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

Ôi mẹ ơi! Bài lm của con khá giống nó nhg may là chưa đang!

Ta có: \(\begin{cases} \dfrac{1 - xy}{x(1 + y^2)} = \dfrac{2}{5} & (1) \\ \dfrac{1 - xy}{y(1 + x^2)} = \dfrac{1}{2} & (2) \end{cases}\)

=>\(\begin{cases}5-5xy=2x+2xy^2 & (1^{\prime})\\ 2-2xy=y+x^2y & (2^{\prime})\end{cases}\)

=>\(\begin{cases}5-5xy=2x+2xy^2 & \\ 1-xy=\frac12\left(y+x^2y\right) & \end{cases}\)

Thay vào (1), ta được: \(\dfrac{\frac{1}{2}y(1 + x^2)}{x(1 + y^2)}=\dfrac{2}{5}\)

=>\(5y(1+x^2)=4x(1+y^2)\)

=>\(5y+5yx^2=4x+4xy^2\)

=>\(5yx^2-4xy^2+5y-4x=0\) (3)

Ta có: \(\begin{cases}5-5xy=2x+2xy^2 & \\ 2-2xy=y+x^2y & \end{cases}\)

=>10-10xy=4x+\(4xy^2\) và 10-10xy=5y+\(5x^2y\)

=>\(5y+5x^2y=4x+4xy^2\)

=>\(4x+4xy^2-5y-5x^2y=0\)

=>\(4xy^2-5yx^2+4x-5y=0\) (4)

Từ (3),(4) suy ra \((5yx^2 - 4xy^2 + 5y - 4x) - (4xy^2 - 5yx^2 + 4x - 5y) = 0\)

=>\(10yx^2-8xy^2+10y-8x=0\)

=>\(5yx^2-4xy^2+5y-4x=0\)

=>\(5y(x^2+1)=4x(y^2+1)\)

=>\(\dfrac{x^2+1}{x}=\dfrac{4}{5}\cdot\dfrac{y^2+1}{y}\)

=>\(x+\dfrac{1}{x}=\dfrac{4}{5}\left(y+\dfrac{1}{y}\right)\)

Đặt \(u=x+\frac{1}{x};v=y+\frac{1}{y}\)

Ta sẽ có: \(u=\frac45v\)

Ta có: \(\dfrac{1 - xy}{x(1 + y^2)}=\dfrac{2}{5}\)

=>\(\dfrac{\frac{1}{x} - y}{1 + y^2} = \dfrac{2}{5}\)

Khi y=1/2 thì ta sẽ có: \(y+\frac{1}{y}=\frac12+2=\frac52\)

=>\(x+\frac{1}{x}=\frac45\cdot\frac52=2\)

=>\(\frac{x^2+1}{x}=2\)

=>\(x^2+1=2x\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

Thay x=1;y=1/2 vào (1), ta được:

\(\frac{1-1\cdot\frac12}{1\cdot\left(1+\frac14\right)}=\frac25\)

=>\(\left(1-\frac12\right):\frac54=\frac25\)

=>\(\frac12\cdot\frac45=\frac25\)

=>2/5=2/5(đúng)

Thay x=1;y=1/2 vào (2): \(\dfrac{1 - xy}{y(1 + x^2)}=\dfrac{1}{2}\) , ta được:

\(\left(1-1\cdot\frac12\right):\left\lbrack\frac12\cdot\left(1+1^2\right)\right\rbrack=\frac12\)

=>\(\left(1-\frac12\right):\left(\frac12\cdot2\right)=\frac12\)

=>\(\frac12=\frac12\) (đúng)

Vậy: x=1;y=1/2

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