\(\dfrac{1}{x-1}\) +\(\dfrac{x}{x^3-1}\) .
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23 tháng 12 2018

\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{x^3-1}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)

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

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

\(=\left(\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x+1\right)^2}{2x+1}\)

\(=\dfrac{2x+1}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x+1\right)^2}{2x+1}\)

\(=\dfrac{x+1}{x-1}\)

Vậy \(A=\dfrac{x+1}{x-1}\)

Giả sử tìm được \(x\in Z\) để \(A\in Z\)

\(x\in Z\Leftrightarrow\left\{{}\begin{matrix}x+1\in Z\\x-1\in Z\end{matrix}\right.\)

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

\(\Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)

Ta có các trường hợp :

+) \(x-1=1\Leftrightarrow x=2\)

+) \(x-1=2\Leftrightarrow x=3\)

+) \(x-1=-1\Leftrightarrow x=0\)

+) \(x-1=-2\Leftrightarrow x=-1\)

Vậy..

a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5

=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5

=(x-2)/(2x^2-5x+5)(x-1)

 

22 tháng 7 2018

\(a.\)

\(P=\left[\left(\dfrac{1}{x^2}+1\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)\right].\dfrac{x-1}{x^3}\)

\(P=\left[\left(\dfrac{1}{x^2}+\dfrac{x^2}{x^2}\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+\dfrac{x}{x}\right)\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^2}.\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{x+1}{x}\right)\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^2\left(x^2+2x+1\right)}+\dfrac{2}{x\left(x+1\right)^2}\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2}{x^3+2x^2+x}\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x\left(x^3+2x^2+x\right)}\right].\dfrac{x-1}{x^3}\)

\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x^4+2x^3+x^2}\right].\dfrac{x-1}{x^3}\)

\(P=\dfrac{x^2+1+2x}{x^4+2x^3+x^2}.\dfrac{x-1}{x^3}\)

\(P=\dfrac{x^2+2x+1}{x^2\left(x^2+2x+1\right)}.\dfrac{x-1}{x^3}\)

\(P=\dfrac{1}{x^2}.\dfrac{x-1}{x^3}\)

\(P=\dfrac{x-1}{x^5}\)

22 tháng 7 2018

Làm nốt đi cậu ! Bạn ko làm là tớ làm đó @@

8 tháng 12 2017

a) \(M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)

\(\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)

\(\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)

\(\Leftrightarrow M=\dfrac{2}{1-2x}\)

b) \(M=\dfrac{2}{1-2x}=\dfrac{-2}{3}\)

\(\Rightarrow2.3=\left(1-2x\right).\left(-2\right)\)

\(\Rightarrow6=-2+4x\)

\(\Rightarrow4x=6-\left(-2\right)\)

\(\Rightarrow4x=6+2\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=8:4\)

\(\Rightarrow x=2\)

Vậy \(M=\dfrac{-2}{3}\) thì \(x=2\)

c) Để \(M=\dfrac{2}{1-2x}\in Z\) \(\Leftrightarrow2⋮1-2x\)

\(\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.\)

\(x\in Z\)

\(\Rightarrow x\in\left\{1;0\right\}\)

Vậy \(x=1\) hoặc \(x=0\) thì \(M\in Z\)

8 tháng 12 2017

a) M = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

= \(\left(\dfrac{1}{1-x}+\dfrac{2}{1+x}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{x^2-1}{1-2x}\)

= \(\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

= \(\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)\(=\dfrac{-2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

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

=\(\dfrac{2}{1-2x}\)

b) M = \(\dfrac{-2}{3}\Leftrightarrow\dfrac{2}{1-2x}=\dfrac{-2}{3}\)

=> 2 . 3 = -2 (1 - 2x) (tích chéo)

=> 6 = -2 + 4x

=> 6 + 2 - 4x = 0

=> 8 - 4x = 0

=> 4x = 8

=> x = 2 (thỏa mãn đkxđ)

Vậy để M = \(\dfrac{-2}{3}\) thì x = 2

21 tháng 7 2018

\(a,\dfrac{1}{x^2-x}+\dfrac{2x}{4x^3}-\dfrac{1}{x^2+x+1}\)

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

\(=\dfrac{2x\left(x^2+x+1\right)+\left(x-1\right).\left(x^2+x+1\right)-2x^2.\left(x-1\right)}{2x^2.\left(x-1\right).\left(x^2+x+1\right)}\)

\(=\dfrac{2x^3+2x^2+2x+x^3-1-2x^3+2x^2}{2x^2.\left(x^3-1\right)}\)

\(=\dfrac{4x^2+2x+x^3-1}{2x^5-2x^2}\)

\(=\dfrac{x^3+4x^2+2x-1}{2x^5-2x^2}\)

21 tháng 7 2018

\(b,\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x+1\right).\left(x^2-x+1\right)}\)

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

\(=\dfrac{x+1\left(x+1\right).\left(x^2-x+1\right)-\left(x^2+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x+1+x^3+1-x^2-2}{\left(x+1\right).\left(x^2-x+1\right)}\)

\(=\dfrac{x+0+x^3-x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x\left(1+x^2-x\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)

\(=\dfrac{x}{x+1}\)

17 tháng 1 2018

\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\) ( Chữa đề nhé.)

a) \(ĐKXĐ:x\ne-3;x\ne2\)

\(\text{Với }x\ne-3;x\ne2,\text{ ta có: }A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\\ =\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\\ =\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x-4}{x-2}\\ \text{Vậy }A=\dfrac{x-4}{x-2}\text{ với }x\ne-3;x\ne2\)

b) Lập bảng xét dấu:

x x-4 x-2 x-4 2 4 0 0 x-2 _ _ + _ + + 0 + _ +

\(\Rightarrow\left[{}\begin{matrix}x< 2\\x>4\end{matrix}\right.\)

Vậy để \(A>0\) thì \(x< 2\) hoặc \(x>4\)

c) \(\text{Với }x\ne-3;x\ne2\)

\(\text{Ta có : }A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}\\ =\dfrac{x-2}{x-2}-\dfrac{2}{x-2}=1-\dfrac{2}{x-2}\)

\(\Rightarrow\) Để A nhận giá trị nguyên

thì \(\Rightarrow\dfrac{2}{x-2}\in Z\)

\(\Rightarrow2⋮x-2\\ \Rightarrow x-2\inƯ_{\left(2\right)}\)

\(Ư_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)

Lập bảng giá trị:

\(x-2\) \(-2\) \(-1\) \(1\) \(2\)
\(x\) \(0\left(TM\right)\) \(1\left(TM\right)\) \(3\left(TM\right)\) \(4\left(TM\right)\)

\(\Rightarrow x\in\left\{-2;-1;1;2\right\}\)

Vậy với \(x\in\left\{-2;-1;1;2\right\}\)

thì \(A\in Z\)

17 tháng 1 2018

Câu 2:

a) \(ĐKXĐ:x\ne\dfrac{3}{2};x\ne1\)

\(\text{Với }x\ne\dfrac{3}{2};x\ne1,\text{ ta có : }B=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\\ =\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right]:\left(\dfrac{3\left(1-x\right)}{1-x}+\dfrac{2}{1-x}\right)\\ =\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3-3x+2}{\left(1-x\right)}\\ =\dfrac{\left(-3x+5\right)\cdot\left(1-x\right)}{\left(2x-3\right)\left(x-1\right)\cdot\left(-3x+5\right)}\\ =-\dfrac{1}{2x-3}\)

Vậy \(B=-\dfrac{1}{2x-3}\) với \(x\ne\dfrac{3}{2};x\ne1\)

b) \(\text{Với }x\ne\dfrac{3}{2};x\ne1\)

Để \(B=\dfrac{1}{x^2}\)

\(\text{thì }\Rightarrow\dfrac{-1}{2x-3}=\dfrac{1}{x^2}\\ \Rightarrow2x-3=-x^2\\ \Leftrightarrow2x-3+x^2=0\\ \Leftrightarrow x^2-3x+x-3=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\left(TM\right)\)

Vậy với \(x=-1;x=3\) thì \(B=\dfrac{1}{x^2}\)

a: \(A=\dfrac{2x-5+x^2-4+x^2-9}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2+2x-18}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{2\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x+6}{x-3}\)

b: Để A/2=x+3/x-3 là số nguyên thì \(x-3+6⋮x-3\)

=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{4;51;6;0;9;-3\right\}\)

c: Để A=1/x-1 thì \(\dfrac{2x+6}{x-3}=\dfrac{1}{x-1}\)

=>2x^2-2x+6x-6=x-3

=>2x^2+5x-6-x+3=0

=>2x^2+4x-3=0

hay \(x=\dfrac{-2\pm\sqrt{10}}{2}\)

Giải các phương trình có chứa ẩn ở mẫu sau: a, \(\dfrac{x-3}{x-2}+\dfrac{x+2}{x}=2\) b, \(\left(x-2\right)\left(\dfrac{2}{3}x-6\right)=0\) d, \(\dfrac{x}{x+1}-\dfrac{2x-3}{x-1}=\dfrac{2x+3}{x^2-1}\) f, \(\dfrac{x-1}{x}+\dfrac{x-2}{x+1}=2\) g, \(\dfrac{x}{x-1}+\dfrac{x-1}{x}=2\) h, \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\) i, \(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\) j, \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\) k, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x-3}=1\) l,...
Đọc tiếp

Giải các phương trình có chứa ẩn ở mẫu sau:

a, \(\dfrac{x-3}{x-2}+\dfrac{x+2}{x}=2\)

b, \(\left(x-2\right)\left(\dfrac{2}{3}x-6\right)=0\)

d, \(\dfrac{x}{x+1}-\dfrac{2x-3}{x-1}=\dfrac{2x+3}{x^2-1}\)

f, \(\dfrac{x-1}{x}+\dfrac{x-2}{x+1}=2\)

g, \(\dfrac{x}{x-1}+\dfrac{x-1}{x}=2\)

h, \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)

i, \(\dfrac{2}{x+1}-\dfrac{3}{x-1}=5\)

j, \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

k, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x-3}=1\)

l, \(\dfrac{2}{x+1}-\dfrac{1}{xx-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

m, \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

n, \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

o, \(\dfrac{x-2}{x+2}+\dfrac{3}{x-2}=\dfrac{x^2-11}{x^2-4}\)

p, \(\dfrac{x+4}{x+1}+\dfrac{x}{x-1}=\dfrac{2x^2}{x^2-1}\)

z, \(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)

q, \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)

r, \(\dfrac{1}{x-3}+2=\dfrac{5}{x-1}+x\)

s, \(\dfrac{2}{x^2+4x-21}=\dfrac{3}{x-3}\)

3
1 tháng 5 2017

ai giải giúp mk vs đg cần gấp

7 tháng 5 2018

nhân cả 2 vế với 2 rồi bunhia

6 tháng 4 2018

câu c là \(\dfrac{1}{2}\)(x+y+z) nhé, mih chép nhầm