Giải Phương trình
\(\sqrt{x-3}+\sqrt{5-x}=2\)
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Những câu hỏi liên quan
TN
31 tháng 10 2016
Bài 1:
Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\) hpt thành:
\(\hept{\begin{cases}S^2-P=3\\S+P=9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S^2-P=3\\S=9-P\end{cases}}\Leftrightarrow\left(9-P\right)^2-P=3\)
\(\Leftrightarrow\orbr{\begin{cases}P=6\Rightarrow S=3\\P=13\Rightarrow S=-4\end{cases}}\).Thay 2 trường hợp S và P vào ta tìm dc
\(\hept{\begin{cases}x=3\\y=0\end{cases}}\)và\(\hept{\begin{cases}x=0\\y=3\end{cases}}\)
CH
Cô Hoàng Huyền
Admin
VIP
1 tháng 11 2016
Câu 3: ĐK: \(x\ge0\)
Ta thấy \(x-\sqrt{x-1}=0\Rightarrow x=\sqrt{x-1}\Rightarrow x^2-x+1=0\) (Vô lý), vì thế \(x-\sqrt{x-1}\ne0.\)
Khi đó \(pt\Leftrightarrow\frac{3\left[x^2-\left(x-1\right)\right]}{x+\sqrt{x-1}}=x+\sqrt{x-1}\Rightarrow3\left(x-\sqrt{x-1}\right)=x+\sqrt{x-1}\)
\(\Rightarrow2x-4\sqrt{x-1}=0\)
Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow2\left(t^2+1\right)-4t=0\Rightarrow t=1\Rightarrow x=2\left(tm\right)\)
BG
2
31 tháng 10 2021
=>\(x^2+9-12\sqrt{x^2-25}=13x+5-12\sqrt{x^2-25}\)
<=> \(x^2-13x+4=0\)
........
31 tháng 10 2021
\(=>x^2+11-12\sqrt{x^2-25}=13x+25-12\sqrt{x^2-25}\)
\(< =>x^2-13x-14=0\)
\(< =>\left(x+1\right)\left(x-14\right)=0\)
..............
MN
2
13 tháng 8 2021
ĐK:\(x\ge\dfrac{5}{2}\)
Ta có:\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=7.2\)
\(\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+6}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)
\(\Leftrightarrow2\sqrt{2x-5}=10\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow2x-5=25\Leftrightarrow2x=30\Leftrightarrow x=15\left(tm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 8 2021
ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)
\(\Leftrightarrow\sqrt{2x-5+2\sqrt{2x-5}+1}+\sqrt{2x-5+6\sqrt{2x-5}+3}=14\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)
\(\Leftrightarrow2.\sqrt{2x-5}+4=14\)
\(\Leftrightarrow\sqrt{2x-5}=5\)
\(\Leftrightarrow x=15\)
NL
0
NK
2
28 tháng 8 2021
`sqrt{x^2-25}-6=3sqrt{x+5}-2sqrt{x-5}(x>=5)`
`<=>sqrt{(x-5)(x+5)}+2sqrt{x-5}=3sqrt{x+5}+6`
`<=>sqrt{x-5}(sqrt{x+5}+2)=3(sqrt{x+5}+2)`
`<=>(sqrt{x+5}+2)(sqrt{x-5}-3)=0`
Vì `sqrt{x+5}+2>0`
`<=>sqrt{x-5}-3=0`
`<=>sqrt{x-5}=3`
`<=>x-5=9<=>x=14(tm)`
Vậy `x=14`
28 tháng 8 2021
\(\sqrt{x^2-25}-6=3\sqrt{x+5}-2\sqrt{x-5}\\ \Leftrightarrow\sqrt{\left(x-5\right)\left(x+5\right)}-6-3\sqrt{x+5}+2\sqrt{x-5}=0\\ \Leftrightarrow\left(2\sqrt{x-5}+\sqrt{\left(x-5\right)\left(x+5\right)}\right)-\left(3\sqrt{x+5}+6\right)=0\Leftrightarrow\sqrt{x-5}\left(2+\sqrt{x+5}\right)-3\left(2+\sqrt{x+5}\right)=0\\ \Leftrightarrow\left(\sqrt{x-5}-3\right)\left(2+\sqrt{x-5}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=3\\\sqrt{x-5}=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=9\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=14\)
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: \(x\ge0\)
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)
LC
2
VT
12 tháng 10 2019
điệu kiện \(\hept{\begin{cases}x\ge0\\2-x\ge0;3-x\ge0;5-x\ge0\end{cases}< =>0\le x\le2;}\)
ta có 2x = \(2\sqrt{2-x}\sqrt{3-x}+2\sqrt{3-x}\sqrt{5-x}+2\sqrt{5-x}\sqrt{2-x}\)
<=> 2x = \(\sqrt{2-x}\left(\sqrt{3-x}+\sqrt{5-x}\right)+\sqrt{3-x}\left(\sqrt{5-x}+\sqrt{2-x}\right)\)+\(\sqrt{5-x}\left(\sqrt{2-x}+\sqrt{3-x}\right)\)
<=> 2x = \(\sqrt{2-x}\left(x-\sqrt{2-x}\right)+\sqrt{3-x}\left(x-\sqrt{3-x}\right)+\sqrt{5-x}\left(x-\sqrt{5-x}\right)\)
<=> 2x = x (\(\sqrt{2-x}+\sqrt{3-x}+\sqrt{5-x}\)) - (2-x +3-x + 5-x)
<=> 2x= x.x - 10 +3x <=> x2+x-10 = 0 <=> \(\orbr{\begin{cases}x=\frac{-1+\sqrt{41}}{2}\left(loai\right)\\x=\frac{-1-\sqrt{41}}{2}\left(loai\right)\end{cases}}\) cả 2 nghiệm đều không thỏa mãn \(0\le x\le2\)
=> phương trình vô nghiệm
29 tháng 2 2020
ĐK: \(x\le2\)
pt <=> \(2=2-x+\sqrt{2-x}\sqrt{3-x}+\sqrt{3-x}\sqrt{5-x}+\sqrt{5-x}\sqrt{2-x}.\)
<=> \(2=\sqrt{2-x}\left(\sqrt{2-x}+\sqrt{3-x}\right)+\sqrt{5-x}\left(\sqrt{2-x}+\sqrt{3-x}\right).\)
<=> \(2=\left(\sqrt{2-x}+\sqrt{3-x}\right)\left(\sqrt{5-x}+\sqrt{2-x}\right).\)
<=> \(2\left(\sqrt{5-x}-\sqrt{2-x}\right)=3\left(\sqrt{2-x}+\sqrt{3-x}\right)\)( vì \(\sqrt{5-x}-\sqrt{2-x}\ne0;\forall x\inℝ\))
<=> \(2\sqrt{5-x}=5\sqrt{2-x}+3\sqrt{3-x}\)
<=> \(4\left(5-x\right)=25\left(2-x\right)+9\left(3-x\right)+30\sqrt{\left(2-x\right)\left(3-x\right)}\)
<=> \(-57+30x=30\sqrt{\left(2-x\right)\left(3-x\right)}\)
<=> \(\hept{\begin{cases}30x-57\ge0\\900x^2-3420x+3249=900x^2-4500x+5400\end{cases}}\)
<=> \(\hept{\begin{cases}x\ge\frac{57}{30}\\x=\frac{239}{120}\end{cases}}\Leftrightarrow x=\frac{239}{120}\)tmđk
15 tháng 9 2021
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
HP
15 tháng 9 2021
a, ĐK: \(x\ge11\)
\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)
\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)
\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)
\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)
Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{}8\)
\(\Rightarrow\) phương trình vô nghiệm.
15 tháng 9 2021
\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)
\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)
\(đk:3\le x\le5\)
PT\(\Rightarrow x-3+5-x+2\sqrt{\left(x-3\right)\left(5-x\right)}=2\)
\(\Leftrightarrow2\sqrt{\left(x-3\right)\left(5-x\right)}=0\Leftrightarrow\orbr{\begin{cases}x-3=0\\5-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}}\)
GIÚP MÌNH NỐT BÀI NÀY NHÉ
\(\sqrt{x-4}+\sqrt{6-x}=x^2-10x+27\)