giải pt \(\frac{120}{x}+\frac{120}{x+10}=7\)7x x + 10...
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giải pt \(\frac{120}{x}+\frac{120}{x+10}=7\)
7x x + 10
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2S
0
NT
1
D
6 tháng 8 2019
\(ĐK:x\ne0;x\ne-6\)
⇔ \(\frac{720\left(x+6\right)}{6x\left(x+6\right)}=\frac{6x\left(x+6\right)}{6x\left(x+6\right)}+\frac{x\left(x+6\right)}{6x\left(x+6\right)}+\frac{6x\left(120-x\right)}{6x\left(x+6\right)}\)
\(\Rightarrow720x+4320=6x^2+36x+x^2+6x+720x-6x^2\)
\(\Leftrightarrow6x^2+36x+x^2+6x+720x-6x^2-720x-4320=0\)
\(\Leftrightarrow x^2+42x-4320=0\)
\(\Leftrightarrow x^2+90x-48x-4320=0\)
\(\Leftrightarrow\left(x+90\right)\left(x-48\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+90=0\\x-48=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-90\\x=48\end{matrix}\right.\) ( tm )
5 tháng 3 2020
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{9x}{x^2-7x+10}=10\)
\(\Leftrightarrow\frac{3x^2-15x-x^2+2x+9x}{\left(x-2\right)\left(x-5\right)}=10\)
\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)
\(\Leftrightarrow8x^2-66+100=0\)
\(\Leftrightarrow4x^2-33x+50=0\)
\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)
HP
6 tháng 3 2020
b) [(x-7)(x-2)][(x-4)(x-5)]=72
<=> (x2-9x+14)(x2-9x+20)=72
Đặt x2-9x+17=a
=> (a+3)(a-3)=72
<=> a2-9=72
<=> a2=81
=> a=\(\left\{9;-9\right\}\)
TH1: a=9
=> x2-9x+17=9
<=> x2-9x+8=0
<=> (x-1)(x-8)=0
=> x=\(\left\{1;8\right\}\)
TH2: a=-9
=> x2-9x+17=-9
<=> x2-9x+26=0
<=> x2-9x+20,25+5,75=0
<=> (x-4,5)2+5,75=0
=> x\(\in\varnothing\)
Vậy x=\(\left\{1;8\right\}\)
KN
5 tháng 3 2020
\(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)
\(\Rightarrow\frac{3x^2-15x-x^2+2x+9x}{x^2-7x+10}=10\)
\(\Rightarrow\frac{2x^2-4x}{x^2-7x+10}=10\)
\(\Rightarrow2x^2-4x=10x^2-70x+100\)
\(\Rightarrow8x^2-66x+100=0\)
Ta có \(\Delta=66^2-4.8.100=1156,\sqrt{\Delta}=34\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{66+34}{16}=\frac{25}{4}\\x=\frac{66-34}{16}=2\end{cases}}\)
5 tháng 3 2020
a) \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)
<=> \(\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\frac{9x}{\left(x-2\right)\left(x-5\right)}=10\)
<=> \(\frac{3x^2-15x-x^2+2x+9x}{\left(x-5\right)\left(x-2\right)}=10\)
<=> \(\frac{2x^2-4x}{\left(x-5\right)\left(x-2\right)}=10\)
<=> \(\frac{2x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=10\)
<=> \(2x=10\left(x-5\right)\)
<=> 2x - 10x = -50
<=> -8x = -50
<=>x = 6,25
Vậy S = {6,25}
b) (x - 7)(x - 2)(x - 4)(x - 5) = 72
<=> (x2 - 9x + 14)(x2 - 9x + 20) = 72
Đặt x2 - 9x + 14 = t <=> t(t + 6) = 72
<=> t2 + 6t - 72 = 0
<=> t2 + 12t - 6t - 72 = 0
<=> (t + 12)(t - 6) = 0
<=> \(\orbr{\begin{cases}t+12=0\\t-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x^2-9x+14+12=0\\x^2-9x+14-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x-9x+20,25\right)+5,75=0\\x^2-9x+8=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x-4,5\right)^2+5,75=0\left(vn\right)\\x^2-x-8x+8=0\end{cases}}\)
<=> (x - 1)(x - 8) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\\x=8\end{cases}}\)
Vậy S = {1; 8}
29 tháng 10 2016
C, CHO 7X=3Y VA X -Y =16
=> \(\frac{x}{3}=\frac{y}{7}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\)
=> \(\hept{\begin{cases}x=-4.3\\y=-4.7\end{cases}\Rightarrow\hept{\begin{cases}x=-12\\y=-28\end{cases}}}\)
bạn viết lại đề đi đè gì mà sai hết
T
3
TA
27 tháng 6 2021
ĐK: ` x \ne 10; x \ne 0`
`120/(x-10)-3/5=120/x`
`<=>120/(x-10)-120/x=3/5`
`<=>1/(x-10) - 1/x= 1/200`
`<=> (x-x+10)/(x(x-10)) = 1/200`
`<=> 10/(x(x-10))= 1/200`
`<=> x^2-10=2000`
`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)
Vậy `S={50;-40}`.
27 tháng 6 2021
`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`
`<=>40/(x-10)-1/5=40/x`
`<=>200x-x(x-10)=200(x-10)`
`<=>200x-200x+2000-x^2+10x=0`
`<=>x^2-10x-2000=0`
`Delta'=25+2000=2025`
`<=>x_1=50,x_2=-40`
Vậy `S={50,-40}`
PT
giải hệ pt sau: \(\hept{\begin{cases}-x+y=-24\\\frac{120}{x}-\frac{120}{y}=\frac{5}{6}\end{cases}}\)
0
NT
2
D
3 tháng 8 2019
\(\Leftrightarrow\frac{200\left(x+20\right)}{2x\left(x+20\right)}-\frac{240x}{2x\left(x+20\right)}=\frac{x\left(x+20\right)}{2x\left(x+20\right)}\) đk: x\(\ne0\) , x \(\ne-20\)
\(\Rightarrow200x+4000-240x=x^2+20x\)
\(\Leftrightarrow-x^2-60x+4000=0\)
\(\Leftrightarrow x^2+60x-4000=0\)
\(\Leftrightarrow x^2+100x-40x-4000=0\)
\(\Leftrightarrow\left(x+100\right)\left(x-40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+100=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-100\left(tmđk\right)\\x=40\left(tmđk\right)\end{matrix}\right.\)
Vậy S\(=\left\{-100;40\right\}\)
3 tháng 8 2019
\(\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2}\)
\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}=\frac{1}{2},x\ne0,x\ne-20\)
\(\Leftrightarrow\frac{100}{x}-\frac{120}{x+20}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{200\left(x+20\right)-240x-x\left(x+20\right)}{2x\left(x+20\right)}=0\)
\(\Leftrightarrow\frac{200x+4000-240x-x^2-20x}{2x\left(x+20\right)}=0\)
\(\Leftrightarrow-60x+4000-x^2=0\)
\(\Leftrightarrow-x^2-60x+4000=0\)
\(\Leftrightarrow x^2+60x-4000=0\)
\(\Leftrightarrow\frac{-60\pm\sqrt{60^2}-4.1\left(-4000\right)}{2}\)
\(\Leftrightarrow\frac{-60\pm\sqrt{3600+16000}}{2}\)
\(\Leftrightarrow\frac{-60\pm\sqrt{19600}}{2}\)
\(\Leftrightarrow\frac{-60\pm140}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{-60+140}{2}\\\frac{-60-140}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-100\end{matrix}\right.,x\ne0,x\ne-20\)
1 tháng 4 2020
a) \(\frac{3}{7}x-1=\frac{1}{7}x\left(3x-7\right)\)
<=> \(3x-7=x\left(3x-7\right)\)
<=> \(\left(3x-7\right)-x\left(3x-7\right)=0\)
<=> \(\left(3x-7\right)\left(1-x\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{7}{3}\\x=1\end{cases}}\)
Vậy S = { 7/3; 1}
b) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
<=> \(\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
<=> \(\left(3x-1\right)\left(x^2-7x+12\right)=0\)
<=> \(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
<=> \(\left(3x-1\right)\left(x\left(x-3\right)-4\left(x-3\right)\right)=0\)
<=> \(\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
<=> x = 1/3 hoặc x = 3 hoặc x = 4.
Vậy S = { 1/3; 3; 4}
\(\frac{120}{x}+\frac{120}{x+10}=7\)
\(\Leftrightarrow\frac{120\left(x+10\right)}{x\left(x+10\right)}+\frac{120x}{x\left(x+10\right)}=\frac{7x\left(x+10\right)}{x\left(x+10\right)}\)
\(\Leftrightarrow120\left(x+10\right)+120x=7x\left(x+10\right)\)
\(\Leftrightarrow120x+1200+120x=7x^2+70x\)
\(\Leftrightarrow240x+1200=7x^2+70x\)
\(\Leftrightarrow240x+1200-7x^2-70x=0\)
\(\Leftrightarrow170x+1200-7x^2=0\)
\(\Leftrightarrow-7x^2+170x+1200=0\)
\(\Leftrightarrow7x^2+40x-210x-1200=0\)
\(\Leftrightarrow x\left(7x+40\right)-30\left(7x+40\right)=0\)
\(\Leftrightarrow\left(x-30\right)\left(7x+40\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=30\\7x=-40\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30\\x=-\frac{40}{7}\end{cases}}}\)