 = (x-2)(x+2)
đkxđ của C là x khác 2 và trừ 2
\(\frac{x^3}{x^2-4}\)- \(\frac{x}{x-2}\)- \(\frac{2}{x+2}\)= \(\frac{x^3}{\left(x-2\right)\left(x+2\right)}\)- \(\frac{x}{x-2}\)- \(\frac{2}{x+2}\)
= \(\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
= \(\frac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
= \(\frac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)= \(\frac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)= x- 1
để C = 0 => x-1 = 0
=> x= 1 ( thỏa mãn điều kiện xác định)
c, để C dương
=> x-1 dương
=> x-1 >0
=> x>1
a) Để biểu thức xác định \(\Rightarrow\hept{\begin{cases}x^2-4\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Rightarrow x\ne2;-2\)
Vậy ...
b) \(C=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-\left(x^2+2x\right)-\left(2x-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^3-x^2\right)-\left(4x-4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
Để C = 0 \(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy ...
c) Để C > 0 thì \(x-1>0\Rightarrow x>1\)
Vậy ...
làm ơn nhanh với ạ
Lớp 8 trình bày kiểu khác, thôi thì cứ tạm cách này vậy >>:
Để A \(\in\) Z
\(\Rightarrow\)x3-x2+2 \(⋮\) x-1
\(\Rightarrow\)x2(x-1)+2 \(⋮\) x-1
\(\Rightarrow\)2 \(⋮\) x-1
\(\Rightarrow\)x-1 \(\in\) Ư(2)
\(\Rightarrow\)x-1 \(\in\) {\(\pm\)1; \(\pm\)2}
Lập bảng:
Vậy x \(\in\) {-1;1;-2;2}