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ND
1
24 tháng 10 2021
\(ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\\ \Leftrightarrow\sqrt{15x}\left(\dfrac{5}{3}-1-\dfrac{1}{3}\right)=2\\ \Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\Leftrightarrow\sqrt{15x}=6\Leftrightarrow15x=36\\ \Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)
NT
1
2 tháng 11 2021
\(ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{2}{3}\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\\ \Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\Leftrightarrow\sqrt{15x}=6\\ \Leftrightarrow15x=36\Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)
NH
2
4 tháng 9 2022
Đặt \(\sqrt{15x}=a\)
Pt sẽ là \(\dfrac{5}{3a}-a+11=\dfrac{1}{3a}\)
=>\(\dfrac{4}{3a}=a-11\)
\(\Leftrightarrow3a^2-33a-4=0\)
=>\(a=11.12\)
=>căn 15x=11,12
=>15x=123,6544
hay \(x\simeq8,24\)
25 tháng 10 2018
a) \(\sqrt{\left(2x-1\right)^2}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\dfrac{1}{3}\sqrt{15x}\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\)
\(\Leftrightarrow\sqrt{15x}=6\)
\(\Leftrightarrow15x=6^2\Leftrightarrow15x=36\)
\(\Rightarrow x=\dfrac{5}{12}\)
MH
1
QD
26 tháng 6 2019
B=\(\frac{3\sqrt{x}+4}{3\sqrt{x}-2}-\frac{42\sqrt{x}+34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{(3\sqrt{x}+4)(5\sqrt{x}+7)-42\sqrt{x}-34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{15x+20\sqrt{x}+21\sqrt{x}+28-42\sqrt{x}-34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{15x-\sqrt{x}-6}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+3\right)}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{5\sqrt{x}+3}{5\sqrt{x}+7}\)
1 tháng 7 2019
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
1 tháng 7 2019
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
PA
31 tháng 7 2017
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
\(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\)
\(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-\frac{1}{3}\sqrt{15x}-2=0\)
\(\left(\frac{5}{3}-1-\frac{1}{3}\right)\sqrt{15x}-2=0\)
\(\left(\frac{5}{3}-\frac{3}{3}-\frac{1}{3}\right)\sqrt{15x}-2=0\)
\(\frac{1}{3}\sqrt{15x}-2=0\)
\(\left(\frac{1}{3}\sqrt{15x}\right)^2-2^2=0\)
\(\frac{1}{9}.15x-4=0\)
\(\frac{1}{9}.15x=4\)
\(15x=4:\frac{1}{9}\)
\(15x=4.9\)
\(15x=36\)
\(x=\frac{12}{5}\)
vay nghiem cua phuong trinh da cho la: \(x=\frac{12}{5}\)