tìm m để phương trình, bất phương trình sau có nghiệm:
a. \(\frac{3x^2-1}{\sqrt{2x-1}}=\sqrt{2x-1}+mx\)
b. \(\sqrt{x-1}+4m\sqrt[4]{x^2-3x+2}+\left(m+3\right)\sqrt{x-2}=0\)
c. \(\sqrt{4-x}+\sqrt{x+5}\ge m\)
d. \(m\sqrt{2x^2+9}< x+m\)
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a, ĐK: \(x\le-1,x\ge3\)
\(pt\Leftrightarrow2\left(x^2-2x-3\right)+\sqrt{x^2-2x-3}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x^2-2x-3}+3\right).\left(\sqrt{x^2-2x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x-3}=-\dfrac{3}{2}\left(l\right)\\\sqrt{x^2-2x-3}=1\end{matrix}\right.\)
\(\Leftrightarrow x^2-2x-3=1\)
\(\Leftrightarrow x^2-2x-4=0\)
\(\Leftrightarrow x=1\pm\sqrt{5}\left(tm\right)\)
b, ĐK: \(-2\le x\le2\)
Đặt \(\sqrt{2+x}-2\sqrt{2-x}=t\Rightarrow t^2=10-3x-4\sqrt{4-x^2}\)
Khi đó phương trình tương đương:
\(3t-t^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=0\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2+x}-2\sqrt{2-x}=0\\\sqrt{2+x}-2\sqrt{2-x}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2+x=8-4x\\2+x=17-4x+12\sqrt{2-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(tm\right)\\5x-15=12\sqrt{2-x}\left(1\right)\end{matrix}\right.\)
Vì \(-2\le x\le2\Rightarrow5x-15< 0\Rightarrow\left(1\right)\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{6}{5}\)
1.
ĐKXĐ: \(x\ge\dfrac{3+\sqrt{41}}{4}\)
\(\Leftrightarrow x^2+x-1+2\sqrt{x\left(x^2-1\right)}=2x^2-3x-4\)
\(\Leftrightarrow x^2-4x-3-2\sqrt{\left(x^2-x\right)\left(x+1\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x}=a>0\\\sqrt{x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow a^2-3b^2-2ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-3b\right)=0\)
\(\Leftrightarrow a=3b\)
\(\Leftrightarrow\sqrt{x^2-x}=3\sqrt{x+1}\)
\(\Leftrightarrow x^2-x=9\left(x+1\right)\)
\(\Leftrightarrow...\) (bạn tự hoàn thành nhé)
2.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=a\ge0\) pt trở thành:
\(x^3+3\left(x^2-4a^2\right)a=0\)
\(\Leftrightarrow x^3+3ax^2-4a^3=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+2a\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=x\\2a=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=x\left(x\ge0\right)\\2\sqrt{x+1}=-x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=x+1\\x^2=4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2-4x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=2-2\sqrt{2}\end{matrix}\right.\)
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 :<
a/ ĐKXĐ: \(x>\frac{1}{2}\)
\(\Leftrightarrow\frac{3x^2-1}{\sqrt{2x-1}}-\sqrt{2x-1}=mx\)
\(\Leftrightarrow\frac{3x^2-2x}{\sqrt{2x-1}}=mx\Leftrightarrow\frac{3x-2}{\sqrt{2x-1}}=m\)
Đặt \(\sqrt{2x-1}=a>0\Rightarrow x=\frac{a^2+1}{2}\Rightarrow\frac{3a^2-1}{2a}=m\)
Xét hàm \(f\left(a\right)=\frac{3a^2-1}{2a}\) với \(a>0\)
\(f'\left(a\right)=\frac{12a^2-2\left(3a^2-1\right)}{4a^2}=\frac{6a^2+2}{4a^2}>0\)
\(\Rightarrow f\left(a\right)\) đồng biến
Mặt khác \(\lim\limits_{a\rightarrow0^+}\frac{3a^2-1}{2a}=-\infty\); \(\lim\limits_{a\rightarrow+\infty}\frac{3a^2-1}{2a}=+\infty\)
\(\Rightarrow\) Phương trình đã cho luôn có nghiệm với mọi m
b/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\sqrt[4]{\left(x-1\right)^2}+4m\sqrt[4]{\left(x-1\right)\left(x-2\right)}+\left(m+3\right)\sqrt[4]{\left(x-2\right)^2}=0\)
Nhận thấy \(x=2\) không phải là nghiệm, chia 2 vế cho \(\sqrt[4]{\left(x-2\right)^2}\) ta được:
\(\sqrt[4]{\left(\frac{x-1}{x-2}\right)^2}+4m\sqrt[4]{\frac{x-1}{x-2}}+m+3=0\)
Đặt \(\sqrt[4]{\frac{x-1}{x-2}}=a\) pt trở thành: \(a^2+4m.a+m+3=0\) (1)
Xét \(f\left(x\right)=\frac{x-1}{x-2}\) khi \(x>0\)
\(f'\left(x\right)=\frac{-1}{\left(x-2\right)^2}< 0\Rightarrow f\left(x\right)\) nghịch biến
\(\lim\limits_{x\rightarrow2^+}\frac{x-1}{x-2}=+\infty\) ; \(\lim\limits_{x\rightarrow+\infty}\frac{x-1}{x-2}=1\) \(\Rightarrow f\left(x\right)>1\Rightarrow a>1\)
\(\left(1\right)\Leftrightarrow m\left(4a+1\right)=-a^2-3\Leftrightarrow m=\frac{-a^2-3}{4a+1}\)
Xét \(f\left(a\right)=\frac{-a^2-3}{4a+1}\) với \(a>1\)
\(f'\left(a\right)=\frac{-2a\left(4a+1\right)-4\left(-a^2-3\right)}{\left(4a+1\right)^2}=\frac{-4a^2-2a+12}{\left(4a+1\right)^2}=0\Rightarrow a=\frac{3}{2}\)
\(f\left(1\right)=-\frac{4}{5};f\left(\frac{3}{2}\right)=-\frac{3}{4};\) \(\lim\limits_{a\rightarrow+\infty}\frac{-a^2-3}{4a+1}=-\infty\)
\(\Rightarrow f\left(a\right)\le-\frac{3}{4}\Rightarrow m\le-\frac{3}{4}\)