Giaỉ pt: \(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 12 2020
a, \(\dfrac{6-x}{4x-3}=\dfrac{2}{4x-3}\)
ĐKXĐ: \(x\ne\dfrac{3}{4}\)
PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(6-x\right)\left(4x-3\right)}{4x-3}=\dfrac{2\left(4x-3\right)}{4x-3}\)
\(\Rightarrow6-x=2\)
\(\Leftrightarrow x=4\)(thỏa mãn ĐKXĐ)
21 tháng 12 2020
b, \(\dfrac{3-x}{2x-3}+x-1=\dfrac{-4}{2x-3}\)
ĐKXĐ: \(x\ne\dfrac{3}{2}\)
PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(3-x\right)\left(2x-3\right)}{2x-3}+\left(x+1\right)\left(2x-3\right)=\dfrac{-4\left(2x-3\right)}{2x-3}\)
\(\Rightarrow3-x+2x-3x+2x-3=-8x+12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\dfrac{3}{2}\)(không thỏa mãn ĐKXĐ)
Vậy \(x\in\varnothing\).
26 tháng 8 2021
d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)
\(\Leftrightarrow8x+4-6+6x\ge12-3x\)
\(\Leftrightarrow14x+3x\ge12+2=14\)
\(\Leftrightarrow x\ge\dfrac{14}{17}\)
e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)
\(\Leftrightarrow6x+12+4x-8< 6x-9\)
\(\Leftrightarrow4x< -9+8-12=-13\)
hay \(x< -\dfrac{13}{4}\)
17 tháng 2 2022
a: \(\Leftrightarrow30\left(x-3\right)-16=9\left(x-1\right)+72\)
\(\Leftrightarrow30x-90-16=9x-9+72\)
=>30x-106=9x+63
=>21x=169
hay x=169/21
b: =>4x+20=2x-3
=>2x=-23
hay x=-23/2
11 tháng 7 2023
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
9 tháng 8 2023
b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
14 tháng 2 2022
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
=>x=0(nhận) hoặc x=3(loại)
14 tháng 2 2022
đk : x khác -1 ; 3
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x\Leftrightarrow2x^2-2x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\Leftrightarrow2x\left(x-3\right)=0\Leftrightarrow x=0;x=3\left(ktm\right)\)
30 tháng 4 2023
a: =>\(\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)
=>x^2-3x-4x=-x^2-x
=>x^2-7x+x^2+x=0
=>2x^2-6x=0
=>x=0(nhận) hoặc x=3(loại)
b: =>\(\dfrac{2x-3-3x-15}{x+5}>=0\)
=>\(\dfrac{-x-18}{x+5}>=0\)
=>x+18/x+5<=0
=>-18<=x<-5
KZ
30 tháng 4 2023
\(\dfrac{x}{2x+1}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\) (ĐKXĐ: \(x\ne3;x\ne-1\)
\(\Leftrightarrow\dfrac{x}{2x+1}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=-\dfrac{x}{2\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\dfrac{2.2x}{2\left(x-3\right)\left(x+1\right)}=-\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x^2-3x-4x=-x^2-x\)
\(\Leftrightarrow x^2-7x=-x^2-x\)
\(\Leftrightarrow x^2+x^2-7x+x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
*TM: Thỏa mãn, KTM: Ko thỏa mãn
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
\(\dfrac{2x-3}{x+5}\ge3\) (ĐKXĐ: \(x\ne-5\)
\(\Leftrightarrow\dfrac{2x-3}{x+5}-3\ge0\)
\(\Leftrightarrow\dfrac{2x-3}{x+5}-\dfrac{3x+15}{x+5}\ge0\)
\(\Leftrightarrow\dfrac{2x-3-3x-15}{x+5}\ge0\)
\(\Leftrightarrow\dfrac{-x-18}{x+5}\ge0\)
\(\Leftrightarrow-18\le x\le-5\)
\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\) (ĐKXĐ: \(x\ne0;x\ne\dfrac{1}{2}\))
\(\)\(\Leftrightarrow\dfrac{x+3}{x}=\dfrac{2\left(x+1\right)}{2x-1}\Leftrightarrow\left(x+3\right)\left(2x-1\right)=2x\left(x+1\right)\)
\(\Leftrightarrow2x^2+6x-x-3=2x^2+2x\)
\(\Leftrightarrow2x^2-2x^2+6x-x-2x=3\)
\(\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)\)
\(\Rightarrow S=\left\{1\right\}\)
\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=x\left(2x+2\right)\)
\(\Leftrightarrow2x^2-x+6x-3=2x^2+2x\)
\(\Leftrightarrow2x^2+5x-3-2x^2-2x=0\)
\(\Leftrightarrow3x-3=0\)
\(\Leftrightarrow3\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)