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5 tháng 5 2017

(1) <=> x=-1+y (3)

từ (2) và (3) suy ra:

\(\dfrac{2}{y-1}+\dfrac{3}{y}=2\)

<=>\(\dfrac{2y}{y\left(y-1\right)}+\dfrac{3\left(y-1\right)}{y\left(y-1\right)}=2\)

<=> \(\dfrac{2y+3y-3}{y\left(y-1\right)}=2\)

<=>\(\dfrac{5y-3}{y\left(y-1\right)}=2\)

<=> 5y-3=2y(y-1)

<=> 5y-3=\(2y^2-2y\)

<=>\(2y^2-7y-3=0\)

rồi bạn giải như bình thường là ra

5 tháng 5 2017

ĐKXĐ:x khác 0 y khác 0

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

<=>\(\left\{{}\begin{matrix}y=x+1\left(1\right)\\2y+3x=2xy\left(2\right)\end{matrix}\right.\)

Thay 1 vào 2 ta có:

2(x+1)+3x=2x(x+1)

<=>5x+2=2x2+2x

<=>2x2-3x+2=0

<=>2x2-3x+\(\dfrac{9}{8}\)+\(\dfrac{7}{8}\)=0

<=>2(x-\(\dfrac{3}{4}\))2+\(\dfrac{7}{8}\)=0(vô lí do \(2\left(x-\dfrac{3}{4}\right)^2\ge0\forall x\))

Vậy hệ vô nghiệm

ĐKXĐ: x<>0; y<>0

\(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{6}{x}+\dfrac{3}{y}=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{x}=4\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\\dfrac{1}{y}=-1-\dfrac{2}{x}=-1-2:\dfrac{-1}{4}=-1+8=7\end{matrix}\right.\)

=>x=-1/4 và y=1/7

D
datcoder
CTVVIP
7 tháng 10 2023

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

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

Hệ phương trình trở thành \(\left\{{}\begin{matrix}5a+3b=1\\2a+b=-1\end{matrix}\right.\)

 \(\Rightarrow\left\{{}\begin{matrix}b=-1-2a\\5a+3\left(-1-2a\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-1-2a\\-a-3=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=-4\\b=-1-2.\left(-4\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=7\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}a=\dfrac{1}{x}=-4\\b=\dfrac{1}{y}=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\left(tm\right)\\y=\dfrac{1}{7}\left(tm\right)\end{matrix}\right.\)

Vậy HPT có nghiệm \(x=-\dfrac{1}{4}\) và \(y=\dfrac{1}{7}\)

NV
29 tháng 1 2024

a.

ĐKXĐ: \(x\ne\pm y\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=2\\2u+3v=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+3v=6\\2u+3v=5\\\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=2-u\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-4x+7=x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-5x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

AH
Akai Haruma
Giáo viên
16 tháng 12 2021

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

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

23 tháng 1 2022

ĐK:   \(x\ne0\) ; \(y\ne0\)

Hệ phương trình tương đương với:

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=4\\\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=8\end{matrix}\right.\)

Đặt  \(S=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)\)

         \(P=\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\)

Mà   \(S^2\ge4P\)

Ta có:      \(\left\{{}\begin{matrix}S=4\\S^2-2P=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S=4\\P=4\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=4\\\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)=4\end{matrix}\right.\)

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

          

30 tháng 9 2023

Bài 1:

Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{2x-2}\\v=\dfrac{1}{y-1}\end{matrix}\right.\) (ĐK: \(x,y\ne1\))  

Hệ trở thành:

\(\Leftrightarrow\left\{{}\begin{matrix}u-v=2\\3u-2v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3u-3v=6\\3u-2v=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}v=-5\\u=2+-5\end{matrix}\right.\)

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

Trả lại ẩn của hệ pt:

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x=\dfrac{5}{6}\end{matrix}\right.\left(tm\right)\)

30 tháng 12 2017

Đặt ẩn phụ nhé

\(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b=< =>\int_{2a-3b=1}^{a+b=3}< =>\int_{2.\left(3-b\right)-3b=1}^{,a=3-b}< =>\int_{b=1}^{a=2}\)

<=>\(\dfrac{1}{x+y}=2;\dfrac{1}{x-y}=1< =>\int_{x-y=1}^{x+y=2}< =>\int_{y=0,5}^{x=1,5}\)

31 tháng 12 2017

Đặt :

\(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)

Ta có hệ phương trình :

\(\left\{{}\begin{matrix}u+v=3\\2u-3v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2u+2v=6\\2u-3v=1\end{matrix}\right.\)

\(\Leftrightarrow5v=5\Leftrightarrow v=1\)

Thay \(v=1\) vào phương trình thứ nhất ta đc :

\(u+1=3\Leftrightarrow u=2\)

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

\(\Leftrightarrow2y=-\dfrac{1}{2}\Rightarrow y=-\dfrac{1}{4}\)

Thay \(y=-\dfrac{1}{4}\) vào phương trình thứ 2 ta được :

\(x+\dfrac{1}{4}=1\Leftrightarrow x=\dfrac{3}{4}\)

Vậy \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)

13 tháng 2 2022

\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)

\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)

\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)