\(\sqrt{x+3}+x^2+4x=7\)
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22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
HN
25 tháng 2 2017
1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
H
17 tháng 6 2019
3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
14 tháng 10 2023
a: ĐKXĐ: x>=1
\(\dfrac{1}{2}\sqrt{x-1}-\sqrt{4x-4}+3=0\)
=>\(3+\dfrac{1}{2}\sqrt{x-1}-2\sqrt{x-1}=0\)
=>\(3-\dfrac{3}{2}\sqrt{x-1}=0\)
=>\(\dfrac{3}{2}\sqrt{x-1}=3\)
=>\(\sqrt{x-1}=2\)
=>x-1=4
=>x=5(nhận)
b: \(\sqrt{x^2-4x+4}+x-2=0\)
=>\(\sqrt{\left(x-2\right)^2}=-x+2\)
=>|x-2|=-(x-2)
=>x-2<=0
=>x<=2
c:
ĐKXĐ: 7-x>=0
=>x<=7
\(\sqrt{7-x}+1=x\)
=>\(\sqrt{7-x}=x-1\)
=>\(\left\{{}\begin{matrix}x-1>=0\\7-x=x^2-2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1< =x< =7\\x^2-2x+1-7+x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1< =x< =7\\x^2-x-6=0\end{matrix}\right.\Leftrightarrow x=3\)
30 tháng 10 2023
a: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2-4x+4}=7\)
=>\(\sqrt{\left(x-2\right)^2}=7\)
=>|x-2|=7
=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)
b: ĐKXĐ: x>=-3
\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)
=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)
=>\(3\sqrt{x+3}=6\)
=>\(\sqrt{x+3}=2\)
=>x+3=4
=>x=1(nhận)
TQ
11 tháng 6 2021
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
15 tháng 10 2021
a: Ta có: \(\sqrt{x^2-x+3}+7=10\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)
\(\Leftrightarrow x^2-4x+8=4\)
\(\Leftrightarrow x-2=0\)
hay x=2
