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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}`.
`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}`
\(\left\{{}\begin{matrix}x-y=10\\\dfrac{-120\left(x-y\right)}{xy}=\dfrac{2}{5}\end{matrix}\right.\) \(\Rightarrow\dfrac{-1200}{xy}=\dfrac{2}{5}\Rightarrow xy=-3000\)
Ta được hệ: \(\left\{{}\begin{matrix}x-y=10\\xy=-3000\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=y+10\\xy=-3000\end{matrix}\right.\)
Thay pt trên vào dưới:
\(\left(y+10\right).y=-3000\Rightarrow y^2+10y+3000=0\)
\(\Rightarrow\) pt vô nghiệm
Vậy hệ đã cho vô nghiệm
\(\left\{{}\begin{matrix}X+44=Y\\\dfrac{120}{X}+\dfrac{11}{30}=\dfrac{120}{Y}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}X=Y-44\\3600Y+11XY=3600X\end{matrix}\right.\)
\(3600Y+11\left(Y-44\right)Y=3600\left(Y-44\right)\\ =11Y^2-484Y+158400 =0\)
\(\Delta'=\left(-242\right)^2-158400.11=-1683836\)
=> DO \(\Delta'>0\) nên pt vô nghiệm
Đặt \(\dfrac{1}{y-1}=a\), hpt tở thành
\(\left\{{}\begin{matrix}\dfrac{5}{x+1}+a=10\\\dfrac{1}{x-2}+3a=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{x+1}+3a=30\left(1\right)\\\dfrac{1}{x-1}+3a=18\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\), ta được:
\(\dfrac{15}{x+1}-\dfrac{1}{x-1}=12\\ \Leftrightarrow\dfrac{15x-15-x-1}{\left(x-1\right)\left(x+1\right)}=12\\ \Leftrightarrow12x^2-12=14x-16\\ \Leftrightarrow12x^2-14x+4=0\\ \Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Với \(x=\dfrac{1}{2}\Leftrightarrow\dfrac{10}{3}+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{10y-7}{3\left(y-1\right)}=10\)
\(\Leftrightarrow30y-30=10y-7\Leftrightarrow y=\dfrac{23}{20}\)
Với \(x=\dfrac{2}{3}\Leftrightarrow3+\dfrac{1}{y-1}=10\Leftrightarrow\dfrac{1}{y-1}=7\Leftrightarrow7y-7=1\Leftrightarrow y=\dfrac{8}{7}\)
Vậy \(\left(x;y\right)=\left\{\left(\dfrac{1}{2};\dfrac{23}{20}\right);\left(\dfrac{2}{3};\dfrac{8}{7}\right)\right\}\)
\(\left\{{}\begin{matrix}\dfrac{120}{x}=\dfrac{80}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\)(1)
Đặt \(a=\dfrac{1}{x}\);\(b=\dfrac{1}{y}\)
Vậy (1)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}120a=80b\\104b-1=96a\left(2\right)\end{matrix}\right.\)
Ta có \(120a=80b\Leftrightarrow b=\dfrac{3}{2}a\)
Thay \(b=\dfrac{3}{2}a\) vào (2)\(\Leftrightarrow104.\dfrac{3}{2}a-1=96a\Leftrightarrow156a-1=96a\Leftrightarrow60a=1\Leftrightarrow a=\dfrac{1}{60}\)
Vậy \(b=\dfrac{3}{2}.a=\dfrac{3}{2}.\dfrac{1}{60}=\dfrac{1}{40}\)
Vậy \(\left\{{}\begin{matrix}a=\dfrac{1}{60}\\b=\dfrac{1}{40}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=60\\y=40\end{matrix}\right.\)
Vậy (x;y)=(60;40)
\(\left\{{}\begin{matrix}\dfrac{3}{x}=\dfrac{2}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{96}{x}=\dfrac{64}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\) \(\Rightarrow\dfrac{104}{y}-1=\dfrac{64}{y}\)
\(\Rightarrow\dfrac{40}{y}=1\Rightarrow y=40\)
\(\Rightarrow x=\dfrac{3y}{2}=60\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(60;40\right)\)
1.\(A=\left(\sqrt{3}+1\right)\sqrt{\dfrac{14-6\sqrt{3}}{5+\sqrt{3}}}=\left(\sqrt{3}+1\right)\sqrt{\dfrac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\dfrac{44\left(2-\sqrt{3}\right)}{22}}=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
2.1.a) \(x^2=\left(x-1\right)\left(3x-2\right)\Leftrightarrow x^2=3x^2-5x+2\Leftrightarrow2x^2-5x+2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
b) \(9x^4+5x^2-4=0\Leftrightarrow9x^4+9x^2-4x^2-4=0\)
\(\Leftrightarrow9x^2\left(x^2+1\right)-4\left(x^2+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(9x^2-4\right)=0\)
mà \(x^2+1>0\Rightarrow9x^2=4\Rightarrow x^2=\dfrac{4}{9}\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2) Gọi số xe lúc đầu của đội là a(xe) \(\left(a\in N,a>0\right)\)
Theo đề,ta có: \(\left(a-2\right)\left(\dfrac{120}{a}+3\right)=120\Leftrightarrow120+3a-\dfrac{240}{a}-6=120\)
\(\Leftrightarrow\dfrac{3a^2-6a-240}{a}=0\Rightarrow3a^2-6a-240=0\Rightarrow a^2-2a-80=0\)
\(\Leftrightarrow\left(a+8\right)\left(a-10\right)=0\) mà \(a>0\Rightarrow a=10\)
`đk:x ne 2,y ne 1/2`
ĐẶt `a=1/(x-2),b=1/(2y-1)`
`hpt<=>` $\begin{cases}a+5b=3\\3a-b=1\\\end{cases}$
`<=>` $\begin{cases}3a+15b=9\\3a-b=1\\\end{cases}$
`<=>` $\begin{cases}16b=8\\a=3-5b\\\end{cases}$
`<=>` $\begin{cases}b=\dfrac12\\a=\dfrac12\\\end{cases}$
`<=>` $\begin{cases}x-2=2\\2y-1=2\\\end{cases}$
`<=>` $\begin{cases}x=4\\y=\dfrac32\\\end{cases}$
Đk: \(x\ne2;y\ne\dfrac{1}{2}\)
Đặt \(a=\dfrac{1}{x-2},b=\dfrac{1}{2y-1}\) (a,b khác 0)
Có hệ: \(\left\{{}\begin{matrix}a+5b=3\\3a-b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+5b=3\\15a-5b=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}16a=8\\3a-b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=3a-1=\dfrac{1}{2}\end{matrix}\right.\)(tm)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{1}{2}\\\dfrac{1}{2y-1}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3}{2}\end{matrix}\right.\)(tm)
\(\dfrac{120}{x}+\dfrac{120}{x-10}=\dfrac{3}{5}\left(dkxd:x>0,x\ne10\right)\)
\(\Leftrightarrow\dfrac{120}{x}+\dfrac{120}{x-10}-\dfrac{3}{5}=0\)
\(\Leftrightarrow\dfrac{120.5\left(x-10\right)+5.120x-3x\left(x-10\right)}{5x\left(x-10\right)}=0\)
\(\Leftrightarrow600x-6000+600x-3x^2+30x=0\)
\(\Leftrightarrow-3x^2+1230x-6000=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\approx405\\x\approx5\end{matrix}\right.\)\(\left(tmdk\right)\)
Vậy ...
ĐKXĐ: x ≠ 0; x ≠ 10 em ơi