\(\sqrt{X^2}\) =7
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![](https://rs.olm.vn/images/avt/0.png?1311)
c: Ta có: \(\sqrt{x^2}=x\)
\(\Leftrightarrow\left|x\right|=x\)
hay \(x\ge0\)
d: Ta có: \(\sqrt{x^2-2x+1}=x-1\)
\(\Leftrightarrow\left|x-1\right|=x-1\)
hay \(x\ge1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,ĐK:x\ge-7\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)
Đạt \(\sqrt{x+7}=a\ge0\)
\(PT\Leftrightarrow\sqrt{\left(a+1\right)^2}+\sqrt{a^2-a-6}=4\\ \Leftrightarrow a+1+\sqrt{a^2-a-6}=4\\ \Leftrightarrow\sqrt{a^2-a-6}=3-a\\ \Leftrightarrow a^2-a-6=a^2-6a+9\\ \Leftrightarrow5a=15\Leftrightarrow a=3\\ \Leftrightarrow\sqrt{x+7}=3\\ \Leftrightarrow x+7=9\\ \Leftrightarrow x=2\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(7-x+2\sqrt{x}=\left(2+\sqrt{x}\right)\sqrt{7-x}\)
\(\Leftrightarrow\sqrt{\left(7-x\right)^2}+2\sqrt{x}=2\sqrt{7-x}+\sqrt{x}\cdot\sqrt{7-x}\)
=>\(\sqrt{7-x}\cdot\left(\sqrt{7-x}-\sqrt{x}\right)=2\left(\sqrt{7-x}-\sqrt{x}\right)\)
=>7-x=4
=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)ĐK:\(x\ge-7\)
Đặt \(t=\sqrt{x+7}\left(t\ge0\right)\)
\(\Rightarrow t+1-4=\sqrt{t^2-t-6}\)
\(\Leftrightarrow t^2-6t+9=t^2-t-6\left(t\ge3\right)\)
\(\Leftrightarrow5t=15\)
\(\Leftrightarrow t=3\left(TM\right)\)\(\Rightarrow x=2\left(tm\right)\)
S={2}
b)ĐK:\(x\ge2\)
pt\(\Leftrightarrow\sqrt{x-2+2\sqrt{x-2}+2}-\sqrt{x-2-2\sqrt{x-2}+2}=-2\)
Đặt t= can(x-2)(t>=0)
Đến đây bạn giải tiếp nhé!
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