Question

[123456]

Câu 1: Tìm m để \(mx^2-2mx-1\le0,\forall x\in\left[0;3\right]\)

Câu 2: Giải bất phương trình:

a) \(2\left(x-1\right)\sqrt{x^2+2x-1}\le x^2-2x-1\)

b) \(\frac{3-2\sqrt{x^2+3x+2}}{1-2\sqrt{x^2-x+1}}>1\)

c)\(\frac{x^2-x}{\sqrt{x^4+3x^2}-2x}\le1\)

d)\(\sqrt{x-2}-2\ge\sqrt{2x-5}-\sqrt{x+1}\)

e) \(\sqrt{x+1}-\sqrt{3x^2-4x-15}+\sqrt{x-3}>0\)

Answer 1

[Hanako-kun]

\\(\\sqrt{2}x-y=0\\)

\n

Answer 2

[Hanako-kun]

Câu 1:

Xét \(m=0\Rightarrow f\left(x\right)=0-0-1\le0\left(lđ\right)\)

Xét \(m>0\Rightarrow f\left(x\right)\le0\Leftrightarrow x_1\le0< 3\le x_2\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(0\right)\le0\\f\left(3\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le0\left(lđ\right)\\9m-6m-1\le0\end{matrix}\right.\Leftrightarrow m\le\frac{1}{3}\Rightarrow0< m\le\frac{1}{3}\)

Xét \(m< 0\Rightarrow f\left(x\right)\le0\)

Chia làm 3 TH:

TH₁: \(\Delta< 0\Leftrightarrow m\left(m+1\right)< 0\Leftrightarrow-1< m< 0\)

TH₂: \(\Delta=0\Rightarrow m\left(m+1\right)=0\Leftrightarrow\left[{}\begin{matrix}m=0\left(l\right)\\m=-1\end{matrix}\right.\)

TH₃: \(\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}0\le x_1< x_2\\x_1< x_2\le3\end{matrix}\right.\end{matrix}\right.\)

\(\Delta>0\Leftrightarrow m< -1\)

\(0\le x_1< x_2\Leftrightarrow\left\{{}\begin{matrix}f\left(0\right)\le0\\\frac{x_1+x_2}{2}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le0\left(lđ\right)\\\frac{2m}{m}>0\left(lđ\right)\end{matrix}\right.\)

\(x_1< x_2\le3\Leftrightarrow\left\{{}\begin{matrix}f\left(3\right)\le0\\\frac{x_1+x_2}{2}< 3\left(lđ\right)\end{matrix}\right.\)

Vậy \(m\in\left[-1;\frac{1}{3}\right]\)

Có gì sai sót bảo mình ạ :<