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6 tháng 5 2019

a) \(\frac{2x+1}{x-1}\)=\(\frac{5\left(x-1\right)}{x+1}\):dkxd x\(\ne\)\(\pm\)1

\(\Rightarrow\)(2x+1)(x+1)=5(x-1)2

\(\Leftrightarrow\)2x2+2x+x+1=5(x2-2x+1)

\(\Leftrightarrow\)2x2+2x+x+1=5x2-10x+5

\(\Leftrightarrow\)2x2+2x+x+1-5x2+10x-5=0

\(\Leftrightarrow\)-3x2+13x-4=0

\(\Leftrightarrow\)-3x2+12x+1x-4=0

\(\Leftrightarrow\)-4x(x-4)+(x-4)=0

\(\Leftrightarrow\)(x-4)(-4x+1)=0

\(\Leftrightarrow\)x-4=0 hoac -4x+1=0

\(\Leftrightarrow\)x=4(tmdkxd) \(\Leftrightarrow\)x=1/4(tmdkxd)

vay s={4;1/4}

b)\(\frac{x}{x-1}\)-\(\frac{2x}{x^{ }2^{ }-1}\)=0 dkxd x\(\ne\)\(\pm\)1

\(\Leftrightarrow\)\(\frac{x\left(X+1\right)-2x^{ }}{\left(x-1\right)\left(x+1\right)}\)=0

\(\Rightarrow\)x2+x-2x=0

\(\Leftrightarrow\)x2-x=0

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

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

\(\Leftrightarrow\)x=0(tmdkxd)\(\Leftrightarrow\)x=1(ktmdkxd)

vay s={0}

c.\(\frac{1}{x-2}\)+3=\(\frac{x-3}{2-x}\) dkxd x\(\ne\)2

\(\Leftrightarrow\)\(\frac{1}{x-2}\)+3=\(\frac{-\left(x-3\right)}{x-2}\)

\(\Leftrightarrow\)\(\frac{1+3\left(x-2\right)}{x-2}\)=\(\frac{-x+3}{x-2}\)

\(\Rightarrow\)1+3x-6=-x+3

\(\Leftrightarrow\)4x=8

\(\Leftrightarrow\)x=2(ktmdkxd)

vay s=\(\varnothing\)

chuc ban hoc totbanh

6 tháng 5 2019

a.\(\frac{2x+1}{x-1}\) = \(\frac{5\left(x-1\right)}{x+1}\)
\(\leftrightarrow\) 2x+1 = 5x - 5
\(\leftrightarrow\) 2x - 5= -1-5
\(\leftrightarrow\) -3x = -6
\(\leftrightarrow\) x =2

Vậy S=\(\left\{2\right\}\)
b.\(\frac{x}{x-1}\) - \(\frac{2x}{x^2-1}\) =0

\(\leftrightarrow\) \(\frac{x}{x-1}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\)= 0 (ĐK : x\(_{\ne}\) -1 và 1)

\(\leftrightarrow\)\(\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\) =0

\(\leftrightarrow\) x2 + x -2x = 0

\(\leftrightarrow\)(x2 + x) -2x =0

\(\leftrightarrow\)x(x+1) -2x =0

\(\leftrightarrow\) x =0 -> x=0
x+1 =0 -> x = -1(Loại)
-2x = 0 -> x= 2(TM)
Vậy x =\(\left\{0,2\right\}\)
(BẠN NHỚ COI LẠI CÁI CÂU TRẢ LỜI Ở CUỐI MỖI BÀI NHA ,MÌNH KO CHẮC CÂU TRẢ LỜI ĐÓ )

11 tháng 2 2018

a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)

\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)

\(\Leftrightarrow6x+6+12x-8=x-7\)

\(\Leftrightarrow6x+12x-x=-7-6+8\)

\(\Leftrightarrow17x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{17}\)

Vậy .........................

b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)

\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)

\(\Leftrightarrow2x^2-x^2+x+15-21=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2-2x+3x-6=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)

Vậy .........................

P/s: các câu còn lại tương tự, bn tự giải nha

12 tháng 2 2018

làm hộ mình câu còn lại đi :))

23 tháng 2 2019

a) Đk : \(x\ne0;\ne1\)

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

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

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

\(\Rightarrow\dfrac{0}{x-1}=0\)

=> Phương trình có vô số nghiệm x

b) Đk : \(x\ne2;x\ne3\)

\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)

\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)

=0

\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)

\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)

=> Phương trình vô nghiệm

c)

\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)

\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)

\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)

=> PTVN

d) Thôi tự làm đi, câu này dễ :Vvv

e)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40

\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)

\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

Đặt

\(x^2+6x+7=t\)

Phương trình tương đương

\(\left(t-1\right)\left(t+1\right)=40\)

\(t^2=41\)

\(\)\(t=\pm\sqrt{41}\)

Thay vào tìm x.

24 tháng 2 2019

Thanks ;)

19 tháng 7 2018

undefined

3 tháng 3 2017

a. \(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

c. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

1 tháng 5 2018

a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)

=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)

=>\(x^2+2x+1-x^2+2x-1=16\)

=>4x=16=>x=4

b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)

=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)

=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)

=>\(12-x^2-3x-2+x^2+5x-14=0\)

=>2x-4=0=>2x=4=>x=2

c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)

=>\(12=x^3+8+x^2-2x+4\)

=>\(x^3+x^2-2x=0\)

=>\(x^3-x+x^2-x=0\)

1 tháng 5 2018

c)=>\(x\left(x^2-1\right)+x\left(x-1\right)=0\)

=>\(x\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

=>\(x\left(x-1\right)\left(x+2\right)=0\)

=>x=?

28 tháng 10 2018

a)\(\dfrac{2\left(2x-1\right)-\left(2x+1\right)+4}{4x^2-1}\)

\(=\dfrac{4x-2-2x-1+4}{4x^2-1}=\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{1}{2x-1}\)

câu b đề đúng ko vậy

28 tháng 10 2018

Cam ơn bn nhahaha

30 tháng 3 2018

Hỏi đáp Toán

30 tháng 3 2018

Dài quá c ơi :<

28 tháng 4 2018

câu nào cũng ghi lại đề nha

a) \(x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b)\(x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)

28 tháng 4 2018

d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)

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

\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )

\(\Leftrightarrow4x-8=0\Rightarrow x=2\)

đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)

\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))

\(\Leftrightarrow8-x-8x+56-1=0\)

\(\Leftrightarrow-9x+63=0\)

\(\Leftrightarrow x=7\)

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

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