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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
\(ĐK:x\le-3;x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(VT=\sqrt{3x^2-6x+19}+\sqrt{x^2-2x+26}\)
\(=\sqrt{3\left(x-1\right)^2+16}+\sqrt{\left(x-1\right)^2+25}\ge4+5=9\)
\(VP=8-x^2+2x=9-\left(x-1\right)^2\le9\)
Dấu = xảy ra khi \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK: \(x\ge\dfrac{5}{3}\)
Ta có: \(\sqrt{2x+5}=2+\sqrt{3x-5}\)
\(\Leftrightarrow2x+5=4+3x-5+4\sqrt{3x-5}\)
\(\Leftrightarrow6-x=4\sqrt{3x-5}\) ĐK: x≤6
\(\Leftrightarrow36-12x+x^2=48x-80\)
\(\Leftrightarrow x^2-60x+116=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-58\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=58\end{matrix}\right.\)
So với điều kiện thì phương trình có nghiệm duy nhất là x = 2
\(ĐK:x\ge\dfrac{5}{3}\\ PT\Leftrightarrow\left(\sqrt{2x+5}-3\right)-\left(\sqrt{3x-5}-1\right)=0\\ \Leftrightarrow\dfrac{2x-4}{\sqrt{2x+5}+3}-\dfrac{3x-6}{\sqrt{3x-5}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{2}{\sqrt{2x+5}+3}-\dfrac{3}{\sqrt{3x-5}+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{2}{\sqrt{2x+5}+3}=\dfrac{3}{\sqrt{3x-5}+1}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{3x-5}+2=3\sqrt{2x+5}+9\\ \Leftrightarrow2\sqrt{3x-5}=7+3\sqrt{2x+5}\\ \Leftrightarrow4\left(3x-5\right)=49+9\left(2x+5\right)+42\sqrt{2x+5}\\ \Leftrightarrow12x-20=49+18x+45+42\sqrt{2x+5}\\ \Leftrightarrow-6x-144=42\sqrt{2x+5}\)
Vì \(x\ge\dfrac{5}{3}>0\Leftrightarrow-6x-144< 0< 42\sqrt{2x+5}\)
Do đó (1) vô nghiệm
Vậy PT có nghiệm \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK \(x\ge-3\)
PT <=> \(x^3+5x^2+6x+2=4\sqrt{x+3}+2\sqrt{2x+7}\)
<=> \(2\left(x+3-2\sqrt{x+3}\right)+\left(x+5-2\sqrt{2x+7}\right)+x^3+5x^2+3x-9=0\)
+ Với x=-3 =>thỏa mãn
+Với \(x>-3\) ta liên hợp
\(2.\frac{x^2+2x-3}{x+3+2\sqrt{x+3}}+\frac{x^2+2x-3}{x+5+2\sqrt{2x+7}}+\left(x+3\right)\left(x^2+2x-3\right)=0\)
<=> \(\left(x^2+2x-3\right)\left(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3\right)=0\)
Do \(x>-3\)=> \(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3>0\)
=> \(x=1\)(TMĐKXĐ)
Vậy \(x=1;x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) chắc là nhóm lại thui để sau mk làm:v
b)\(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
Đk: tự lm nhé :v
\(pt\Leftrightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}-\left(\sqrt{2x-1}-\sqrt{3}\right)=2x^2-8\)
\(\Leftrightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2x-1-3}{\sqrt{2x-1}+\sqrt{3}}=2\left(x^2-4\right)\)
\(\Leftrightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}=2\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow\frac{\frac{-2\left(x-2\right)}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}-2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)\right)=0\)
Dễ thấy: \(\frac{\frac{-2}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}-\frac{2}{\sqrt{2x-1}+\sqrt{3}}-2\left(x+2\right)< 0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge2\)
\(\dfrac{\left(\sqrt{3x-5}-\sqrt{x-2}\right)\left(\sqrt{3x-5}+\sqrt{x-2}\right)}{\sqrt{3x-5}+\sqrt{x-2}}=\dfrac{2x-3}{3}\)
\(\Leftrightarrow\dfrac{2x-3}{\sqrt{3x-5}+\sqrt{x-2}}=\dfrac{2x-3}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\Rightarrow x=\dfrac{3}{2}\left(ktm\right)\\\sqrt{3x-5}+\sqrt{x-2}=3\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow\sqrt{3x-5}-2+\sqrt{x-2}-1=0\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\sqrt{3x-5}+2}+\dfrac{x-3}{\sqrt{x-2}+1}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\dfrac{3}{\sqrt{3x-5}+2}+\dfrac{1}{\sqrt{x-2}+1}\right)=0\)
\(\Leftrightarrow x-3=0\) (do \(\dfrac{3}{\sqrt{3x-5}+2}+\dfrac{1}{\sqrt{x-2}+1}>0;\forall x\ge2\))
\(\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(x+1\right)^2=\sqrt{2\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow16\left(x+1\right)^4=2\left(x^4+x^2+1\right)\)
\(\Leftrightarrow\left(x^2+3x+1\right)\left(7x^2+11x+7\right)=0\)
\(\sqrt{\frac{x+56}{16}+\sqrt{x-8}}=\frac{x}{8}\)
\(\Leftrightarrow2\sqrt{x+56+16\sqrt{x-8}}=x\)
\(\Leftrightarrow2\sqrt{\left(\sqrt{x-8}+8\right)^2}=x\)
\(\Leftrightarrow2\sqrt{x-8}+16=x\)
\(\Leftrightarrow x=24\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
ĐKXĐ: x>4
Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)
\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)
\(\Leftrightarrow8x+6x=8-15\)
\(\Leftrightarrow14x=-7\)
hay \(x=-\dfrac{1}{2}\)(loại)
2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)
\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
ĐK: \(x\ge8\)
PT \(\Leftrightarrow\sqrt{x-8}-\sqrt{2x-3}=0\)
\(\Leftrightarrow3x-11=2\sqrt{2x^2-19x+24}\)
\(\Leftrightarrow9x^2-66x+121=4x^2-76x+96\)
\(\Leftrightarrow5x^2+10x+25=0\)
\(\Leftrightarrow x\in\varnothing\)
Vậy phương trình vô nghiệm