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1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
a,dk x>0
\(\Leftrightarrow\)\(\dfrac{\left(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}\right)\left(\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}\right)}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}=3x\)
\(\Leftrightarrow x\left(\dfrac{x+2}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}-3\right)=0\)
\(\Rightarrow\dfrac{x+2}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}=3\)
\(\Rightarrow\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\)
kh vs dé bài ta có hệ \(\left\{{}\begin{matrix}\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\\\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\end{matrix}\right.\)
cộng vs nhau ta có
\(2\sqrt{2x^2+x+1}=3x+\dfrac{x+2}{2}\)
\(\Leftrightarrow3\sqrt{2x^2+x+1}=5x+1\)
giải ra ta có x=1(tm) x=-8/7 (l)
b, dk tu xd nhé 
\(\Leftrightarrow\dfrac{\left(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\right)\left(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\right)}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-2x=0\)
\(\Leftrightarrow2x\left(\dfrac{1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=1\left(l\right)\end{matrix}\right.\)
ns \(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}>1\)
\(\Rightarrow x=0\left(tm\right)\)
a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)
\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)
a') (tiếp)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)
Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)
Với mọi \(x\ge4\), ta có:
\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)
\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)
\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)
Do đó phương trình (1) vô nghiệm.
Vậy phương trình đã cho vô nghiệm.
@Nguyễn Huy Thắng@Mysterious Person@bảo nam trần@Lightning Farron@Thiên Thảo@Sky SơnTùng