60/5+x=60/x-1 giải pt
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NC
31 tháng 8 2020
Bài làm:
1) đk: \(x\ne0;x\ne-5\)
Ta có: \(\frac{30}{x}-\frac{30}{x+5}=1\)
\(\Leftrightarrow\frac{30\left(x+5\right)-30x}{x\left(x+5\right)}=1\)
\(\Leftrightarrow x^2+5x=150\)
\(\Leftrightarrow x^2+5x-150=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)
2) đk: \(x\ne0;x\ne-2\)
Ta có: \(\frac{60}{x}-\frac{60}{x+2}=1\)
\(\Leftrightarrow\frac{60\left(x+2\right)-60x}{x\left(x+2\right)}=1\)
\(\Leftrightarrow x^2+2x=120\)
\(\Leftrightarrow x^2+2x-120=0\)
\(\Leftrightarrow\left(x-10\right)\left(x+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
31 tháng 8 2020
\(\frac{30}{x}-\frac{30}{x+5}=1\)( ĐKXĐ : \(x\ne0;x\ne-5\))
<=> \(30\left(\frac{1}{x}-\frac{1}{x+5}\right)=1\)
<=> \(30\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}\right)=1\)
<=> \(30\left(\frac{5}{x\left(x+5\right)}\right)=1\)
<=> \(\frac{5}{x\left(x+5\right)}=\frac{1}{30}\)
<=> \(5\cdot30=x\left(x+5\right)\)
<=> \(x^2+5x-150=0\)
<=> \(x^2+15x-10x-150=0\)
<=> \(x\left(x+15\right)-10\left(x+15\right)=0\)
<=> \(\left(x-10\right)\left(x+15\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)( tmđk )
Vậy S = { 10 ; -15 }
\(\frac{60}{x}-\frac{60}{x+2}=1\)( ĐKXĐ : \(x\ne0;x\ne-2\))
<=> \(60\left(\frac{1}{x}-\frac{1}{x+2}\right)=1\)
<=> \(60\left(\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}\right)=1\)
<=> \(60\left(\frac{2}{x\left(x+2\right)}\right)=1\)
<=> \(\frac{2}{x\left(x+2\right)}=\frac{1}{60}\)
<=> \(2\cdot60=x\left(x+2\right)\)
<=> \(x^2+2x-120=0\)
<=> \(x^2+12x-10x-120=0\)
<=> \(x\left(x+12\right)-10\left(x+12\right)=0\)
<=> \(\left(x-10\right)\left(x+12\right)=0\)
<=> \(\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)
Vậy S = { 10 ; -12 }
9 tháng 8 2023
b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)
=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)
=>x*(x+20)=400*6=2400
=>x^2+20x-2400=0
=>(x+60)(x-40)=0
=>x=-60 hoặc x=40
c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1(nhận)
21 tháng 4 2017
\(\dfrac{1}{x}+\dfrac{1}{x+50}=\dfrac{1}{60}\left(x\ne0;x\ne-5\right)\)
\(pt\Leftrightarrow\dfrac{x+50}{x\left(x+50\right)}+\dfrac{x}{x\left(x+50\right)}=\dfrac{1}{60}\)
\(\Leftrightarrow\dfrac{2x+50}{x\left(x+50\right)}=\dfrac{1}{60}\Leftrightarrow x\left(x+50\right)=60\left(2x+50\right)\)
\(\Leftrightarrow x^2+50x=120x+3000\)
\(\Leftrightarrow x^2-70x-3000=0\)
\(\Leftrightarrow x^2-100x+30x-3000=0\)
\(\Leftrightarrow x\left(x-100\right)+30\left(x-100\right)=0\)
\(\Leftrightarrow\left(x+30\right)\left(x-100\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+30=0\\x-100=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-30\\x=100\end{matrix}\right.\)
TN
1
30 tháng 1 2019
\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{5}{y}=\dfrac{2}{3}\\\dfrac{5}{x}+\dfrac{4}{y}=\dfrac{41}{60}\end{matrix}\right.\left(I\right)\)
Đặt \(:\left\{{}\begin{matrix}t=\dfrac{1}{x}\\u=\dfrac{1}{y}\end{matrix}\right.\)
\(\left(I\right):\left\{{}\begin{matrix}4t+5u=\dfrac{2}{3}\\5t+4u=\dfrac{41}{60}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}20t+25u=\dfrac{10}{3}\\20t+16u=\dfrac{41}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9u=\dfrac{3}{5}\\20t+16u=\dfrac{41}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{1}{15}\\t=\dfrac{1}{12}\end{matrix}\right.\)
Với \(:\left\{{}\begin{matrix}t=\dfrac{1}{12}\\u=\dfrac{1}{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{12}\\\dfrac{1}{y}=\dfrac{1}{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=12\\y=15\end{matrix}\right.\)
Vậy nghiệm hệ phương trình là \(\left(12;15\right)\)
HT
2
ND
4
3 tháng 9 2022
b: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-4x^2+11x-2}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+4x+4+4x^2-11x+2=0\)
\(\Leftrightarrow5x^2-7x+6=0\)
hay \(x\in\varnothing\)
c: \(\Leftrightarrow\left(3x^2+2\right)^2-5x\left(3x^2+2\right)=0\)
=>3x^2-5x+2=0
=>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
NT
1
3 tháng 8 2019
\(\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10}\)
\(\Leftrightarrow\frac{60}{x}=\frac{30}{x-6}+\frac{30}{x+10},Đkxđ:x\ne0,6,-10\)
\(\Leftrightarrow\frac{60}{x}-\frac{30}{x-6}-\frac{30}{x+10}=0\)
\(\Leftrightarrow\frac{60\left(x-6\right)\left(x+10\right)-30x\left(x+10\right)=30\left(x-6\right)}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{\left(60x-360\right)\left(x+10\right)-30x^2-300x-30x^2+180x}{x\left(x-6\right)\left(x+10\right)}\)
\(\Leftrightarrow\frac{60x^2+600x-360x-3600-30x^2-300x-30x^2+180}{x\left(x-6\right)\left(x=10\right)}=0\)
\(\Leftrightarrow\frac{120x-3600}{x\left(x-6\right)\left(x+10\right)}=0\)
\(\Leftrightarrow120x-3600=0\)
\(\Leftrightarrow120x=3600\)
\(\Leftrightarrow x=30;x\ne0;x\ne6,x\ne-10\)
\(\frac{60}{5}+x=\frac{60}{x-1}\) (x khác 1)
\(\Leftrightarrow12+x=\frac{60}{x-1}\)
\(\Leftrightarrow\frac{x^2-x-60}{x-1}=12\)\(\Leftrightarrow x^2-x-60=12x-12\)
\(\Leftrightarrow x^2+11x-48=0\)
ĐỂ PT CÓ NGHIỆM THÌ
\(\Delta=11^2+4.48=121+192=323>0\)
giải tiếp là ra