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PT có 2 nghiệm phân biệt \(\Leftrightarrow\Delta=\left(2m-3\right)^2-4\left(m-3\right)=9>0\)
Vậy PT có 2 nghiệm phân biệt với mọi m
Ta có \(\left[{}\begin{matrix}x_1=\dfrac{2m-3+3}{2}=m\\x_2=\dfrac{2m-3-3}{2}=m-3\end{matrix}\right.\)
Ta thấy \(m>m-3\) nên \(1< m-3< m< 6\Leftrightarrow4< m< 6\)
Vậy \(4< m< 6\) thỏa yêu cầu đề
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(x^4+y^4-1=xy\left(3-2xy\right)\)
\(\Leftrightarrow x^4+y^4-1=3xy-2x^2y^2\)
\(\Leftrightarrow x^4+2x^2y^2+y^4=3xy+1\)
\(\Leftrightarrow\left(x^2+y^2\right)^2=3xy+1\)
Vì \(\left(x^2+y^2\right)^2\ge0\forall x;y\)
\(\Rightarrow3xy+1\ge0\)
\(\Leftrightarrow xy\ge-\frac{1}{3}\)
\(\Leftrightarrow P\ge-\frac{1}{3}\)
Dấu "=" tại x = y = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
thi cấp tỉnh mà với có 1 số bài thi vào chuyên đại học với cấp 3 nữa
Bài 2: Ta có:
\(\left(2x+5y+1\right)\left(2020^{\left|x\right|}+y+x^2+x\right)=105\) là số lẻ
\(\Rightarrow\left\{{}\begin{matrix}2x+5y+1\\2020^{\left|x\right|}+y+x^2+x\end{matrix}\right.\) đều lẻ
\(\Rightarrow y⋮2\)\(\Rightarrow2020^{\left|x\right|}⋮̸2\Leftrightarrow\left|x\right|=0\Leftrightarrow x=0\).
Thay vào tìm được y...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\orbr{x=\frac{\pm\sqrt{5}-3}{4}}\)
\(b\hept{\begin{cases}x=5\\y=4\end{cases}}\)
2)\(\Leftrightarrow\left(x^3-x^2y\right)+\left(y^3-xy^2\right)=5\)
\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(y-x\right)=5\)
\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)=5\)
\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)=5\)
TH1\(\hept{\begin{cases}x-y=1\\x^2-y^2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}\left(N\right)}}\)
TH2\(\hept{\begin{cases}x-y=5\\x^2-y^2=1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)
TH3\(\hept{\begin{cases}x-y=-1\\x^2-y^2=-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}\left(N\right)}}\)
TH4\(\hept{\begin{cases}x-y=-5\\x^2-y^2=-1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)
Vậy......
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-1}=-5\\\dfrac{1}{2x-2}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=-\dfrac{1}{5}\\2x-2=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x=\dfrac{5}{6}\end{matrix}\right.\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) a là 1 nghiệm \(\Rightarrow\sqrt{2}a^2+a-1=0\Leftrightarrow2a^4=\left(1-a\right)^2=a^2-2a+1\)
\(\Rightarrow2a^4-2a+3=a^2-2a+1-2a+3=\left(a-2\right)^2\)
\(\sqrt{2\left(2a^4-2a+3\right)}+2a^2=\sqrt{2}\left(a-2\right)+2a^2\)(1)
mà \(\sqrt{2}a^2+a-1=0\Rightarrow2a^2+\sqrt{2}a-\sqrt{2}=0\)
(1)= \(2a^2+\sqrt{2}a-2\sqrt{2}=-\sqrt{2}\)
...
b) find nghiệm nguyên dương:
\(Pt\Leftrightarrow x^2+2y^2+2xy-2\left(x+2y\right)+1=0\)
\(\Leftrightarrow x^2+2x\left(y-1\right)+\left(2y^2-4y+1\right)=0\)\(\Delta'=\left(y-1\right)^2-\left(2y^2-4y+1\right)=-y^2+2y\ge0\)
\(\Leftrightarrow0\le y\le2\) kết hợp \(y\in N\)=> ....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\\left(x-1\right)^2+\left(y-1\right)^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\\left(x+y-2\right)^2-2\left(x-1\right)\left(y-1\right)=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=v\\x+y-2=u\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv=6\\u^2-2v=5\end{matrix}\right.\) \(\Rightarrow u^2-\dfrac{12}{u}=5\)
\(\Rightarrow u^3-5u-12=0\)
\(\Leftrightarrow\left(u-3\right)\left(u^2+3u+4\right)=0\)
\(\Leftrightarrow u=3\Rightarrow v=2\)
\(\Rightarrow\left\{{}\begin{matrix}x+y-2=3\\\left(x-1\right)\left(y-1\right)=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5-x\\\left(x-1\right)\left(y-1\right)=2\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(5-x-1\right)=2\)
\(\Leftrightarrow...\) em tự hoàn thành bài toán