Pt <=>  \(\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{1}{550}\)

<=>  \(\dfrac{\left(x+5\right)-x}{x\left(x+5\right)}=\dfrac{1}{550}\)

<=> \(\dfrac{5}{x\left(x+5\right)}=\dfrac{1}{550}\)

<=> \(x^2+5x=2750\)

<=> \(x^2+5x-2750=0\)

<=> \(\left(x^2+5x+2,5^2\right)-52,5^2=0\) (bước này hơi tắt xíu nha :<)

<=> \(\left(x+2,5\right)^2-52,5^2=0\)

<=> \(\left(x+55\right)\left(x-50\right)=0\)

<=> \(\left[{}\begin{matrix}x+55=0\\x-50=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-55\\x=50\end{matrix}\right.\)

Vậy nghiệm của phương trình là x \(\in\left\{-55;50\right\}\)

\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\)

\(\dfrac{1100\left(x+5\right)-1100x}{x\left(x+5\right)}=\dfrac{2x\left(x+5\right)}{x\left(x+5\right)}\)

\(1100x+5500-1100x=2x^2+10x\)

\(2x^2+10x-5500=0\)

Δ' \(=5^2-2\left(-5500\right)\)

Δ'\(=11025\)

\(\left[{}\begin{matrix}x=50\\x=-55\end{matrix}\right.\)

=>2x(x+5)=1100(x+5)-1100x

=>2x(x+5)=5500

=>2x^2+10x-5500=0

=>x=50 hoặc x=-55

đk : x khác 0 ; -5 

\(1100x+5500-1100x=2x\left(x+5\right)\)

\(\Leftrightarrow2x^2+10x-5500=0\Leftrightarrow x=50;x=-55\)(tm) 

= - 55 tm

gọi HPT trên là (1)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{x+y}{xy}=\dfrac{9}{2}\\xy+\dfrac{1}{xy}=\dfrac{5}{2}\end{matrix}\right.\)

Đặt x+y=a;xy=b(b#0).HPT trở thành:

\(\left\{{}\begin{matrix}a+\dfrac{a}{b}=\dfrac{9}{2}\left(!\right)\\b+\dfrac{1}{b}=\dfrac{5}{2}\left(!!\right)\end{matrix}\right.\)

Giải PT (!!) ta được \(b_1=2;b=\dfrac{1}{2}\)

TH1: Với b=2 thay vào (!)=>a=3

=> x+y=3 và xy=2 => x=2;y=1.

TH2: Với b=1/2 thay vào (!)=> a=3/2

=> x+y=3/2 và xy=1/2 => x=1 và y=1/2.

Vậy \(\left(x;y\right)=\left\{\left(2;1\right);\left(1;\dfrac{1}{2}\right)\right\}\)

1.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

a)

\(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)

Đặt x+y = S, xy = P,ta có hệ

\(\left\{{}\begin{matrix}S+P=17\\S^2-P=13\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}P=S-17\\S^2-S+4=0\end{matrix}\right.\)

\(S^2-S+4>0\)

=> Hệ phương trình vô nghiệm