ĐKXĐ: \(x\ne_-^+y;y\ne0\)

Từ PT thứ 2 ta có:\(\dfrac{96}{x-y}+\dfrac{72}{x-y}=\dfrac{24}{y}\)

<=>\(\dfrac{168}{x-y}=\dfrac{24}{y}\)

<=>\(\dfrac{168}{x-y}=\dfrac{168}{7y}\)

<=>x-y=7y

<=>x=8y

Thay x=8y vào PT thứ nhất:

\(\dfrac{96}{8y+y}+\dfrac{96}{8y-y}=14\)

<=>\(\dfrac{32}{3y}+\dfrac{96}{7y}=14\)

<=>32.7y+96.3y=294y²

<=>512y=294y²

<=>y=\(\dfrac{256}{147}\left(Doy\ne0\right)\)

=>x=8y=\(\dfrac{2048}{147}\)

Vậy...

=>3/x=2/y và 96/x+1=104/y

=>2x=3y và 96/x+1=104/y

=>x/3=y/2=k và 96/x+1=104/y

=>x=3k; y=2k

\(\dfrac{96}{x}+1=\dfrac{104}{y}\)

=>\(\dfrac{96}{3k}+1=\dfrac{104}{2k}\)

=>\(\dfrac{32}{k}+1=\dfrac{52}{k}\)

=>20/k=1

=>k=20

=>x=60; y=40

\(\left\{{}\begin{matrix}\dfrac{120}{x}=\dfrac{80}{y}\\\dfrac{104}{y}-1=\dfrac{96}{x}\end{matrix}\right.\)(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 \(\left(x;y\right)=\left(60;40\right)\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-448y^3=-3x+6y\\96=385x^2-16y^2\end{matrix}\right.\)

\(\Rightarrow96\left(x^3-448y^3\right)=\left(-3x+6y\right)\left(385x^2-16y^2\right)\)

\(\Leftrightarrow\left(x-4y\right)\left(417x^2+898xy+3576y^2\right)=0\)

\(\Leftrightarrow x-4y=0\)

\(\Leftrightarrow x=4y\)

Thế vào \(385x^2-16y^2=96\)

\(\Rightarrow...\)

b.

ĐKXĐ: \(x+y\ne0\)

\(\left\{{}\begin{matrix}\left(3x^3-y^3\right)\left(x+y\right)=1\\1=x^2+y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x^3-y^3\right)\left(x+y\right)=1\\1=\left(x^2+y^2\right)^2\end{matrix}\right.\)

\(\Rightarrow\left(3x^3-y^3\right)\left(x+y\right)=\left(x^2+y^2\right)^2\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)\left(2x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)

Thế vào \(x^2+y^2=1\)...

Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix} 14a-10b=9\\ 3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 14a-10b=9\\ 15a+10b=20\end{matrix}\right.\)

$\Rightarrow (14a-10b)+(15a+10b)=9+20$

$\Leftrightarrow 29a=29\Leftrightarrow a=1$.

$b=\frac{4-3a}{2}=\frac{1}{2}$

Vậy: \(\left\{\begin{matrix} \frac{1}{x-y+2}=1\\ \frac{1}{x+y-1}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y+2=1\\ x+y-1=2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x-y=-1\\ x+y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{72}{x}+\dfrac{54}{y}=6\\x-y=6\end{matrix}\right.\left(x,y>0\right)< =>\left\{{}\begin{matrix}x=6+y\left(1\right)\\\dfrac{72}{\left(6+y\right)}+\dfrac{54}{y}=6\left(2\right)\end{matrix}\right.\)(x,y>0,y\(\ne-6\))

giải pt(2) \(\dfrac{72}{\left(6+y\right)}+\dfrac{54}{y}=6< =>\dfrac{72y+54\left(6+y\right)}{y\left(6+y\right)}=6\)

\(< =>\dfrac{126y+324}{y\left(6+y\right)}=6=>126y+324=6y\left(6+y\right)\)

\(< =>126y+324=36y+6y^2\)

\(< =>-6y^2+90y+324=0\)

\(\Delta=90^2-4\left(-6\right).324=15876>0\)

=>x1=\(\dfrac{-90+\sqrt{15876}}{2\left(-6\right)}=-3\left(loai\right)\)

x2=\(\dfrac{-90-\sqrt{15876}}{2\left(-6\right)}=18\left(TM\right)\)

=>x=x2=18 thay vào pt(1)=>x=6+18=24

vậy (x,y)=(24,18)

Đặt \(\left[{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\).

Ta có hệ: \(\left[{}\begin{matrix}a+b=\dfrac{1}{16}\\3a+6b=\dfrac{1}{4}\end{matrix}\right.\)

`<=>` \(\left[{}\begin{matrix}a=\dfrac{1}{24}\\b=\dfrac{1}{48}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=24\\y=48\end{matrix}\right.\)

Vậy `(x;y)=(24;48)`.

Giải:

Ta có: \(\dfrac{2}{x}=\dfrac{3}{y}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Mà \(xy=96\)

\(\Rightarrow2k3k=96\)

\(\Rightarrow6k^2=96\)

\(\Rightarrow k^2=16\)

\(\Rightarrow\left[{}\begin{matrix}k=4\\k=-4\end{matrix}\right.\)

+) \(k=4\Rightarrow x=8,y=12\)

+) \(k=-4\Rightarrow x=-8,y=-12\)

Vậy cặp số \(\left(x;y\right)\) là: \(\left(8;12\right);\left(-8;-12\right)\)

Giải:

Ta có: \(\dfrac{2}{x}=\dfrac{3}{y}\)

\(\Rightarrow\dfrac{2}{3}=\dfrac{x}{y}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=t\)

\(\Rightarrow x=2y;y=3t\)

\(\Rightarrow x.y=96\)

\(\Rightarrow2t.3t=96\)

\(\Rightarrow6t^2=96\)

\(\Rightarrow t^2=16\)

\(\Rightarrow t=\pm4\)

Với \(t=4\) ta có:

\(x=4.2\)

\(\Rightarrow x=8\)

\(y=4.3\)

\(\Rightarrow y=12\)

Với \(t=-4\) tương tự ta cũng có:

\(x=-4.2\)

\(\Rightarrow x=-8\)

\(y=-4.3\)

\(\Rightarrow y=-12\)

Vậy .....