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TT
0
AT
26 tháng 4 2018
heoheo lần sau bạn đánh = kí hiệu đi :(((
a/ \(\dfrac{x}{3}+\dfrac{2x-1}{6}=\dfrac{1}{2}\)
\(\Leftrightarrow2x+2x-1=3\)
<=> 4x = 4 <=> x = 1
Vậy x = 1
b/ \(\dfrac{3x+1}{2}+\dfrac{x-1}{3}=\dfrac{x-9}{6}\)
\(\Leftrightarrow3\left(3x+1\right)+2\left(x-1\right)=x-9\)
\(\Leftrightarrow9x+3+2x-2=x-9\)
\(\Leftrightarrow10x=-10\Leftrightarrow x=-1\)
Vậy pt có nghiệm x = -1
c/ \(\dfrac{x-1}{x-2}=\dfrac{x+3}{x+2}\) ĐKXĐ: \(x\ne\pm2\)
<=> \(\left(x-1\right)\left(x+2\right)=\left(x+3\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+2x-x-2=x^2-2x+3x-6\)
\(\Leftrightarrow0x=-4\left(voly\right)\)
Vậy pt vô nghiệm
d/ \(\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}=2\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-3\\x\ne-\dfrac{1}{3}\end{matrix}\right.\)
pt <=> \(\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}+\dfrac{\left(x-3\right)\left(3x+1\right)}{\left(3x+1\right)\left(x+3\right)}=\dfrac{2\left(3x+1\right)\left(x+3\right)}{\left(3x+1\right)\left(x+3\right)}\)
=> (3x-1)(x+3) + (x-3)(3x+1) = 2(3x+1)(x+3)
\(\Leftrightarrow3x^2+8x-3+3x^2-8x-3=6x^2+20x+6\)
\(\Leftrightarrow-20x=12\Leftrightarrow x=-\dfrac{3}{5}\left(tm\right)\)
Vậy pt có nghiệm x=....
e/ như ý d
PD
1
GH
12 tháng 2 2018
a) \(4x-3=11-3x\)
\(\Leftrightarrow4x+3x=11+3\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
Vậy .............
b) \(x^3-4x^2+3x=0\)
\(\Leftrightarrow x\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x\left(x^2-x-3x+3\right)=0\)
\(\Leftrightarrow x\left[x\left(x-1\right)-3\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=3\end{matrix}\right.\)
Vậy .................
P/s: câu c bn gõ lại dc ko
4 tháng 6 2022
c: \(\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)
\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)
\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}=\dfrac{x^2-10x+25}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\)
e: \(\dfrac{4x^2-3x+17}{x^3-1}+\dfrac{2x-1}{x^2+x+1}+\dfrac{6}{1-x}\)
\(=\dfrac{4x^2-3x+17+\left(2x-1\right)\left(x-1\right)-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-2x^2-9x+11+2x^2-3x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-12}{x^2+x+1}\)
GH
2 tháng 3 2018
\(\dfrac{2x}{x-1}+\dfrac{3x-2}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\left(ĐKXĐ:x\ne1;x\ne-2\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\)
\(\Rightarrow2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)=6\)
\(\Leftrightarrow2x^2+4x+3x^2-3x-2x+2-6=0\)
\(\Leftrightarrow5x^2-x-4=0\)
\(\Leftrightarrow5x^2-5x+4x-4=0\)
\(\Leftrightarrow5x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=\dfrac{-4}{5}\left(n\right)\end{matrix}\right.\)
Vậy .....................
M
2 tháng 3 2018
ĐK: $x\ne 1; x\ne -2$.
\(\dfrac{2x}{x-1}+\dfrac{3x-2}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}\\ \Rightarrow2x\left(x+2\right)+\left(3x-2\right)\left(x-1\right)=6\\ \Leftrightarrow2x^2+4x+3x^2-3x-2x+2=6\\ \Leftrightarrow5x^2-x-4=0\\ \Leftrightarrow5x^2-5x+4x-4=0\\ \Leftrightarrow5x\left(x-1\right)+4\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)
$\Leftrightarrow x-1=0$ hoặc $5x+4=0$
$\Leftrightarrow x=1$ (loại) hoặc $x=\dfrac{-4}5$
OS
1
9 tháng 10 2019
a)(x+1 phần 3).(x^2-1 phần 3x + 1 phần 9)
b) (x-2y)^2 - (x+2y)^2
mình chép lại đề
ĐKXĐ: \(x\ne-1\) , \(x\ne2\), \(x\ne-2\)
\(\frac{2}{x+1}-\frac{1}{x+2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2\left(x+2\right)\left(x-2\right)-\left(x+1\right)\left(x-2\right)-\left(3x-11\right)\left(x+2\right)=0\)
\(\Rightarrow2\left(x^2-4\right)-\left(x^2-x-2\right)-\left(3x^2-5x-22\right)=0\)
\(\Rightarrow2x^2-8-x^2+x+2-3x^2+5x+22=0\)
\(\Rightarrow-2x^2+6x+16=0\)
\(\Rightarrow-x^2+3x+8=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{41}}{2}\\x=\frac{3-\sqrt{41}}{2}\end{cases}}\)