Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2y+2y+x=4xy< =>xy\left(x+3\right)=4xy< =>x+3=4< =>x=1\)
Thế x=1 vào 1 trong 2 phương trình => y=1
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+y}+6x-3y=6\\\dfrac{3}{x+y}+2x-4y=1\end{matrix}\right.\)
\(\Rightarrow4x+y=5\Rightarrow y=5-4x\)
Thế vào phương trình đầu:
\(\dfrac{1}{x+5-4x}+2x-\left(5-4x\right)=2\)
\(\Leftrightarrow\dfrac{1}{5-3x}+6x-7=0\)
\(\Leftrightarrow\left(6x-7\right)\left(5-3x\right)+1=0\)
\(\Leftrightarrow...\)
\(1.\left(x\ne\pm1\right)\Rightarrow pt\Leftrightarrow\left(x-m\right)\left(x-1\right)=\left(x+1\right)\left(x-2\right)\)
\(\Leftrightarrow x^2-x\left(m+1\right)+m=x^2-x-2\)
\(\Leftrightarrow-x\left(m+1\right)+m=-x-2\)
\(\Leftrightarrow x=\dfrac{m+2}{m}\left(m\ne0\right)\)
\(pt-có-ngo-duy-nhất\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m+2}{m}\ne1\\\dfrac{m+2}{m}\ne-1\end{matrix}\right.\)\(\Leftrightarrow m\ne-1\)
\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ne-1\end{matrix}\right.\)
\(2.\left\{{}\begin{matrix}x^2+8y^2=12\left(1\right)\\x^3+2xy^2+12y=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^3+2xy^2+y\left(x^2+8y^2\right)=0\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2y\left(3\right)\\x^2-xy+4y^2=\left(x-\dfrac{y}{2}\right)^2+\dfrac{15}{4}y^2=0\left(4\right)\end{matrix}\right.\)
\(\left(3\right)\left(1\right)\Rightarrow4y^2+8y^2=12\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)
với \(x=y=0\) không là nghiệm của hệ pt
với \(x=y\ne0\Rightarrow\left(4\right)>0\Rightarrow\left(4\right)-vô-nghiệm\)
\(\Rightarrow\left(x;y\right)=\left\{\left(-2;1\right);\left(2;-1\right)\right\}\)
\(1,\Leftrightarrow\left(x-m\right)\left(x-1\right)=x^2-x-2\\ \Leftrightarrow x^2-x-mx+m-x^2+x+2=0\\ \Leftrightarrow mx=m+2\)
PT có nghiệm duy nhất \(\Leftrightarrow m\ne0\)
\(2,\Leftrightarrow\left\{{}\begin{matrix}x^2y+8y^3=12y\\x^3+2xy^2+12y=0\end{matrix}\right.\)
Thế \(PT\left(1\right)\rightarrow PT\left(2\right)\Leftrightarrow x^3+2xy^2+x^2y+8y^3=0\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-2xy+4y^2\right)+xy\left(x+2y\right)=0\\ \Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2y\\\left(x-\dfrac{1}{2}y\right)^2+\dfrac{15}{4}y^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2y\\\left\{{}\begin{matrix}x-\dfrac{1}{2}y=0\\y^2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2y\\x=y=0\end{matrix}\right.\)
Thay \(x=y=0\Leftrightarrow0+0=12\left(loại\right)\)
Thay \(x=-2y\Leftrightarrow4y^2+8y^2=12y^2=12\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(-2;1\right);\left(2;-1\right)\right\}\)
b.
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) hệ tương đương:
\(\left\{{}\begin{matrix}\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\\\dfrac{1}{x^3}+y^3=19\end{matrix}\right.\)
Đặt \(\left(\dfrac{1}{x};y\right)=\left(u;v\right)\) ta được: \(\left\{{}\begin{matrix}uv^2+u^2v=-6\\u^3+v^3=19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3uv^2+3u^2v=-18\\u^3+v^3+19\end{matrix}\right.\)
Cộng vế với vế:
\(\left(u+v\right)^3=1\Rightarrow u+v=1\)
Thay vào \(u^2v+uv^2=-6\Rightarrow uv=-6\)
Theo Viet đảo, u và v là nghiệm của:
\(t^2-t-6=0\) \(\Rightarrow\left[{}\begin{matrix}t=-2\\t=3\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(-2;3\right);\left(3;-2\right)\)
\(\Rightarrow\left(\dfrac{1}{x};y\right)=\left(-2;3\right);\left(3;-2\right)\)
\(\Rightarrow\left(x;y\right)=\left(-\dfrac{1}{2};3\right);\left(\dfrac{1}{3};-2\right)\)
a.
ĐKXĐ: \(x\ne3\)
- Với \(x\ge0\) pt trở thành:
\(\dfrac{x^2-x-12}{x-3}=2x\Rightarrow x^2-x-12=2x^2-6x\)
\(\Leftrightarrow x^2-5x+12=0\) (vô nghiệm)
- Với \(x< 0\) pt trở thành:
\(\dfrac{x^2+x-12}{x-3}=2x\Rightarrow\dfrac{\left(x-3\right)\left(x+4\right)}{x-3}=2x\)
\(\Rightarrow x+4=2x\Rightarrow x=4>0\) (ktm)
Vậy pt đã cho vô nghiệm
Cộng vế với vế:
\(x^2+2xy+y^2+x+y=12\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)-12=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=-4\\x+y=3\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=-4\\xy=5-\left(x+y\right)=9\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm: \(t^2-4t+9=0\) (vô nghiệm)
TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=5-\left(x+y\right)=2\end{matrix}\right.\)
Theo Viet đảo, x và y là nghiệm:
\(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
(Pt trên là pt (1), pt dưới là pt (2))
Đk: \(x;y\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3=2x^3+x^2y\\3=2y^3+xy^2\end{matrix}\right.\)
\(\Rightarrow2\left(x^3-y^3\right)+\left(x^2y-xy^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+2xy+y^2\right)=0\)\(\Leftrightarrow\left(x-y\right)\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
TH1: \(x=y\) thay vào pt (1) \(\Rightarrow\dfrac{3}{y^2}=2y+y\)
\(\Leftrightarrow3=3y^3\) \(\Leftrightarrow y=1\) \(\Rightarrow x=y=1\) (TM)
TH2:\(x=-y\) thay vào pt (1) \(\Rightarrow\dfrac{3}{y^2}=-2y+y\)
\(\Leftrightarrow\dfrac{3}{y^2}=-1\left(L\right)\)
Vậy (x;y)=(1;1)
ĐKXĐ: ...
Cộng vế với vế: \(3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)=3\left(x+y\right)\Rightarrow x+y=\dfrac{1}{x^2}+\dfrac{1}{y^2}\)
Trừ vế cho vế:
\(3\left(\dfrac{1}{x^2}-\dfrac{1}{y^2}\right)=x-y\)
\(\Leftrightarrow-3\left(\dfrac{x-y}{xy}\right)\left(\dfrac{x+y}{xy}\right)=x-y\)
\(\Leftrightarrow\left(x-y\right)\left(1+\dfrac{3\left(x+y\right)}{x^2y^2}\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(1+\dfrac{3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)}{x^2y^2}\right)=0\)
\(\Leftrightarrow x-y=0\) (do \(1+\dfrac{3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)}{x^2y^2}>0\))
Thế vào pt đầu:
\(\dfrac{3}{x^2}=3x\Leftrightarrow x^3=1\Leftrightarrow x=y=1\)
Đặt \(x+\dfrac{1}{x}=a;y+\dfrac{1}{y}=b\left(\left|a\right|\ge2;\left|b\right|\ge2\right)\)
\(\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x^3+\dfrac{1}{x^3}\right)+\left(y^3+\dfrac{1}{y^3}\right)=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)^3-3\left(y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\a^3+b^3=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(a+b\right)^3-3ab\left(a+b\right)=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\125-15ab=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=9-m\end{matrix}\right.\)
\(\Rightarrow a,b\) là nghiệm của phương trình \(t^2-5t+9-m=0\left(1\right)\)
a, Nếu \(m=3\), phương trình \(\left(1\right)\) trở thành
\(t^2-5t+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y+\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\y^2-3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\pm\sqrt{5}}{2}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=3\\y+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3\pm\sqrt{5}}{2}\\y=1\end{matrix}\right.\)
Vậy ...
b, \(\left(1\right)\Leftrightarrow t=\dfrac{5\pm\sqrt{4m-11}}{2}\left(m\ge\dfrac{11}{4}\right)\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\pm\sqrt{4m-11}}{2}\\b=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5\pm\sqrt{4m-11}}{2}\\y+\dfrac{1}{y}=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-\left(5\pm\sqrt{4m-11}\right)+2=0\left(2\right)\\2y^2-\left(5\mp\sqrt{4m-11}\right)+2=0\end{matrix}\right.\)
Yêu cầu bài toán thỏa mãn khi phương trình \(\left(2\right)\) có nghiệm dương
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(5\pm\sqrt{4m-11}\right)^2-16\ge0\\\dfrac{5\pm\sqrt{4m-11}}{2}>0\\1>0\end{matrix}\right.\)
\(\Leftrightarrow...\)