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30 tháng 7 2021
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
6 tháng 10 2020
1) đk: \(x\ge1\)
Ta có: \(\sqrt{x-1}-\sqrt{2x\left(x-1\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}=\sqrt{2x\left(x-1\right)}\)
\(\Leftrightarrow x-1=2x^2-2x\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Vậy x = 1
2) đk: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{5x^2}=2x-1\)
\(\Leftrightarrow5x^2=\left(2x-1\right)^2\)
\(\Leftrightarrow5x^2=4x^2-4x+1\)
\(\Leftrightarrow x^2+4x-1=0\)
\(\Leftrightarrow\left(x+2\right)^2-5=0\)
\(\Leftrightarrow\left(x+2-\sqrt{5}\right)\left(x+2+\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2+\sqrt{5}\left(ktm\right)\\x=-2-\sqrt{5}\left(ktm\right)\end{cases}}\)
=> PT vô nghiệm
3) đk: \(x\ge-1\)
Ta có: \(\sqrt{x+1}+\sqrt{9x+9}=4\)
\(\Leftrightarrow\sqrt{x+1}+3\sqrt{x+1}=4\)
\(\Leftrightarrow4\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=1\)
\(\Rightarrow x=0\)
6 tháng 10 2020
4) đk: \(x\ge2\)
Ta có: \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{x\left(x-2\right)}\)
\(\Leftrightarrow x-2=x\left(x-2\right)\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Vậy x = 2
6) đk: \(x\ge-\frac{7}{5}\)
Ta có: \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\frac{2x-3}{x-1}=2\)
\(\Leftrightarrow2x-3=2x-2\)
\(\Leftrightarrow0x=1\) vô lý
=> PT vô nghiệm
NH
2
26 tháng 1 2022
\(a,\left(đk:x\ge0\right)\)
\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)
\(x>0\)
\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)
\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)
\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)
\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)
DT
1
21 tháng 2 2023
a: =>(x^2-1)(x^2-4)=0
=>(x-1)(x+1)(x-2)(x+2)=0
=>\(x\in\left\{1;-1;2;-2\right\}\)
b: =>2x^4-4x^2+x^2-2=0
=>(x^2-2)(2x^2+1)=0
=>x^2-2=0
=>\(x=\pm\sqrt{2}\)
c: =>(căn x-6)(căn x+1)=0
=>căn x-6=0
=>x=36
MH
6 tháng 9 2021
Thay xx=√0,7 vào biểu thức ta được :
5√0,7^3 − 2√0,7^2 + 2,5√0,7 − 2,6 / √0,7^2 + 3√0,7 − 2,7
=3,5√0,7 − 1,4 + 2,5√0,7 − 2,6 / 0,7 + 3√0,7 −2,7
=6√0,7−4 / −2+3√0,7
=2
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
a/ ĐXĐK: ...
\(\Leftrightarrow9x^2-1-x-8x\sqrt{x+1}=0\)
\(\Leftrightarrow x^2-x-1+8x\left(x-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x^2-x-1+\frac{8x\left(x^2-x-1\right)}{x+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\Rightarrow x=...\\\frac{-8x}{x+\sqrt{x+1}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-8x=x+\sqrt{x+1}\)
\(\Leftrightarrow-9x=\sqrt{x+1}\) (\(x\le0\))
\(\Leftrightarrow81x^2-x-1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1-5\sqrt{13}}{162}\\x=\frac{1+5\sqrt{13}}{162}>0\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
d/
\(\Leftrightarrow3x^2+2\left(x^2+x+1\right)-5x\sqrt{x^2+x+1}=0\)
Đặt \(\sqrt{x^2+x+1}=a\)
\(\Leftrightarrow3x^2-5ax+2a^2=0\)
\(\Leftrightarrow\left(x-a\right)\left(3x-2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a\\3x=2a\end{matrix}\right.\) (\(x\ge0\))
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=x\\2\sqrt{x^2+x+1}=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=x^2\\2\left(x^2+x+1\right)=9x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\7x^2-2x-2=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1+\sqrt{15}}{7}\)
14 tháng 10 2021
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
31 tháng 8 2020
1. \(x^3-6x^2+10x-4=0\)
<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
<=> \(\left(x-2\right)\left(x^2-4x+2\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)
Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)
=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)
\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)
31 tháng 8 2020
1) Ta có: \(x^3-6x^2+10x-4=0\)
\(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)
+ \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
+ \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=2\)
\(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,5858;2;3,4142\right\}\)
10 tháng 7 2021
Ta có: \(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{2\sqrt{x}+1}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)