Gpt: \(1+\dfrac{1}{\sqrt{x^2-1}}=\dfrac{35}{12x}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
1/ \(\dfrac{5}{3}\le x\le\dfrac{7}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x-5}=a>0\\\sqrt{7-3x}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=2\\17-6x=2b^2+3\\6x-7=2a^2+3\end{matrix}\right.\)
Mặt khác theo BĐT Bunhiacốpxki:
\(a+b=\sqrt{3x-5}+\sqrt{7-3x}\le\sqrt{\left(1+1\right)\left(3x-5+7-3x\right)}=2\)
\(\Rightarrow0< a+b\le2\)
Ta được hệ pt:
\(\left\{{}\begin{matrix}a^2+b^2=2\\\left(2b^2+3\right).a+\left(2a^2+3\right)b=2+8ab\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\2ab^2+3a+2a^2b+3b-8ab-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-2\\2ab\left(a+b\right)+3\left(a+b\right)-8ab-2=0\end{matrix}\right.\)
\(\Rightarrow\left(\left(a+b\right)^2-2\right)\left(a+b\right)+3\left(a+b\right)-4\left(a+b\right)^2+6=0\)
\(\Leftrightarrow\left(a+b\right)^3-4\left(a+b\right)^2+\left(a+b\right)+6=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b=-1< 0\left(l\right)\\a+b=2\\a+b=3>2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a+b=2\) , dấu "=" xảy ra khi và chỉ khi:
\(3x-5=7-3x\Rightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
2/ ĐKXĐ: \(x\ne\pm2\)
\(\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-\left(\dfrac{15}{x^2-4}+5\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-5.\left(\dfrac{x^2-1}{x^2-4}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{x^2-1}{x^2-4}\right)-4\left[\left(\dfrac{x^2-1}{x^2-4}\right)-\left(\dfrac{x+1}{x-2}\right)^2\right]=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)-4\left(\dfrac{x+1}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}-\dfrac{4\left(x+1\right)}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}=\dfrac{4\left(x+1\right)}{x-2}\\\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=4\left(x^2+3x+2\right)\\x^2-3x+2=x^2+3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2+15x+6=0\\6x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5+\sqrt{17}}{2}\\x=\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
30 tháng 1 2021
Lời giải:ĐK: $\cos 3x>\frac{-1}{2}$
PT $\Rightarrow 4\sin ^2\frac{x}{2}-\sqrt{3}\cos 2x-1-2\cos ^2(x-\frac{3\pi}{4})=0$
$\Leftrightarrow 2(1-\cos x)-\sqrt{3}\cos 2x-2+[1-2\cos ^2(x-\frac{3\pi}{4})]=0$
$\Leftrightarrow -2\cos x-\sqrt{3}\cos 2x-cos (2x-\frac{3\pi}{2})=0$
$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\cos (2x-\frac{3\pi}{2})=0$
$\Leftrightarrow 2\cos x+\sqrt{3}\cos 2x+\sin 2x=0$
$\Leftrightarrow \cos x+\frac{\sqrt{3}}{2}\cos 2x+\frac{1}{2}\sin 2x=0$
$\Leftrightarrow \cos x-\cos (2x+\frac{5\pi}{6})=0
$\Leftrightarrow \cos x=\cos (2x+\frac{5\pi}{6})$
$\Rightarrow x+2k\pi =2x+\frac{5}{6}\pi$ hoặc $-x+2k\pi =2x+\frac{5}{6}\pi$
Vậy......
11 tháng 4 2023
\(\left(x^2-3x+2\right)\sqrt{\dfrac{x+3}{x-1}}=-\dfrac{1}{2}x^3+\dfrac{15}{2}x-11\left(1\right)\)
Đk: \(\sqrt{\dfrac{x+3}{x-1}}\ge0\Leftrightarrow\left[{}\begin{matrix}x>1\\x\le-3\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-2\left(x-1\right)\left(x-2\right)\sqrt{\dfrac{x+3}{x-1}}=x^3-15x+22\)
\(\Rightarrow-2\sqrt{\left(x-1\right)\left(x+3\right)}.\left(x-2\right)=\left(x-2\right)\left(x^2+2x-11\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\-2\sqrt{\left(x-1\right)\left(x+3\right)}=x^2+2x-11\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow-2\sqrt{x^2+2x-3}=\left(x^2+2x-3\right)-8\)
Đặt \(a=\sqrt{x^2+2x-3}\left(a\ge0\right)\). Từ phương trình (2) suy ra:
\(a^2+2a-8=0\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-4\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+2x-3}=2\Leftrightarrow x^2+2x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1+2\sqrt{2}\left(nhận\right)\\x=-1-2\sqrt{2}\left(nhận\right)\end{matrix}\right.\)
Thử lại ta có \(x=2\) và \(x=-1+2\sqrt{2}\) là 2 nghiệm của phương trình (1).
11 tháng 4 2023
\(\Leftrightarrow2\left(x^2-3x+2\right)\cdot\sqrt{\dfrac{x+3}{x-1}}=-x^3+15x-22\)
\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)\cdot\dfrac{\sqrt{\left(x+3\right)\left(x-1\right)}}{x-1}=-x^3+2x^2-2x^2+4x+11x-22\)
\(\Leftrightarrow2\left(x-2\right)\sqrt{\left(x+3\right)\left(x-1\right)}=\left(x-2\right)\left(-x^2-2x+11\right)\)
\(\Leftrightarrow\left(x-2\right)\left(\sqrt{4\left(x^2+2x-3\right)}+x^2+2x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\left(1\right)\\2\sqrt{x^2+2x-3}+x^2+2x-11=0\left(2\right)\end{matrix}\right.\)
(1) =>x=2
(2): Đặt \(\sqrt{x^2+2x-3}=a\left(a>=0\right)\)
=>2a+a^2-8=0
=>(a+4)(a-2)=0
=>a=2
=>x^2+2x-3=4
=>x^2+2x-7=0
=>\(x=-1\pm2\sqrt{2}\)
HN
2
AH
Akai Haruma
Giáo viên
28 tháng 4 2023
Lời giải:
ĐKXĐ:.......
$PT\Leftrightarrow \frac{4}{x}-x=\sqrt{2x-\frac{5}{x}}-\sqrt{x-\frac{1}{x}}$
$\Leftrightarrow \frac{4}{x}-x = \frac{(2x-\frac{5}{x})-(x-\frac{1}{x})}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}$
$\Leftrightarrow \frac{4}{x}-x = \frac{x-\frac{4}{x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}$
$\Leftrightarrow (\frac{4}{x}-x)\left[1+\frac{1}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}\right]=0$
Hiển nhiên biểu thức trong ngoặc vuông luôn dương nên $\frac{4}{x}-x=0$
$\Rightarrow 4-x^2=0$
$\Leftrightarrow x=\pm 2$
Thử lại thấy $x=2$ thỏa mãn.
Vậy.......
28 tháng 4 2023
\(\Leftrightarrow x-\dfrac{4}{x}=\sqrt{x-\dfrac{1}{x}}-\sqrt{2x-\dfrac{5}{x}}\)
\(x-\dfrac{4}{x}=\dfrac{\dfrac{4}{x}-x}{\sqrt{x-\dfrac{1}{x}}+\sqrt{2x-\dfrac{5}{x}}}\)
x-4/x>0
=>4/x-x<0
=>Loại
x-4/x<0
=>4/x-x>0
=>Mâu thuẫn
=>Loại
Do đó, chỉ có 1 trường hợp là x-4/x=0
=>x=2
PQ
0
\(1+\dfrac{1}{\sqrt{x^2-1}}=\dfrac{35}{12x}\left(x< -1;1< x\right)\)
Với \(x< -1\) thì pt vô nghiệm
Xét \(x>1\)
\(PT\Leftrightarrow x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}\left(nhân.x.2.vế\right)\\ \Leftrightarrow x^2+\dfrac{x^2}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}=\dfrac{1225}{144}\\ \Leftrightarrow\dfrac{x^4}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}=\dfrac{1225}{144}\\ \Leftrightarrow\left(\dfrac{x^2}{\sqrt{x^2-1}}\right)^2+\dfrac{2x^2}{\sqrt{x^2-1}}-\dfrac{1225}{144}=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2}{\sqrt{x^2-1}}=\dfrac{25}{12}\left(tm\right)\\\dfrac{x^2}{\sqrt{x^2-1}}=-\dfrac{49}{12}\left(ktm\right)\end{matrix}\right.\Leftrightarrow\dfrac{x^4}{x^2-1}=\dfrac{625}{144}\\ \Leftrightarrow144x^4-625x^2+625=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=\dfrac{5}{4}\left(tm\right)\\x=-\dfrac{5}{4}\left(tm\right)\\x=-\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)