LT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
a. ĐKXĐ: $x\geq 2$ hoặc $x=1$
PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)
b.
PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$
$\Leftrightarrow |x-2|=|2x-3|$
\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
c. ĐKXĐ: $x=2$ hoặc $x\geq 3$
PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$
\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)
d.
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
EN
Giải phương trình:
a)\(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{3}+x\)
b)\(\sqrt{x-3+2\sqrt{x-4}}=2\sqrt{x-4}+1\)
2
10 tháng 7 2021
a)Pt\(\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=x+\sqrt{3}\)
\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)
\(\Leftrightarrow x+\sqrt{3}\ge0\)\(\Leftrightarrow x\ge-\sqrt{3}\)
Vậy...
b)Đk:\(x\ge4\)
Pt\(\Leftrightarrow\sqrt{\left(x-4\right)+2\sqrt{x-4}+1}=2\sqrt{x-4}+1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+1\right)^2}=1+2\sqrt{x-4}\)
\(\Leftrightarrow\sqrt{x-4}+1=2\sqrt{x-4}+1\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Leftrightarrow x=4\) (tm)
Vậy...
10 tháng 7 2021
a) Ta có: \(\sqrt{x^2+2x\sqrt{3}+3}=x+\sqrt{3}\)
\(\Leftrightarrow\left|x+\sqrt{3}\right|=x+\sqrt{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=x+\sqrt{3}\left(x\ge-\sqrt{3}\right)\\x+\sqrt{3}=-x-\sqrt{3}\left(x< -\sqrt{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge-\sqrt{3}\\x=-\sqrt{3}\left(loại\right)\end{matrix}\right.\Leftrightarrow x\ge-\sqrt{3}\)
30 tháng 9 2021
b: Ta có: \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
\(\Leftrightarrow\sqrt{x^2-1}=2\)
\(\Leftrightarrow x^2-1=4\)
hay \(x\in\left\{\sqrt{5};-\sqrt{5}\right\}\)
HP
30 tháng 9 2021
a. \(x+\sqrt{x^2-4x+4}=\dfrac{1}{2}\)
<=> \(x+\sqrt{\left(x-2\right)^2}=\dfrac{1}{2}\)
<=> \(x+\left|x-2\right|=\dfrac{1}{2}\)
<=> \(\left[{}\begin{matrix}x+x-2=\dfrac{1}{2}\\x+\left[-\left(x-2\right)\right]=\dfrac{1}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=\dfrac{5}{2}\\x-x+2=\dfrac{1}{2}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\0=\dfrac{-3}{2}\left(VLí\right)\end{matrix}\right.\)
Vậy nghiệm của PT là \(S=\left\{\dfrac{5}{4}\right\}\)
b. \(\sqrt{9x^2-9}+\sqrt{4x^2-4}=\sqrt{16x^2-16}+2\)
<=> \(\sqrt{9\left(x^2-1\right)}+\sqrt{4\left(x^2-1\right)}=\sqrt{16\left(x^2-1\right)}+2\)
<=> \(3\sqrt{x^2-1}+2\sqrt{x^2-1}-4\sqrt{x^2-1}=2\)
<=> \(\left(3+2-4\right)\sqrt{x^2-1}=2\)
<=> \(\sqrt{x^2-1}=2\)
<=> x2 - 1 = 4
<=> x2 = 5
<=> x = \(\sqrt{5}\)
23 tháng 8 2021
d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
23 tháng 8 2021
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
NH
2
26 tháng 1 2022
\(a,\left(đk:x\ge0\right)\)
\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)
\(x>0\)
\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)
\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)
\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)
\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)
29 tháng 6 2021
`a)sqrt{x^2-6x+9}=2`
`<=>sqrt{(x-3)^2}=2`
`<=>|x-3|=2`
`**x-3=2`
`<=>x=5`
`**x-3=-2`
`<=>x=1`
Vậy `S={1,5}`
`b)sqrt{4x-20}+sqrt{x-5}-1/3sqrt{9x-45}=4`
đk:`x>=5`
`pt<=>2sqrt{x-5}+sqrt{x-5}-1/3*3*sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4<=>x=9`
Vậy `S={9}`
AH
Akai Haruma
Giáo viên
29 tháng 6 2021
Lời giải:
a.
PT $\Leftrightarrow \sqrt{(x-3)^2}=2$
$\Leftrightarrow |x-3|=2$
$\Leftrightarrow x-3=\pm 2$
$\Leftrightarrow x=1$ hoặc $x=5$
b. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4(x-5)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9(x-5)}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)
LD
27 tháng 7 2021
Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)
LD
27 tháng 7 2021
a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)
b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)
\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))
\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)
c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
14 tháng 10 2021
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
I
1
15 tháng 10 2023
a: ĐKXĐ: \(\left\{{}\begin{matrix}x+6>=0\\x-2>=0\end{matrix}\right.\Leftrightarrow x>=2\)
\(\sqrt{x+6}-\sqrt{x-2}=2\)
=>\(\left(\sqrt{x+6}-\sqrt{x-2}\right)^2=4\)
=>\(x+6+x-2-2\sqrt{\left(x+6\right)\left(x-2\right)}=4\)
=>\(2\sqrt{\left(x+6\right)\left(x-2\right)}=2x+4-4=2x\)
=>\(\sqrt{\left(x+6\right)\left(x-2\right)}=x\)
=>\(\left\{{}\begin{matrix}x>=0\\\left(x+6\right)\left(x-2\right)=x^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=2\\x^2+4x-12=x^2\end{matrix}\right.\)
=>x=3
b: ĐKXĐ: \(x-3>=0\)
=>x>=3
\(2\sqrt{x-3}-2x+3=0\)
=>\(\sqrt{4x-12}=2x-3\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\4x-12=4x^2-12x+9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\4x^2-12x+9-4x+12=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\4x^2-16x+21=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)