giải phương trình:
\(\dfrac{60}{x-10}-\dfrac{60}{x}=\dfrac{3}{10}\)
K
Khách
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Những câu hỏi liên quan
3 tháng 2 2021
\(\dfrac{x-130}{20}\)+\(\dfrac{x-100}{25}\)+\(\dfrac{x-60}{30}\)+\(\dfrac{x-10}{35}\)=10
⇔\(\dfrac{2625\left(x-130\right)}{52500}\)+\(\dfrac{2100\left(x-100\right)}{52500}\)+\(\dfrac{1750\left(x-60\right)}{52500}\)+\(\dfrac{1500\left(x-10\right)}{52500}\)=\(\dfrac{525000}{52500}\)
⇔2625\(x\)-341250+2100\(x\)-210000+1750\(x\)-105000+1500\(x\)-15000=525000
⇔ 7975\(x\) = 1196250
⇔ \(x\) = \(\dfrac{1196250}{7975}\)
⇔\(x \) = 150
CV
1
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 5 2023
\(\dfrac{100}{x}-\dfrac{100}{x+10}=\dfrac{30}{60}=0,5\left(ĐKXĐ:x\ne0;x\ne-10\right)\\ \Leftrightarrow\dfrac{100\left(x+10\right)-100x}{x\left(x+10\right)}=\dfrac{0,5x\left(x+10\right)}{x\left(x+10\right)}\\ \Leftrightarrow100x-100x+1000=0,5x^2+5x\\ \Leftrightarrow0,5x^2+5x-1000=0\\ \Leftrightarrow0,5x^2-20x+25x-1000=0\\ \Leftrightarrow0,5x.\left(x-40\right)+25.\left(x-40\right)=0\\ \Leftrightarrow\left(0,5x+25\right)\left(x-40\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,5x+25=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-50\\x=40\end{matrix}\right.\\ Vậy:S=\left\{-50;40\right\}\)
27 tháng 11 2018
\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)
\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)
\(\Rightarrow11\left(x-2\right)=-20x+40\)
\(\Rightarrow11x-22=-20x+40\)
\(\Rightarrow11x+20x=22+40\)
\(\Rightarrow31x=62\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
KC
2
15 tháng 2 2022
Nhg đây là giải pt ạ. Nếu đã là giải phương trình ko đc nhân chéo đâu ạ
AH
Akai Haruma
Giáo viên
23 tháng 8 2023
Lời giải:
PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$
$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$
Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$
$\Rightarrow x+100=0$
$\Leftrightarrow x=-100$ (tm)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 11 2018
\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}...\dfrac{9.10}{9^2}.\dfrac{10.11}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)
\(\Leftrightarrow\dfrac{1.2^2.3^2.4^2...10^2.11}{1^2.2^2.3^2....10^2}\left(x-2\right)+20\left(x+1\right)=60\)
\(\Leftrightarrow11\left(x-2\right)+20\left(x+1\right)=60\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
CV
1
Y
5 tháng 6 2023
\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)
\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)
\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)
\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)
\(\Leftrightarrow-3x^2+1230x-6000=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)
Vậy ...
12 tháng 1 2021
ĐKXĐ: \(x\neq 0\).
Đặt \(\dfrac{x}{3}-\dfrac{4}{x}=t\).
PT đã cho tương đương:
\(3t^2+8-10t=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=\dfrac{4}{3}\end{matrix}\right.\).
Với t = 2 ta có \(\dfrac{x}{3}-\dfrac{4}{x}=2\Leftrightarrow\dfrac{x^2-12}{3x}=2\Leftrightarrow x^2-6x-12=0\Leftrightarrow x=\pm\sqrt{21}+3\).
Với t = \(\frac{4}{3}\) ta có \(\dfrac{x}{3}-\dfrac{4}{x}=\dfrac{4}{3}\Leftrightarrow\dfrac{x^2-12}{3x}=\dfrac{4}{3}\Leftrightarrow x^2-12=4x\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\).
Vậy...
BL
Giải phương trình \(1+\dfrac{2}{x-2}=\dfrac{-10}{x+3}+\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)
1
HP
16 tháng 11 2021
\(1+\dfrac{2}{x-2}=\dfrac{-10}{x+3}+\dfrac{50}{\left(2-x\right)\left(x+3\right)}\left(ĐK:x\ne2;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{\left(2-x\right)\left(x+3\right)}{\left(2-x\right)\left(x+3\right)}-\dfrac{2}{2-x}=\dfrac{-10\left(2-x\right)}{\left(2-x\right)\left(x+3\right)}+\dfrac{50}{\left(2-x\right)\left(x+3\right)}\)
\(\Leftrightarrow2x+6-x^2-3x-2=-20+10x+50\)
\(\Leftrightarrow-x^2+2x-3x-10x+6-2+20-50=0\)
\(\Leftrightarrow-x^2-11x-26=0\)
\(\Leftrightarrow-\left(x^2+2x-13x+26\right)=0\)
\(\Leftrightarrow x\left(x+2\right)-13\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow3x\left(x-10\right)=60x-60\left(x-10\right)\)
\(\Leftrightarrow3x\left(x-10\right)=600\)
\(\Leftrightarrow x^2-10x-200=0\)
=>(x-20)(x+10)=0
=>x=20 hoặc x=-10
\(\dfrac{60}{x-10}-\dfrac{60}{x}=\dfrac{3}{10}\)đk : x khác 10 ; 0
\(\Leftrightarrow600x-600\left(x-10\right)=3x\left(x-10\right)\)
\(\Leftrightarrow3x^2-30x-6000=0\Leftrightarrow x=50;x=-40\left(tm\right)\)