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4x-(2x+1)=3-\(\frac{1}{3}+x\)
số 3 trừ cho 1 phần 3 nha
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7 tháng 2 2018
ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)
TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)
\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)
BT
7 tháng 2 2018
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)
<=> x-2=1-2x <=> 3x=3
=> x=1
Đáp số: x=1
11 tháng 8 2015
a) \(\frac{3}{2}-\frac{5}{6}:x=\frac{5}{15}-\frac{3}{15}\)
\(\Leftrightarrow\frac{3}{2}-\frac{5}{6}:x=\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{3}{2}-\frac{2}{15}\)
\(\Leftrightarrow\frac{5}{6}:x=\frac{41}{30}\)
\(\Leftrightarrow x=\frac{5}{6}:\frac{41}{30}\)
\(\Leftrightarrow x=\frac{25}{41}\)
b) \(x-\frac{6}{7}.\frac{14}{8}=\frac{1}{2}-\frac{2}{5}\)
\(\Leftrightarrow x-\frac{3}{2}=\frac{1}{10}\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{3}{2}\)
\(\Leftrightarrow x=\frac{8}{5}\)
c) \(x:\frac{6}{5}+\frac{2}{3}=\frac{7}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{7}{3}-\frac{2}{3}\)
\(\Leftrightarrow x:\frac{6}{5}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{5}{3}.\frac{6}{5}\)
\(\Leftrightarrow x=2\)
NG
3
20 tháng 8 2020
Ta có: \(\left(4,5-2x\right):\frac{3}{4}=1\frac{1}{3}\)
=> \(\left(4,5-2x\right):\frac{3}{4}=\frac{4}{3}\)
=> \(4,5-2x=1\)
=> \(2x=3,5\Rightarrow x=1,75\)
Vậy x=1,75
3 tháng 7 2017
|2x-5|=4
\(\Leftrightarrow\orbr{\begin{cases}2x-5=4\\2x-5=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=9\\2x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{1}{2}\end{cases}}\)
4 tháng 4 2017
\(\left(3\frac{1}{2}+2x\right).2\frac{2}{3}=5\frac{1}{3}\)
<=>\(\left(\frac{7}{2}+2x\right).\frac{8}{3}=\frac{16}{3}\)
<=>\(\frac{28}{3}+\frac{16x}{3}=\frac{16}{3}\)
<=>\(\frac{16x}{3}=\frac{-2}{3}\)
<=>\(16x=-2\)
<=>\(x=\frac{-1}{8}\)
vậy \(x=\frac{-1}{8}\)
b,\(\left|2x+3\right|=5\)
xét x<0,ta co: \(\left|2x+3\right|=5\)<=> \(-2x+3=5\)<=>\(-2x=2\)<=>\(x=-1\)(loại)
xét x>0,ta co:\(\left|2x+3\right|=5\)<=>\(2x+3=5\)<=>\(2x=2\)<=>\(x=1\)
c,\(\frac{x-2}{4}=\frac{5+x}{3}\)
<=>\(\frac{3x-6}{12}=\frac{20+4x}{12}\)
=>\(3x-6=20+4x\)
<=>\(3x-6-20-4x=0\)
<=>\(-x-26=0\)
<=>\(-x=26\)
<=>\(x=-26\)
kl:.......
16 tháng 3 2023
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
29 tháng 6 2021
ĐKXĐ ; \(x\ne\pm1\)
Ta có : \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}+\dfrac{-x^2-3}{x^2-1}=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2-x^2-3=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-x^2-3=0\)
\(\Leftrightarrow-x^2+4x-3=0\)
\(\Leftrightarrow-x^2+3x+x-3=0\)
\(\Leftrightarrow-x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=1\left(L\right)\end{matrix}\right.\)
=> X = 3
Vậy ..
H
16 tháng 3 2023
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
__
`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
9 tháng 5 2019
a) \(f\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5\)
\(g\left(x\right)=x^4+3x^3-\frac{2}{3}x^2-2x-10\)
b) \(f\left(x\right)+g\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5+x^4+3x^3-\frac{2}{3}x^2-2x-10\)
\(=6x^3-x^2-5\)
c) +) Thay x=1 vào đa thức f(x) + g(x) ta được :
\(6.1^3-1^2-5=0\)
Vậy x=1 là nghiệm của đa thức f(x) + g(x)
+) Thay x=-1 vào đa thức f(x) + g(x) ta được :
\(6.\left(-1\right)^3-\left(-1\right)^2-5=-10\)
Vậy x=-1 ko là nghiệm của đa thức f(x) + g(x)
Bên trên có nhé ! Nghĩ tvm thôi làm luôn ko ko biết tìm ở đâu.
\(4x-\left(2x+1\right)=3-\frac{1}{3}+x\)
\(\Leftrightarrow4x-2x-1=\frac{8}{3}+x\Leftrightarrow2x-1=\frac{8}{3}+x\)
\(\Leftrightarrow x=\frac{11}{3}\)
\(4x-\left(2x+1\right)=3-\frac{1}{3}+x\)
\(4x-2x-1-x=3-\frac{1}{3}\)
\(4x-2x-x=\frac{8}{3}+1\)
\(x=\frac{8}{3}+\frac{3}{3}\)
\(x=\frac{11}{3}\)
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