Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)
=>x+1=5
=>x=4
b: =>x^2/10=1,1
=>x^2=11
=>x=căn 11 hoặc x=-căn 11
c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0
=>4x+3=9x+9
=>-5x=6
=>x=-6/5
d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0
=>2x-3=4x-4 và x>=3/2
=->-2x=-1 và x>=3/2
=>x=1/2 và x>=3/2
=>Ko có x thỏa mãn
e: Đặt căn x=a(a>=0)
PT sẽ là a^2-a-5=0
=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)
=>x=(1+căn 21)^2/4=(11+căn 21)/2
a, ĐK: \(x\ge2\)
\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)
Phương trình vô nghiệm.
b, ĐK: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)
\(\Leftrightarrow2x+9=5-4x\)
\(\Leftrightarrow6x=-4\)
hay \(x=-\dfrac{2}{3}\left(nhận\right)\)
b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)
\(\Leftrightarrow2x-1=x-1\)
hay x=0(loại)
c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2+3x=x\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
a. \(\sqrt{2x+9}=\sqrt{5-4x}\)
<=> 2x + 9 = 5 - 4x
<=> 2x + 4x = 5 - 9
<=> 6x = -4
<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)
a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))
Vì hai vế ko âm, bp 2 vế ta được:
2x2 - 3 = 4x - 3
\(\Leftrightarrow\) 2x2 = 4x
\(\Leftrightarrow\) x2 = 2x
\(\Leftrightarrow\) x2 - 2x = 0
\(\Leftrightarrow\) x(x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy S = {2}
b, \(\sqrt{2x-1}=\sqrt{x-1}\) (x \(\ge\) 1)
Vì hai vế ko âm, bp 2 vế ta được:
2x - 1 = x - 1
\(\Leftrightarrow\) x = 0 (KTM)
Vậy x = \(\varnothing\)
c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)
Vì hai vế ko âm, bp 2 vế ta được:
x2 - x - 6 = x - 3
\(\Leftrightarrow\) x2 - 2x - 3 = 0
\(\Leftrightarrow\) x2 - 3x + x - 3 = 0
\(\Leftrightarrow\) x(x - 3) + (x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 1) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)
Vậy S = {3}
d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))
Vì hai vế ko âm, bp 2 vế ta được:
x2 - x = 3x - 5
\(\Leftrightarrow\) x2 - 4x + 5 = 0
\(\Leftrightarrow\) x2 - 4x + 4 + 1 = 0
\(\Leftrightarrow\) (x - 2)2 + 1 = 0
Vì (x - 2)2 \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)2 + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
Chúc bn học tốt!
Nguyễn Lê Phước Thịnh nhờ anh xíu ạ