Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ge\frac{1}{2}\)
Đặt \(t=\sqrt{2x-1}\Leftrightarrow x=\frac{t^2+1}{2}\)(ĐK: \(t\ge0\)) thay vao phương trình ta được:
\(\sqrt{\frac{t^2+1}{2}+4+3t}\)+\(\sqrt{\frac{t^2+1}{2}+12-5t}=7\sqrt{2}\)
\(\Leftrightarrow\sqrt{\frac{t^2+6t+9}{2}}+\sqrt{\frac{t^2-10t+25}{2}}=7\sqrt{2}\)
\(\Leftrightarrow\frac{\sqrt{\left(t+3\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(t-5\right)^2}}{\sqrt{2}}=7\sqrt{2}\)
\(\Leftrightarrow\frac{\left|t+3\right|+\left|t-5\right|}{\sqrt{2}}=7\sqrt{2}\)
\(\Leftrightarrow t+3+\left|t-5\right|=14\)(vì \(t\ge0\Rightarrow t+3>0\))
\(\Leftrightarrow t+\left|t-5\right|=11\)
Xét TH: \(t-5\ge0\Leftrightarrow t\ge5\) thì ta có:
\(t+t-5=11\)
\(\Leftrightarrow2t=16\)
\(\Leftrightarrow t=8\)(chọn)
Xét TH: \(t-5< 0\Leftrightarrow t< 5\) thì ta có:
\(t-t+5=11\)
\(\Leftrightarrow5=11\)(vô lí nên loại)
Lại có: \(t=8\)
\(\Leftrightarrow\sqrt{2x-1}=8\)
\(\Leftrightarrow2x-1=64\)
\(\Leftrightarrow2x=63\)
\(\Leftrightarrow x=\frac{63}{2}=31\frac{1}{2}\)
Vậy nghiệm của phương trình là x=31\(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(\left\{{}\begin{matrix}x+9\ge0\\2x+4\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-9\\2x\ge-4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-9\\x\ge-2\end{matrix}\right.\)
<=> \(x\ge-2\)
Ta có : \(\sqrt{x+9}=5-\sqrt{2x+4}\)
<=> \(\sqrt{x+9}+\sqrt{2x+4}=5\)
<=> \(\left(\sqrt{x+9}+\sqrt{2x+4}\right)^2=5^2\)
<=> \(\left(x+9\right)+2\sqrt{\left(x+9\right)\left(2x+4\right)}+\left(2x+4\right)=25\)
ĐKXĐ : \(x\le4\)
=> \(-2\le x\le4\)
<=> \(x+9+2\sqrt{\left(x+9\right)\left(2x+4\right)}+2x+4=25\)
<=> \(2\sqrt{\left(x+9\right)\left(2x+4\right)}=25-x-9-2x-4\)
<=> \(2\sqrt{\left(x+9\right)\left(2x+4\right)}=12-3x\)
<=> \(\left(2\sqrt{\left(x+9\right)\left(2x+4\right)}\right)^2=\left(12-3x\right)^2\)
<=> \(4\left(x+9\right)\left(2x+4\right)=\left(12-3x\right)^2\)
<=> \(4\left(2x^2+18x+4x+36\right)=144-72x+9x^2\)
<=> \(8x^2+72x+16x+144=144-72x+9x^2\)
<=> \(8x^2+72x+16x+144-9x^2-144+72x=0\)
<=> \(-x^2+160x=0\)
<=> \(x\left(160-x\right)=0\)
<=> \(\left\{{}\begin{matrix}x=0\\160-x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\left(TM\right)\\x=160\left(L\right)\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là x = 0 .
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(\left\{{}\begin{matrix}2x+9\ge0\\4-x\ge0\\3x+1\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}2x\ge-9\\-x\ge-4\\3x\ge-1\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x\ge-\frac{9}{2}\\x\le4\\x\ge-\frac{1}{3}\end{matrix}\right.\)
<=> \(4\ge x\ge-\frac{1}{3}\)
Ta có : \(\sqrt{2x+9}=\sqrt{4-x}+\sqrt{3x+1}\)
<=> \(\left(\sqrt{2x+9}\right)^2=\left(\sqrt{4-x}+\sqrt{3x+1}\right)^2\)
<=> \(2x+9=\left(4-x\right)+2\sqrt{\left(4-x\right)\left(3x+1\right)}+\left(3x+1\right)\)
<=> \(2x+9=4-x+2\sqrt{12x-3x^2+4-x}+3x+1\)
<=> \(2x+9-4+x-3x-1=2\sqrt{12x-3x^2+4-x}\)
<=> \(4=2\sqrt{12x-3x^2+4-x}\)
<=> \(4^2=\left(2\sqrt{12x-3x^2+4-x}\right)^2\)
<=> \(16=4\left(12x-3x^2+4-x\right)\)
<=> \(4=12x-3x^2+4-x\)
<=> \(0=12x-3x^2-x\)
<=> \(0=11x-3x^2\)
<=> \(0=x\left(11-3x\right)\)
<=> \(\left\{{}\begin{matrix}x=0\\11-3x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\-3x=-11\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\x=\frac{11}{3}\end{matrix}\right.\) ( TM )
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : x> -2
\(\sqrt{2x+\sqrt{6x^2+1}}\) = x + 1
=> (\(\sqrt{2x+\sqrt{6x^2+1}}\))2 = (x+1)2
=> 2x+\(\sqrt{6x^2+1}\) = x2+2x+1
=> \(\sqrt{6x^2+1}\) = x2+1
=> 6x2 +1 = (x2+1)(x2+1)
=> 6x2 +1 = x4+2x2+1
=> -x4+4x2 = 0
=> x2(4-x2) = 0
=>x2(2-x)(2+x) = 0
=> x2 =0, 2-x=0 , 2+x =0
=> x=0(TMĐKXĐ)
x=2(TMĐKXĐ)
x= -2 (KTMĐKXĐ)
Vậy ........
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left\{{}\begin{matrix}x+1\ge0\\2x+3\ge0\\x+20\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-\frac{3}{2}\\x\ge-20\end{matrix}\right.\)
\(\sqrt{x+1}+\sqrt{2x+3}=\sqrt{x+20}\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{2x+3}\right)^2=\left(\sqrt{x+20}\right)^2\)
\(\Leftrightarrow x+1+2\sqrt{\left(x+1\right)\left(2x+3\right)}+2x+3=x+20\)
\(\Leftrightarrow3x+4+2\sqrt{\left(x+1\right)\left(2x+3\right)}=x+20\)
\(\Leftrightarrow2\sqrt{\left(x+1\right)\left(2x+3\right)}=-2x+16\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=16-2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}16-2x\ge0\\4\left(2x^2+5x+3\right)=\left(16-2x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\8x^2+20x+12=256-64x+4x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\4x^2+84x-244=0\end{matrix}\right.\)
còn lại bn tự làm nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{2x+1}=x-3\)
→ \(\left(\sqrt{2x+1}\right)^2=\left(x-3\right)^2\)
→ \(2x+1=x^2-6x+9\)
→ \(2x+1-x^2+6x-9=0\)
→ \(-x^2+8x-8=0\rightarrow x^2-8x+8=0\)
→ \(x_1=4+2\sqrt{2}\)
\(x_2=4-2\sqrt{2}\)
ĐK: \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)
\(pt\Leftrightarrow2x+1=\left(x-3\right)^2\\ \Leftrightarrow2x+1=x^2-6x+9\\ \Leftrightarrow x^2-8x+8=0\\ \Leftrightarrow x^2-2.x.4+4^2-4^2+8=0\\ \Leftrightarrow\left(x-4\right)^2-8=0\\ \Leftrightarrow\left(x-4-2\sqrt{2}\right)\left(x-4+2\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4-2\sqrt{2}=0\\x-4+2\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4+2\sqrt{2}\\x=4-2\sqrt{2}\end{matrix}\right.\)
Vậy...............................
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(\left\{{}\begin{matrix}x+4\ge0\\1-x\ge0\\1-2x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-4\\x\le1\\x\le0,5\end{matrix}\right.\)
=> \(-4\le x\le0,5\)
Ta có : \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
<=> \(\left(\sqrt{x+4}-\sqrt{1-x}\right)^2=\left(\sqrt{1-2x}\right)^2\)
<=> \(\left(x+4\right)-2\sqrt{\left(x+4\right)\left(1-x\right)}+\left(1-x\right)=1-2x\)
<=> \(x+4-2\sqrt{\left(x+4\right)\left(1-x\right)}+1-x=1-2x\)
<=> \(-2\sqrt{\left(x+4\right)\left(1-x\right)}=1-2x-4-x-1+x\)
<=> \(-2\sqrt{\left(x+4\right)\left(1-x\right)}=-2x-4\)
<=> \(\sqrt{\left(x+4\right)\left(1-x\right)}=x+2\)
ĐKXĐ : \(x+2\ge0\)
\(x\ge-2\)
=> ĐKXĐ là : \(-2\le x\le0,5\)
<=> \(\left(x+4\right)\left(1-x\right)=\left(x+2\right)^2\)
<=> \(x+4-x^2-4x=x^2+4x+4\)
<=> \(x+4-x^2-4x-x^2-4x-4=0\)
<=> \(-7x-2x^2=0\)
<=> \(x\left(7+2x\right)=0\)
<=> \(\left\{{}\begin{matrix}x=0\\7+2x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\left(TM\right)\\x=-\frac{7}{2}\left(L\right)\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là x = 0 .
\(ĐK:\left\{{}\begin{matrix}x+4\ge0\\1-x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow-4\le x\le\frac{1}{2}\)
Phương trình đc viết dưới dạng:
\(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\Leftrightarrow\sqrt{\left(x+4\right)\left(1-x\right)}=2+x\\ \Leftrightarrow2+x\ge0\\ \left(x+4\right)\left(1-x\right)=\left(2+x\right)^2\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\2x^2+5x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x=0\\x=-\frac{5}{2}\end{matrix}\right.\Leftrightarrow x=0\)
Vậy phương trình có nghiệm \(x=0\)
\(2-\sqrt{x^2+2x+9}=2x+3\)
\(\Rightarrow\sqrt{x^2+2x+9}=-\left(2x+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}-\left(2x+1\right)\ge0\\x^2+2x+9=\left[-\left(2x+1\right)\right]^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-\frac{1}{2}\\x^2+2x+9=4x^2+4x+1\end{matrix}\right.\)
\(\Rightarrow4x^2+4x+1-x^2-2x-9=0\)
\(\Rightarrow3x^2+2x-8=0\)
\(\Rightarrow3x^2+6x-4x-8=0\)
\(\Rightarrow3x\left(x+2\right)-4\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\left(KTMĐK\right)\\x=\frac{4}{3}\left(TMĐK\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình là 4/3