Giải giúp mình với ạ :3
a / \(\sqrt{2x^2+6x+1}=x+2\)
b / \(\left\{{}\begin{matrix}x+\dfrac{1}{y}=2\\\dfrac{x}{y}+\dfrac{1}{y}=2\end{matrix}\right.\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
15 tháng 5 2021
a) \(ĐKXĐ:2x^2+6x+1\ge0\)
Với \(x\ge2\) pt cho trở thành :
\(2x^2+6x+1=x^2+4x+4\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=3\) ( do \(x\ge2\) )
Vậy pt có tập nghiệm \(S=\left\{3\right\}\)
5 tháng 1 2023
a: =>2/x+2/y=2 và 4/x-2/y=1
=>6/x=3 và 1/x+1/y=1
=>x=2 và 1/y=1-1/2=1/2
=>x=2; y=2
b: Đặt 1/x=a; 1/y=b
=>1/3a+1/3b=1/4 và 5/6a+b=2/3
=>a=1/2; b=1/4
=>x=2; y=4
14 tháng 4 2022
a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)
\(ĐK:x;y\ne0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)
hpt trở thành:
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )
Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)
14 tháng 4 2022
b.\(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}+2x=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1+4x}{2}=\dfrac{4}{2}\\2x+y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
CW
17 tháng 1 2018
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
29 tháng 12 2021
\(1,ĐK:x,y\ne0\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2x^2y^2=y^3+1\\2x^2y^2=x^3+1\end{matrix}\right.\\ \Leftrightarrow x^3+1=y^3+1\\ \Leftrightarrow x^3=y^3\Leftrightarrow x=y\)
Thay vào PT 1
\(\Leftrightarrow2x^4=x^3+1\\ \Leftrightarrow2x^4-x^3-1=0\\ \Leftrightarrow2x^4-2x^3+x-1=0\\ \Leftrightarrow\left(2x^3+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^3=-\dfrac{1}{2}\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt[3]{-\dfrac{1}{2}}\\x=y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\sqrt[3]{-\dfrac{1}{2}};\sqrt[3]{-\dfrac{1}{2}}\right);\left(1;1\right)\)
\(2,ĐK:x,y\ge1\\ HPT\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)+\sqrt{y-1}=\dfrac{1}{2}\\2\left(y-1\right)+\sqrt{x-1}=\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}2a^2+b=\dfrac{1}{2}\\2b^2+a=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow2\left(a-b\right)\left(a+b\right)-\left(a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(2a+2b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\2a+2b=1\end{matrix}\right.\)
Với \(a=b\Leftrightarrow x-1=y-1\Leftrightarrow x=y\)
Thay vào \(PT\left(1\right)\Leftrightarrow2x+\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow2\sqrt{x-1}=5-4x\)
\(\Leftrightarrow4x-4=25-40x+16x^2\\ \Leftrightarrow16x^2-44x+29=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y=\dfrac{11+\sqrt{5}}{8}\left(tm\right)\\x=y=\dfrac{11-\sqrt{5}}{8}\left(tm\right)\end{matrix}\right.\)
Với \(2a+2b=1\Leftrightarrow b=\dfrac{1}{2}-a\Leftrightarrow\sqrt{y-1}=\dfrac{1}{2}-\sqrt{x-1}\)
Thay vào \(PT\left(1\right)\Leftrightarrow2x+\dfrac{1}{2}-\sqrt{x-1}=\dfrac{5}{2}\Leftrightarrow2x-2=\sqrt{x-1}\)
\(\Leftrightarrow4x^2-8x+4=x-1\\ \Leftrightarrow4x^2-9x+5=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\Rightarrow y=1\left(tm\right)\\x=1\Rightarrow y=\dfrac{5}{4}\left(tm\right)\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\dfrac{11+\sqrt{5}}{8};\dfrac{11+\sqrt{5}}{8}\right);\left(\dfrac{11-\sqrt{5}}{8};\dfrac{11-\sqrt{5}}{8}\right);\left(\dfrac{5}{4};1\right);\left(1;\dfrac{5}{4}\right)\)
\(a.\sqrt{2x^2+6x+1}=x+2\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\2x^2+6x+1=x^2+4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x=1\\ \Rightarrow S=\left\{1\right\}\)
\(b.\) ĐKXĐ: \(y\ne0\)\(\left(I\right)\Rightarrow x+\dfrac{1}{y}=\dfrac{x}{y}+\dfrac{1}{y}\Leftrightarrow x=\dfrac{x}{y}\Leftrightarrow x\left(1-\dfrac{1}{y}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{1}{2}\\x=1\end{matrix}\right.\left(TM\right)\Rightarrow S=\left\{\left(0;\dfrac{1}{2}\right);\left(1;1\right)\right\}\)