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MH
20 tháng 4 2022
\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{1}{4}-\dfrac{y}{3}+\dfrac{2}{3}=\dfrac{1}{12}\\\dfrac{x}{2}+\dfrac{5}{2}-\dfrac{y}{3}-\dfrac{7}{3}=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{5}{6}\\\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{25}{6}\end{matrix}\right.\) (vô lý)
Vậy HPT vô nghiệm
TA
15 tháng 3 2023
ĐK: `x ne 2; y ne -1`
Đặt `{a=(1/(x-2)),(b=1/(y+1)):}`
Có: `{(2a+b=3),(4a-3b=1):}`
`<=>{(4a+2b=6),(4a-3b=1):}`
`<=>{(2a+b=3),(5b=5):}`
`<=>{(2a+1=3),(b=1):}`
`<=>{(a=1),(b=1):}`
``
`=>{(1/(x-2)=1),(1/(y+1)=1):}`
`<=>{(x-2=1),(y+1=1):}`
`<=>{(x=3),(y=0):}` (TM)
``
Vậy `(x;y)=(3;0)`.
XD
1
4 tháng 3 2022
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{x-y}+3\sqrt{y+1}=12\\\dfrac{1}{x-y}-3\sqrt{y+1}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\3\sqrt{y+1}=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\y+1=4\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(4;3\right)\)
4 tháng 3 2022
Anh/Chị giải chi tiết ra giúp em đc ko ạ. Em ko hiếu lắm ạ
24 tháng 2 2021
1) Ta có: \(\left\{{}\begin{matrix}2\cdot\dfrac{x}{x+2}-\dfrac{y}{y-1}=4\\\dfrac{x}{x+2}-3\cdot\dfrac{y}{y-1}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot\dfrac{x}{x+2}-\dfrac{y}{y-1}=4\\2\cdot\dfrac{x}{x+2}-6\cdot\dfrac{y}{y-1}=-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-7\cdot\dfrac{y}{y-1}=10\\2\cdot\dfrac{x}{x+2}-\dfrac{y}{y-1}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{y-1}=\dfrac{-10}{7}\\2\cdot\dfrac{x}{x+2}+\dfrac{10}{7}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\cdot\dfrac{x}{x+2}=\dfrac{18}{7}\\\dfrac{y}{y-1}=\dfrac{-10}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+2}=\dfrac{9}{7}\\\dfrac{y}{y-1}=\dfrac{-10}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\left(x+2\right)=7x\\-10\left(y-1\right)=7y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x+18-7x=0\\-10y+10-7y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+18=0\\-17y+10=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=-18\\-17y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-9\\y=\dfrac{10}{17}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(-9;\dfrac{10}{17}\right)\)
18 tháng 7 2017
<=>\(\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{2x}{y}+\dfrac{x}{y}=3\left(1\right)\\x+\dfrac{1}{y}+\dfrac{x}{y}=3\left(2\right)\end{matrix}\right.\)
cộng vế với vế của (1) và (2) ta được :
(x+\(\dfrac{1}{y}\))2 +( 1+\(\dfrac{1}{y}\)) = 6
(x +\(\dfrac{1}{y}\))2 +(1+\(\dfrac{1}{y}\)) - 6 = 0
đặt t =x +\(\dfrac{1}{y}\) rồi giải phương trình bậc 2 theo t . tìm ra t thế x theo y vào hệ đã cho ta tìm được x và y .< trước khi làm bài này phải có ĐK y#0>
XO
29 tháng 12 2021
ĐKXĐ : x;y \(\ne0\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=-2\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{1}{x}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x=\dfrac{1}{9}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 11 2018
Đặt \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=a\\y-\dfrac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}=a^2-2\\y^2+\dfrac{1}{y^2}=b^2+2\end{matrix}\right.\)hệ đã cho tương đương:
\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\end{matrix}\right.\) \(\Rightarrow a^2+\left(3-a\right)^2-5=0\Rightarrow a^2-3a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1;b=2\\a=2;b=1\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=1\\y-\dfrac{1}{y}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-x+1=0\left(vn\right)\\y^2-2y-1=0\end{matrix}\right.\) (loại)
TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y-\dfrac{1}{y}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-2x+1=0\\y^2-y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=\dfrac{1-\sqrt{5}}{2}\\y=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy hệ đã cho có 2 cặp nghiệm:
\(\left(x;y\right)=\left(1;\dfrac{1-\sqrt{5}}{2}\right);\left(1;\dfrac{1+\sqrt{5}}{2}\right)\)
24 tháng 11 2018
Đặt \(a=x+\dfrac{1}{x}\Leftrightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow x^2+\dfrac{1}{x^2}=a^2-2\)
\(b=y-\dfrac{1}{y}\Leftrightarrow b^2=y^2+\dfrac{1}{y^2}-2\Leftrightarrow y^2+\dfrac{1}{y^2}=b^2+2\)
Nên \(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=5\Leftrightarrow a^2-2+b^2+2=5\Leftrightarrow a^2+b^2=5\)Vậy ta có hệ phương trình \(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\left(1\right)\end{matrix}\right.\)
Ta có a+b=3\(\Leftrightarrow b=3-a\)
Thay b=3-a vào (1)\(\Leftrightarrow a^2+\left(3-a\right)^2=5\Leftrightarrow a^2+9-6a+a^2=5\Leftrightarrow2a^2-6a+4=0\Leftrightarrow2\left(a^2-3a+2\right)=0\Leftrightarrow a^2-3a+2=0\Leftrightarrow a^2-a-2a+2=0\Leftrightarrow a\left(a-1\right)-2\left(a-1\right)=0\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-1=0\\a-2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}b=2\\b=1\end{matrix}\right.\)
TH1:\(\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{1}{x}=1\\y-\dfrac{1}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-x+1=0\\y^2-2y-1=0\end{matrix}\right.\)
Ta có \(x^2-x+1=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy phương trình (2) vô nghiệm
TH2: \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y-\dfrac{1}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-2x+1=0\\y^2-y-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=\dfrac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Vậy (x,y)={(\(1;\dfrac{1+\sqrt{5}}{2}\));(\(1;\dfrac{1-\sqrt{5}}{2}\))}
10 tháng 2 2022
Đặt 1/x = a ; 1/y = b
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\\dfrac{10}{3}a+10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10a+10b=\dfrac{5}{3}\\\dfrac{10}{3}a+10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{20}{3}a=\dfrac{2}{3}\\b=\dfrac{1}{6}-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{10}\\b=\dfrac{1}{15}\end{matrix}\right.\)
Theo cách đặt x = 10 ; y = 15
10 tháng 2 2022
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3}.\dfrac{1}{x}+\dfrac{10}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3x}+\dfrac{10}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{3x}-\dfrac{1}{y}=\dfrac{1}{6}-\dfrac{1}{10}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}=\dfrac{1}{15}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=30\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{3.10}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{30}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{y}=\dfrac{1}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=15\end{matrix}\right.\)
=>y=x+3 và 1/x+1/y=1/2
=>1/x+1/(x+3)=1/2
=>(x+x+3)/(x^2+3x)=1/2
=>x^2+3x=2(2x+3)
=>x^2-x-6=0
=>(x-3)(x+2)=0
=>x=3 hoặc x=-2
=>y=6 hoặc y=1