x+5/x^2-5x - x-5/2x^2=10x = x+25/2x^2-50
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HC
2
19 tháng 7 2021
\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)
\(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+5}{2\left(x-5\right)\left(x+5\right)}\)
dkxd : x ≠ 0
x ≠ 5
x ≠ -5
MTC : 2x(x - 5)(x + 5)
Quy đồng mẫu thức hai vế của phương trình :
⇒ \(\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}\) = \(\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
Suy ra : 2(x - 5)(x + 5) - (x - 5)(x + 5) = x(x + 25)
\(\Leftrightarrow\) 2(x2 - 25) - (x2 - 25) = x2 + 25x
\(\Leftrightarrow\) 2x2 - 50 - x2 + 25 - x2 - 25x = 0
\(\Leftrightarrow\) -25 - 25x = 0
\(\Leftrightarrow\) -25x = 25
\(\Leftrightarrow\) x = \(\dfrac{25}{-25}=-1\) (thỏa mãn)
Vậy S = \(\left\{-1\right\}\)
Chúc bạn học tốt
19 tháng 7 2021
Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
hay \(x=-\dfrac{5}{3}\)(thỏa ĐK)
KT
2
TT
26 tháng 5 2021
\(x\ne0;x\ne\pm5\)
PT \(\Leftrightarrow\dfrac{x+25}{2\left(x+5\right)\left(x-5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}=0\)
\(\Rightarrow x^2+25x-2x^2-20x-50+x^2-10x+25=0\)
\(\Leftrightarrow-5x-25=0\)
\(\Leftrightarrow x=-5\) (ktm)
Vậy pt vô nghiệm.
26 tháng 5 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\).
\(PT\Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow x\left(x+25\right)-2\left(x+5\right)^2=\left(5-x\right)\left(x-5\right)\)
\(\Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=10x-x^2-25\)
\(\Leftrightarrow-5x=25\Leftrightarrow x=-5\) (loại)
Vậy PT vô nghiệm
15 tháng 5 2020
ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow5x=-25\)
hay x=-5(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)
15 tháng 8 2020
ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)
Ta có: \(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)
\(\Leftrightarrow\frac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}-\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=0\)
Suy ra: \(x^2+25x-2\left(x^2+10x+25\right)+x^2-10x+25=0\)
\(\Leftrightarrow2x^2+15x+25-2x^2-20x-50=0\)
\(\Leftrightarrow-5x-25=0\)
\(\Leftrightarrow-5x=25\)
hay x=-5(loại)
Vậy: \(S=\varnothing\)
12 tháng 8 2021
\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)
\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)
Tick nha
HT
0
NH
2
DD
26 tháng 4 2019
ĐKXĐ : \(x\ne0;x\ne\pm5\)
\(\frac{x+5}{x^2-5x}-\frac{x-5}{2x^2+10x}=\frac{x+25}{2x^2-50}\)
\(\Leftrightarrow\frac{x+5}{x\left(x-5\right)}-\frac{x-5}{2x\left(x+5\right)}=\frac{x+25}{2\left(x-5\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\frac{\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\frac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow2\left(x+5\right)^2-\left(x-5\right)^2=x\left(x+25\right)\)
\(\Leftrightarrow2x^2+20x+50-x^2+10x-25=x^2+25x\)
\(\Leftrightarrow5x+25=0\)
\(\Leftrightarrow x=-5\)(ko t/m ĐKXĐ)
Vậy phương trình vô nghiệm.
28 tháng 7 2021
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
28 tháng 7 2021
c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
`(x+5)/(x^2-5x)-(x-5)/(2x^2+10x)=(x+25)/(2x^2-50)`
ĐK:`x ne 0,x ne 5,x ne -5`
Nhân 2 vế với `2x(x+5)(x-5)` ta có phương trình:
`2(x+5)(x+5)-(x-5)(x-5)=x(x+25)`
`<=>2(x^2+10x+25)-(x^2-10x+25)=x^2+25x`
`<=>x^2+30x+25=x^2+25x`
`<=>5x+25=0`
`<=>5x=-25`
`<=>x=-5(l)`
Vậy pt vô nghiệm