Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)
\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)
\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)
\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)
\(\Leftrightarrow x^2+2x=\dfrac{3+\sqrt{5}}{2}\left(2x-1\right)\)
\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)
\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)
\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)
b. ĐKXĐ: \(x\ge5\)
\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)
\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)
\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
b, ĐK \(x\ge-4\)
PT
<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)
<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)
Giải (2)
=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)
<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)
<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)
<=> \(x^2-7x-4=6\sqrt{x+4}\)
<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)
Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)
=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)
=> \(a^2-b^2+6\left(a-b\right)=0\)
<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)
+ a=b
=> \(x-6=\sqrt{x+4}\)
=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)
+ a+b+6=0
=> \(x+\sqrt{x+4}=0\)(loại)
Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
a,ĐK:\(x\ge\frac{3}{2}\)
\(PT\Leftrightarrow\left(3x+2\right)\sqrt{2x-3}-\left(3x+2\right)-2x^2+8=0\)
\(\Leftrightarrow\left(3x+2\right)\left(\sqrt{2x-3}-1\right)-2\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(3x+2\right).\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}-2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2\left(x-2\right)\left[\frac{3x+2}{\sqrt{2x-3}+1}-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\frac{3x+2}{\sqrt{2x-3}+1}=x+2\left(1\right)\end{matrix}\right.\)
Giải (1)\(\Leftrightarrow3x+2=\sqrt{2x-3}\left(x+2\right)+x+2\)
\(\Leftrightarrow2x=\sqrt{2x-3}\left(x+2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\4x^2=\left(2x-3\right)\left(x^2+4x+4\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\2x^3+x^2-4x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{3}{2}\\\left(x-2\right)\left(2x^2+5x+6\right)=0\end{matrix}\right.\) \(\Leftrightarrow x=2\left(tm\right)\)
Vậy \(x=2\)
b, Đề là \(5\sqrt{x+1}\) hay \(5\sqrt{x+4}\) vậy?
1. đk: \(x\ge5\)
Ta có: \(PT\Leftrightarrow\sqrt{\left(x+1\right)\left(5x+9\right)}=\sqrt{\left(x+4\right)\left(x-5\right)}+5\sqrt{x+1}\)
\(\Leftrightarrow\left(x+1\right)\left(5x+9\right)=x^2+24x+5+10\sqrt{\left(x+1\right)\left(x+4\right)\left(x-5\right)}\)
\(\Leftrightarrow5x^2+14x+9-x^2-24x-5-10\sqrt{\left[\left(x+1\right)\left(x-5\right)\right]\left(x+4\right)}=0\)
\(\Leftrightarrow4x^2-10x+4-10\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}=0\)
\(\Leftrightarrow\left(2x^2-8x-10\right)+\left(3x+12\right)-5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}=0\)
\(\Leftrightarrow2\left(x^2-4x-5\right)+3\left(x+4\right)-5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}=0\)
Đặt \(\hept{\begin{cases}\sqrt{x^2-4x-5}=a\\\sqrt{x+4}=b\end{cases}}\) khi đó:
\(PT\Leftrightarrow2a^2+3b^2-5ab=0\)
\(\Leftrightarrow\left(2a^2-2ab\right)-\left(3ab-3b^2\right)=0\)
\(\Leftrightarrow2a\left(a-b\right)-3b\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-b=0\\2a-3b=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=b\\2a=3b\end{cases}}\)
Nếu: \(a=b\Leftrightarrow\sqrt{x^2-4x-5}=\sqrt{x+4}\)
\(\Leftrightarrow x^2-4x-5=x+4\)
\(\Leftrightarrow x^2-5x-9=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{61}}{2}\right)\left(x-\frac{5-\sqrt{61}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5+\sqrt{61}}{2}=0\\x-\frac{5-\sqrt{61}}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{61}}{2}\left(tm\right)\\x=\frac{5-\sqrt{61}}{2}\left(ktm\right)\end{cases}}\)
Nếu: \(2a=3b\Leftrightarrow2\sqrt{x^2-4x-5}=3\sqrt{x+4}\)
\(\Leftrightarrow4\left(x^2-4x-5\right)=9\left(x+4\right)\)
\(\Leftrightarrow4x^2-25x-56=0\)
\(\Leftrightarrow\left(x-8\right)\left(4x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\left(tm\right)\\x=-\frac{7}{4}\left(ktm\right)\end{cases}}\)
Vậy \(x\in\left\{\frac{5+\sqrt{61}}{2};8\right\}\)
2. đk: \(x\ge\frac{1}{2}\)
Ta có: \(x^2-2x=2\sqrt{2x-1}\)
\(\Leftrightarrow\left(x-1\right)^2-1=2\sqrt{2x-1}\)
Đặt APKHT như sau: \(a-1=\sqrt{2x-1}\)
Khi đó ta có hệ sau: \(\hept{\begin{cases}x^2-2x=2\left(y-1\right)\\y^2-2y=2\left(x-1\right)\end{cases}}\)
Trừ vế trên cho vế dưới của HPT ta được:
\(x^2-2x-y^2+2y=2\left(y-1\right)-2\left(x-1\right)\)
\(\Leftrightarrow x^2-y^2-2x+2y-2y+2x=0\)
\(\Leftrightarrow x^2-y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=0\)
Nếu \(x-y=0\Leftrightarrow x-1=y-1\Leftrightarrow x-1=\sqrt{2x-1}\)
\(\Leftrightarrow x^2-2x+1=2x-1\)
\(\Leftrightarrow x^2-4x+2=0\)
\(\Leftrightarrow\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{2}\left(tm\right)\\x=2-\sqrt{2}\left(ktm\right)\end{cases}}\)
Nếu \(x+y=0\) mà \(x,y>0\) => vô lý
Vậy \(x=2+\sqrt{2}\)