Giải phương trình x-5/x+5-x+5/x-5= x+95/25-x2
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
TN
1
22 tháng 4 2022
a.\(x^2-25=8\left(5-x\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)
b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)
\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
DT
2
7 tháng 3 2021
Ta có: \(\dfrac{x-25}{75}+\dfrac{x-15}{85}+\dfrac{x-5}{95}+\dfrac{x-145}{15}=0\)
\(\Leftrightarrow\dfrac{x-25}{75}-1+\dfrac{x-15}{85}-1+\dfrac{x-5}{95}-1+\dfrac{x-145}{15}+3=0\)
\(\Leftrightarrow\dfrac{x-100}{75}+\dfrac{x-100}{85}+\dfrac{x-100}{95}+\dfrac{x-100}{15}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}\right)=0\)
mà \(\dfrac{1}{75}+\dfrac{1}{85}+\dfrac{1}{95}+\dfrac{1}{15}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
25 tháng 2 2019
\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)
\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)
\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)
\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)
NH
1
18 tháng 7 2021
\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)
\(< =>2x-7=0< =>x=\dfrac{7}{2}\)
b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)
\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)
\(< =>-25x-11=0\)
\(< =>x=-0,44\)
H
5 tháng 4 2023
\(\dfrac{x+5}{x-5}=\dfrac{5}{x^2-5x}+\dfrac{1}{x}\)
\(\Leftrightarrow\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
Ta có : \(\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)
\(\Leftrightarrow\dfrac{x\left(x+5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}+\dfrac{x-5}{x\left(x-5\right)}\)
`=> x (x+5) = 5 +x-5`
`<=> x^2 +5x - 5-x+5=0`
`<=> x^2 +4x =0`
`<=> x(x+4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-4\end{matrix}\right.\)
Vậy phương trình có nghiệm `x=-4`
CT
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
ĐKXĐ: \(x\ne\left\{0;-5\right\}\)
\(\Leftrightarrow\dfrac{11}{x^2}-\left[1-\dfrac{10}{x+5}+\left(\dfrac{5}{x+5}\right)^2+\dfrac{10}{x+5}\right]=0\)
\(\Leftrightarrow\dfrac{11}{x^2}-\left[\left(1-\dfrac{5}{x+5}\right)^2+\dfrac{10}{x+5}\right]=0\)
\(\Leftrightarrow\dfrac{11}{x^2}-\dfrac{10}{x+5}-\left(\dfrac{x}{x+5}\right)^2=0\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{x}{x+5}\right)\left(\dfrac{11}{x}+\dfrac{x}{x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{x}-\dfrac{x}{x+5}=0\\\dfrac{11}{x}+\dfrac{x}{x+5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-5=0\\x^2+11x+55=0\end{matrix}\right.\)
\(\Leftrightarrow...\) (bấm máy)
HT
1
10 tháng 4 2018
\(\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
TL
2
19 tháng 2 2022
a: \(\Leftrightarrow2x^2+6x-x^2-2x+x+2-x^2-6=0\)
=>5x-4=0
hay x=4/5
b: \(\Leftrightarrow\left(x-5\right)\left(x+x+3\right)=0\)
=>(x-5)(2x+3)=0
=>x=5 hoặc x=-3/2
MH
19 tháng 2 2022
a) \(2x\left(x+3\right)-\left(x-1\right)\left(x+2\right)=x^2+6\)
\(2x^2+6x-\left(x^2+2x-x-2\right)=x^2+6\)
\(x^2+5x+2=x^2+6\)
\(x^2+5x+2-x^2-6=0\)
\(5x-4=0\)
\(x=\dfrac{4}{5}\)
b) \(x\left(x-5\right)+\left(x-5\right)\left(x+3\right)=0\)
\(\left(x-5\right)\left(x+x+3\right)=0\)
\(\left(x-5\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{3}{2}\end{matrix}\right.\)
9 tháng 3 2022
\(\Leftrightarrow\dfrac{x-5}{95}-1+\dfrac{x-132}{32}+1=\dfrac{x-131}{31}+1+\dfrac{x-10}{90}-1\)
=>x-100=0
hay x=100
Mình nên chết đi cho rồi
Mình sửa lại tí ở dấu <=> hàng thứ 3
\(\Leftrightarrow\left(x-5\right)^2-\left(x+5\right)^2=-x-95\\ \Leftrightarrow-20x=-x-95\\ \Leftrightarrow-19x=-95\\ \Leftrightarrow x=\frac{-95}{-19}=5\left(loại\right)\)
Vậy \(S=\varnothing\)