cho phân thức \(A=\frac{x^4-2x^2+1}{x^3-3x-2}\)
tìm điều kiện x để A có nghĩa
rút gọn A
tìm x để A <1
K
Khách
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Những câu hỏi liên quan
3 tháng 6 2018
1. Để A có nghĩa thì \(x^3-3x-2\ne0\)
\(\Rightarrow\left(x^3-x\right)-\left(2x-2\right)\ne0\)
\(\Rightarrow x\left(x^2-1\right)-2\left(x-1\right)\ne0\)
\(\Rightarrow x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\ne0\)
\(\Rightarrow\left(x^2+x-2\right)\left(x-1\right)\ne0\)
\(\Rightarrow\left(x^2-1+x-1\right)\left(x-1\right)\ne0\)
\(\Rightarrow\left[\left(x+1\right)\left(x-1\right)+\left(x-1\right)\right]\left(x-1\right)\ne0\)
\(\Rightarrow\left(x-1\right)^2\left(x+2\right)\ne0\)
\(\Rightarrow x\ne1;x\ne-2\)
2. \(A=\frac{x^4-2x^2+1}{x^3-3x-2}=\frac{\left(x^2-1\right)^2}{\left(x-1\right)^2\left(x+2\right)}=\frac{\left[\left(x-1\right)\left(x+1\right)\right]^2}{\left(x-1\right)^2\left(x+2\right)}\)
\(=\frac{\left(x-1\right)^2.\left(x+1\right)^2}{\left(x-1\right)^2\left(x+2\right)}=\frac{\left(x+1\right)^2}{x+2}\)
3/ Để A < 1 \(\Leftrightarrow\frac{\left(x+1\right)^2}{x+2}< 1\Leftrightarrow\left(x+1\right)^2< x+2\)
\(\Leftrightarrow x^2+2x+1< x+2\)
\(\Leftrightarrow x^2+x< 1\)
\(\Leftrightarrow x.\left(x+1\right)< 1\)
Vậy .....
3 tháng 6 2018
1. A có nghĩa khi \(x^3-3x-2\ne0\)
\(\Leftrightarrow x^3+x^2-x^2-x-2x-2\ne0\)
\(\Leftrightarrow x^2\left(x+1\right)-x\left(x+1\right)-2\left(x+1\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-2\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x-2x-2\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow x-2\ne0\)(do \(\left(x+1\right)^2\ge0\)) \(\Leftrightarrow x\ne2\)
2. Ta có :
Tử = \(x^4-2x^2+1=x^4-x^3+x^3-x^2-x^2+x-x+1\)
=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)-\left(x-1\right)\)
=\(\left(x-1\right)\left(x^3+x^2-x-1\right)=\left(x-1\right)\left[x^2\left(x+1\right)-x\left(x+1\right)\right]\)
=\(\left(x-1\right)\left(x+1\right)\left(x^2-1\right)=\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)^2\left(x-1\right)^2\)
Vậy \(A=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x-1\right)^2}{x-2}\)
3. \(A< 1\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}< 1\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Leftrightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)
\(\Leftrightarrow\frac{x^2-3x+3}{x-2}< 0\)ta có \(x^2-3x+3=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{3}{4}=\left(x-\frac{3}{4}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\)(1) \(\Leftrightarrow x-2< 0\Leftrightarrow x< 2\)(Thỏa mãn)
Vậy x<2 thì A<1
T
1
3 tháng 7 2017
1.A=\(\frac{x^4-2x^2+1}{x^3-3x-2}\)
A có nghĩa \(\Leftrightarrow x^3-3x-2\ne0\Leftrightarrow\left(x+1\right)^2\left(x-2\right)\ne0\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
2 .A = \(\frac{x^4-2x^2+1}{x^3-3x-2}\)=\(\frac{\left(x^2-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-2\right)}=\frac{\left(x-1\right)^2}{x-2}\)
A<1\(\Rightarrow\frac{\left(x-1\right)^2}{x-2}-1< 0\Rightarrow\frac{x^2-2x+1-x+2}{x-2}< 0\)
\(\Rightarrow\frac{x^2-3x+3}{x-2}< 0\Rightarrow x-2< 0\)vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy x<2 thỏa mãn yêu cầu A<1
1 tháng 1 2017
Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy
Bài 4:
\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
a) DK x khác +-1
b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)
c) x+1 phải thuộc Ước của 2=> x=(-3,-2,0))
1 tháng 1 2017
1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa
b) \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{x+2}\)
c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)
\(\Leftrightarrow x-2=\left(x+2\right).0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )
=> ko có gía trị nào của x để A=0
KN
24 tháng 1 2020
a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)
+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)
\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)
+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)
+) \(x+1\ne0\Leftrightarrow x\ne-1\)
+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)
\(\Leftrightarrow x\ne0;x\ne-2\)
+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)
Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)
24 tháng 1 2020
a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)
\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)
\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x+1\right)-2x^2-4x+1-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)
b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)
Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x-2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| x | -4 | -1 | 0 | 1 | 3 | 4 | 5 | 8 |
Vậy ............................
17 tháng 2 2021
a. \(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b. \(A=\dfrac{3x+3}{x^2-1}\\ A=\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\\ A=\dfrac{3}{x-1}\)
c. Để \(A=-2\) thì \(\dfrac{3}{x-1}=-2=\dfrac{3}{\dfrac{-3}{2}}\\ \Leftrightarrow x-1=\dfrac{-3}{2}\\ \Leftrightarrow x=\dfrac{-1}{2}\left(\text{t/m ĐKXĐ}\right)\)
Vậy \(x=\dfrac{-1}{2}\) để phân thức nhận giá trị là -2.
TG
17 tháng 2 2021
a) Có: \(x^2-1=\left(x-1\right)\left(x+1\right)\)
ĐKXĐ là x ≠ 1; x ≠ -1
b) \(\dfrac{3x+3}{x^2-1}=\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x-1}\)
c) Theo đề ta có: \(\dfrac{3}{x-1}=2\)
\(\Rightarrow x-1=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{5}{2}\) (T/m ĐK)
9 tháng 2 2021
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne3\end{matrix}\right.\)
Ta có : \(P=\dfrac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{4}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x\left(x-3\right)+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x-x+6}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x-3}{x-3}\)
b, Ta có : \(P=\dfrac{2x-3}{x-3}=\dfrac{2x-6+3}{x-3}=2+\dfrac{3}{x-3}\)
- Để P là số nguyên \(\Leftrightarrow x-3\in\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{4;3;6;0\right\}\)
Vậy ...
9 tháng 2 2021
a ĐKXĐ : \(x\ne2,x\ne3\)
\(\Rightarrow P=\dfrac{2x\left(x-3\right)+4-\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-6x+4-x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2-7x+6}{x^2-5x+6}\)b Ta có P = \(\dfrac{2x^2-7x+6}{x^2-5x+6}=\dfrac{x^2-5x+6+x^2-2x}{x^2-5x+6}=1+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=1+\dfrac{x}{x-3}\)
Để P\(\in Z\) \(\Leftrightarrow1+\dfrac{x}{x-3}\in Z\) \(\Rightarrow\dfrac{x}{x-3}\in Z\) \(\Rightarrow x⋮x-3\) \(\Rightarrow x-3+3⋮x-3\)
\(\Rightarrow3⋮x-3\) \(\Rightarrow\left(x-3\right)\in\left\{-3;-1;1;3\right\}\) \(\Rightarrow x\in\left\{0;2;4;6\right\}\)
Thử lại ta thấy đúng
Vậy...