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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
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(5\pm\sqrt{4m-11}\right)^2-16\ge0\\\dfrac{5\pm\sqrt{4m-11}}{2}>0\\1>0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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Lời giải
$y'=3x^2+1>0$ với mọi $x\in\mathbb{R}$ nên hàm $y=x^3+x$ đồng biến trên $\mathbb{R}$
PT $\Leftrightarrow x^3+x=\sqrt[3]{2x+1}+2x+1$
Đặt $\sqrt[3]{2x+1}=t$ thì:
$x^3+x=t^3+t$
Vì hàm $y=x^3+x$ đồng biến nên $x^3+x=t^3+t\Leftrightarrow x=t$
$\Leftrightarrow x=\sqrt[3]{2x+1}$
$\Leftrightarrow x^3=2x+1$
Giải pt này dễ dàng có $x=-1; \frac{1\pm \sqrt{5}}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\le1\)
Đặt \(\sqrt{1-x}=t\ge0\Leftrightarrow x=1-t^2\)
\(PT\Leftrightarrow6t-\left(1-t^2\right)=5\sqrt{1-t}\\ \Leftrightarrow t^2-\left(1-t\right)+5t-5\sqrt{1-t}=0\\ \Leftrightarrow\left(t-\sqrt{1-t}\right)\left(t+\sqrt{1-t}+5\right)=0\\ \Leftrightarrow t-\sqrt{1-t}=0\left(t+\sqrt{1-t}+5>0\right)\\ \Leftrightarrow t=\sqrt{1-t}\\ \Leftrightarrow t^2=1-t\\ \Leftrightarrow t=\dfrac{\sqrt{5}-1}{2}\Leftrightarrow1-x=\dfrac{3-\sqrt{5}}{2}\\ \Leftrightarrow x=\dfrac{-1\pm\sqrt{5}}{2}\left(tm\right)\)
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(2x – 1)(x + 3) – 3x + 1 ≤ (x – 1)(x + 3) + x2 – 5
⇔ 2x2 + 6x - x – 3 – 3x + 1 ≤ x2 + 3x - x – 3 + x2 – 5
⇔ 2x2 + 2x – 2 ≤ 2x2 + 2x – 8
⇔ 6 ≤ 0 (Vô lý).
Vậy BPT vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Em kiểm tra lại đề bài, pt này chắc chắn là ko giải được