Giải phương trình
\(\sqrt{x^2+x+1}=2x+\)\(\sqrt{x^2-x+1}\)
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
20 tháng 7 2021
a.
ĐKXĐ: \(x>0\)
\(\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\dfrac{\left(x+2\right)\left(x+3\right)}{x}}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-\sqrt{x+3}\right)+\sqrt{\dfrac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\dfrac{4x-x-3}{2\sqrt{x}+\sqrt{x+3}}\right)-\sqrt{\dfrac{x+2}{x}}\left(\dfrac{4x-x-3}{\sqrt{x+3}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)}{2\sqrt{x}+\sqrt{x+3}}\left(\sqrt{x}-\sqrt{\dfrac{x+2}{x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{x+2}{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 7 2021
b.
ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)
\(x+2+x\sqrt{2x+1}=x\sqrt{x+2}+\sqrt{\left(x+2\right)\left(2x+1\right)}\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{2x+1}-\sqrt{x+2}\right)-x\left(\sqrt{2x+1}-\sqrt{x+2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x+2}\right)\left(\sqrt{x+2}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)
18 tháng 5 2021
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
18 tháng 5 2021
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
MA
0
15 tháng 9 2021
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2021
a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)
\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)
\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2021
b. ĐKXĐ: \(x\ge5\)
\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)
\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
B
2
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 2 2021
\(\Leftrightarrow2x^2+2+2\sqrt{\left(x^2+x+1\right)\left(x^2-x+1\right)}=2x^2+4\)
\(\Leftrightarrow\sqrt{x^4+x^2+1}=1\)
\(\Leftrightarrow x^4+x^2=0\)
\(\Leftrightarrow x=0\)
28 tháng 2 2021
`\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=\sqrt{2x^2+4}`
`<=>2x^2+2+2\sqrt{x^4+x^2+1}=2x^2+3`
`<=>\sqrt{x^4+x^2+1}=1`
`<=>x^4+x^2=0`
`<=>x=0`
LT
0
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2022
a.
\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1\le x\le3\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 4 2022
b.
ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)
dat a=\(\sqrt{x^2+x+1}\) b=\(\sqrt{x^2-x+1}\) dk \(a,b\ge0\)
t a co he phuong trinh \(\hept{\begin{cases}a^2-b^2=2x\\a-b=2x\end{cases}}\) \(\Rightarrow a^2-b^2=a-b\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\)
voi a=b \(\sqrt{x^2+x+1}=\sqrt{x^2-x+1}\Rightarrow x^2+x+1=x^2-x+1\)
\(\Rightarrow x=0\)
vs a+b=1 ket hop vs a-b=2x \(\Rightarrow a=\frac{2x+1}{2}\) \(b=\frac{-2x+1}{2}\)
do \(a\ge0,b\ge0\Rightarrow\frac{-1}{2}\le x\le\frac{1}{2}\)
tu \(\sqrt{x^2+x+1}=\frac{2x+1}{2}\Rightarrow x^2+x+1=\frac{4x^2+4x+1}{4}\)
\(\Rightarrow4\left(x^2+x+1\right)=4x^2+4x+1\)
\(\Rightarrow\) ko co no nao tm
kl x=0 la no cua pt da cho