\(\sqrt{4x-1}+\sqrt{4x^2-1}=1\)
giúp t điii
K
Khách
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CC
1
13 tháng 10 2021
\(\sqrt{x-2}=3\left(x\ge2\right)\\ \Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\\ \sqrt{4x^2}+4x+1=3\Leftrightarrow\left|2x\right|=2-4x\\ \Leftrightarrow\left[{}\begin{matrix}2x=2-4x\left(x\ge0\right)\\2x=4x-2\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{3}\)
28 tháng 8 2021
3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
DH
1
NA
10 tháng 7 2018
\(\sqrt{4x-2\sqrt{4x-1}}+\sqrt{4x+2\sqrt{4x-1}}\)(với \(x\ge\dfrac{1}{4}\))
\(=\sqrt{\left(\sqrt{4x-1}-1\right)^2}+\sqrt{\left(\sqrt{4x-1}+1\right)^2}\)
\(=\left(\sqrt{4x-1}-1\right)+\left(\sqrt{4x-1}+1\right)\)
\(=2\sqrt{4x-1}\) (với \(x\ge\dfrac{1}{4}\))
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 6 2019
\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)
\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)
\(\Leftrightarrow4x^2+4x-1=0\)
\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)
\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)
\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
\(M=1+8+1=10\)
11 tháng 12 2018
Bạn ghi lộn đề rồi \(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}\) chứ không phải \(\left(\dfrac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\right)^{2014}\)
11 tháng 12 2018
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2-1}\right)}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\left|\sqrt{2}-1\right|}{2}=\dfrac{\sqrt{2}-1}{2}\)
Vậy ta có \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)Ta lại có \(\left(4x^5+4x^4-x^3+1\right)^{19}=\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=\left(x^3.0+1\right)^{19}=1^{19}=1\)(1)
\(\left(\sqrt{4x^5+4x^4-5x^3+5x+3}\right)^3=\left(\sqrt{4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1+4}\right)^3=\left(\sqrt{x^3\left(4x^2+4x-1\right)-x^2\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\right)^3=\left(\sqrt{x^3.0+x^2.0+0+4}\right)^3=\left(\sqrt{4}\right)^3=2^3=8\left(2\right)\)
\(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}=\left[\dfrac{1-\sqrt{2}.\dfrac{\sqrt{2}-1}{\sqrt{2}}}{\sqrt{2.\dfrac{3-2\sqrt{2}}{4}+\sqrt{2}-1}}\right]^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}}{2}+\sqrt{2}-1}}\right)^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}+2\sqrt{2}-2}{2}}}\right)^{2014}=\left(\dfrac{\dfrac{\dfrac{1}{\sqrt{2}}}{1}}{\sqrt{2}}\right)^{2014}=1^{2014}=1\left(3\right)\)
Cộng (1),(2),(3) theo vế ta được A=1+8+1=10
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=10
VP
3
KK
6 tháng 9 2021
1. \(\sqrt{x^2-4x+3}=x-2\)
<=> x2 - 4x + 3 = (x - 2)2
<=> x2 - 4x + 3 = x2 - 4x + 4
<=> x2 - x2 - 4x + 4x = 1
<=> 0 = 1 (Vô lí)
vậy PT có nghiệm là S = \(\varnothing\)
KK
6 tháng 9 2021
2. \(\sqrt{4x^2-4x+1}=x-1\)
<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)
<=> 2x - 1 = x - 1
<=> 2x - x = -1 + 1
<=> x = 0
Cách 1:
\(\sqrt{4x-1}+\sqrt{4x^2-1}=1\)
\(\Leftrightarrow\left(\sqrt{4x-1}-1\right)+\sqrt{4x^2-1}=0\)
\(\Leftrightarrow\frac{4x-2}{\sqrt{4x-1}+1}+\sqrt{\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\sqrt{2x-1}\left(\frac{2\sqrt{2x-1}}{\sqrt{4x-1}+1}+\sqrt{2x+1}\right)=0\)
\(\Leftrightarrow\sqrt{2x-1}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Cách 2:
Điều kiện: \(x\ge\frac{1}{2}\)
Ta có:
\(VT=\sqrt{4x-1}+\sqrt{4x^2-1}\ge\sqrt{4.\frac{1}{2}-1}+\sqrt{4.\left(\frac{1}{2}\right)^2-1}=1=VP\)
Dấu = xảy ra khi \(x=\frac{1}{2}\)