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HP
17 tháng 9 2021
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
17 tháng 9 2021
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
7 tháng 8 2017
a)\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\)
\(\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\dfrac{x^2+48-49}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{x^2+35-36}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\dfrac{x^2-1}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{x^2-1}{\sqrt{x^2+35}+6}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\left(\sqrt{x-1}+1\right)^3+2\sqrt{x-1}=2-x\)
\(pt\Leftrightarrow\left(\sqrt{x-1}+1\right)^3-1+2\sqrt{x-1}=1-x\)
\(\Leftrightarrow\left(\sqrt{x-1}+1-1\right)\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}-\left(1-x\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+1\right)+2\sqrt{x-1}+x-1=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}\right)=0\)
Dễ thấy: \(\left(\sqrt{x-1}+1\right)^6+\left(\sqrt{x-1}+1\right)^3+3+\sqrt{x-1}>0\)
\(\Rightarrow\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
12 tháng 1 2019
\(pt\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x^2+48}-7\right)\left(\sqrt{x^2+48}+7\right)}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\dfrac{\left(\sqrt{x^2+35}-6\right)\left(\sqrt{x^2+35}+6\right)}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+48}+7}-4\left(x-1\right)-\dfrac{\left(x+1\right)\left(x-1\right)}{\sqrt{x^2+35}+6}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\right)=0\)
Do : \(\dfrac{x+1}{\sqrt{x^2+48}+7}-4-\dfrac{x+1}{\sqrt{x^2+35}+6}\ne0\)
\(\Rightarrow x=1\)
17 tháng 5 2019
VÌ \(\sqrt{x^2+48}-\sqrt{x^2+35}>0\)
=> \(x>\frac{3}{4}\)
Phương trình tương đương
\((x+6-\sqrt{x^2+48})+3\left(x-1\right)+\left(\sqrt{x^2+35}-6\right)=0\)
=> \(\frac{12\left(x-1\right)}{x+6+\sqrt{x^2+48}}+3\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}=0\)
\(\hept{\begin{cases}x=1\\\frac{12}{x+6+\sqrt{x^2+48}}+3+\frac{x+1}{\sqrt{x^2+35}+6}=0\left(2\right)\end{cases}}\)
Phương trình (2) vô nghiệm do x>3/4=> VT>0
29 tháng 7 2021
1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))
\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)
\(\Leftrightarrow\left(x^2-4\right)^2=x^2-4\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)
Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)
2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)
\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)
Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)
\(\Rightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy pt có no x=2
NK
6
15 tháng 9 2021
\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)
HP
15 tháng 9 2021
a, ĐK: \(x\in R\)
\(\sqrt{4x^2-4x+1}=3-x\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\)
\(\Leftrightarrow\left|2x-1\right|=3-x\)
TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)
TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)
\(\sqrt{x^2+48}=4x-3+\sqrt{x^2+35}\Leftrightarrow\sqrt{x^2+48}-7=4x-4+\sqrt{x^2+35}-6\)
\(\Leftrightarrow\frac{x^2+48-49}{\sqrt{x^2+48}+7}=4x-4+\frac{x^2+35-36}{\sqrt{x^2+35}+6}\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+48}+7}=4\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+35}+6}\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+48}+7}-4-\frac{x+1}{\sqrt{x^2+35}+6}\right)=0\)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\).