Giải hệ phương trình sau
a,\(\dfrac{x}{y}-\dfrac{x}{y+12}=1\) và \(\dfrac{x}{y-12}-\dfrac{x}{y}=2\)
b,4(x+y)=5(x-y) và \(\dfrac{40}{x+y}+\dfrac{40}{x-y}=9\)
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a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)
a: Áp dụng tính chất của DTSBN, ta được:
x/5=y/2=(x-y)/(5-2)=9/3=3
=>x=15; y=6
b: =>(x-3)/12=3/(x-3)
=>(x-3)^2=36
=>(x-9)(x+3)=0
=>x=9 hoặc x=-3
c; x/2=y/3
=>x/10=y/15
y/5=z/4
=>y/15=z/12
=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17
=>x=490/17; y=735/17; z=588/17
\(HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+y}-\dfrac{4}{x-y}=0\\\dfrac{40}{x+y}+\dfrac{40}{x-y}=9\end{matrix}\right.\)
Đặt \(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b\) . Khi đó hệ thành :
\(\left\{{}\begin{matrix}5a-4b=0\\40a+40b=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}40a-32b=0\\40a+40b=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-72b=-9\\40a-32b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{10}\\b=\dfrac{1}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\x-y=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(9;1\right)\)
hệ đối xứng trừ theo vế là xong