NA
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
28 tháng 3 2019
ĐKXĐ x ; y > 0
(1) \(\Rightarrow\left(y-x\right)\left(\frac{1}{\sqrt{x}y}+x+2xy\right)=0\)
\(\Rightarrow x=y\)
\(\Rightarrow...\)
#Kaito#
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 6 2019
Câu 1: ĐKXĐ: ...
\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)
\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)
\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)
\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow...\)
Câu 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)
\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)
\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)
\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)
\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)
28 tháng 6 2019
\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)
\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)
\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)
\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)
\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)
\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)
Tìm được mỗi nghiệm thôi à :v
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2021
- Với \(xy=0\) không phải nghiệm
- Với \(xy\ne0\) hệ tương đương
\(\left\{{}\begin{matrix}3x-2=\dfrac{1}{y^3}\\x^3+2=\dfrac{3}{y}\end{matrix}\right.\)
Đặt \(\dfrac{1}{y}=z\Rightarrow\left\{{}\begin{matrix}3x-2=z^3\\x^3+2=3z\end{matrix}\right.\)
\(\Rightarrow x^3+3x=z^3+3z\)
\(\Leftrightarrow x^3-z^3+3\left(x-z\right)=0\)
\(\Leftrightarrow\left(x-z\right)\left(x^2+zx+z^2+3\right)=0\)
\(\Leftrightarrow x=z\)
Thế vào \(x^3+2=3z\Rightarrow x^3+2=3x\)
\(\Leftrightarrow x^3-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-2\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
NM
0
26 tháng 9 2021
\(\left\{{}\begin{matrix}x^3-3x^2-9x+22=y^3+3y^2-9y\left(1\right)\\x^2+y^2-x+y=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)
PT (1)\(\Leftrightarrow\left(x-y\right)^3+3xy\left(x-y\right)-3\left(x^2+y^2\right)-9\left(x-y\right)=-22\)
\(\Leftrightarrow\left(x-y\right)^3+3xy\left(x-y\right)-3\left(x-y\right)^2-6xy-9\left(x-y\right)=-22\)
PT (2)\(\Leftrightarrow\left(x-y\right)^2-\left(x-y\right)+2xy=\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}a=x-y\\b=xy\end{matrix}\right.\)
Hệ tt \(\left\{{}\begin{matrix}a^3+3ab-3a^2-6b-9a=-22\\a^2-a+2b=\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}a^3+3ab-3a^2-6b-9a=-22\\b=\dfrac{1-2a^2+2a}{4}\end{matrix}\right.\)
\(\Rightarrow a^3+3a\left(\dfrac{1-2a^2+2a}{4}\right)-3a^2-6\left(\dfrac{1-2a^2+2a}{4}\right)-9a=-22\)
\(\Leftrightarrow-2a^3+6a^2-45a+82=0\)
\(\Leftrightarrow a=2\)\(\Rightarrow b=-\dfrac{3}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=2\\xy=-\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}y=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy...
Đặt \(\hept{\begin{cases}\left|3x-1\right|=a\\x-y=b\end{cases}}\Rightarrow\hept{\begin{cases}2a+b=\frac{7}{3}\\a+3b=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=\frac{1}{3}\end{cases}}\)
hay ta có : \(\hept{\begin{cases}\left|3x-1\right|=1\\x-y=\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\Rightarrow y=-\frac{1}{3}\\x=\frac{2}{3}\Rightarrow y=\frac{1}{3}\end{cases}}\)
\(\hept{\begin{cases}2\left|3x-1\right|+x-y=\frac{7}{3}\\\left|3x-1\right|+3\left(x-y\right)=2\end{cases}}\)(1)
Với x < 1/3 \(\left(1\right)\Leftrightarrow\hept{\begin{cases}2-6x+x-y=\frac{7}{3}\\1-3x+3x-3y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}-5x-y=\frac{1}{3}\\-3y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(tm\right)\\y=-\frac{1}{3}\end{cases}}\)
Với x >= 1/3 \(\left(1\right)\Leftrightarrow\hept{\begin{cases}6x-2+x-y=\frac{7}{3}\\3x-1+3x-3y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}7x-y=\frac{13}{3}\\2x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=\frac{10}{3}\\2x-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{2}{3}\left(tm\right)\\y=\frac{1}{3}\end{cases}}\)