\(2x+3+\sqrt{4x^2+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}\left(x\ge-\frac{1}{4}\right)\)
\(\Leftrightarrow2\left(x+2\right)-1+\sqrt{\left(x+2\right)\left(4x+1\right)}=2\sqrt{x+2}+\sqrt{4x+1}\)
\(\Leftrightarrow4\left(x+2\right)-2+2\sqrt{x+2}.\sqrt{4x+1}=4\sqrt{x+2}+2\sqrt{4x+1}\)
Đặt \(\hept{\begin{cases}2\sqrt{x+2}=a\left(a\ge0\right)\\\sqrt{4x+1}=b\left(b\ge0\right)\end{cases}\Rightarrow}a^2-b^2=4\left(x+2\right)-4x-1=7\)\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=7\)(1)
\(pt:a^2-2+ab=2a+2b\)
\(\Leftrightarrow a\left(a+b\right)-2\left(a+b\right)=2\)
\(\Leftrightarrow\left(a-2\right)\left(a+b\right)=2\)(2)
Nhân chéo 2 vế của (1) với (2) được
\(7\left(a-2\right)\left(a+b\right)=2\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow7\left(a-2\right)=2\left(a-b\right)\left(Do\left(a+b\right)>0\right)\)
\(\Leftrightarrow7a-14=2a-2b\)
\(\Leftrightarrow5a=14-2b\)
\(\Leftrightarrow10\sqrt{x+2}=14-2\sqrt{4x+1}\)
\(\Leftrightarrow5\sqrt{x+2}=7-\sqrt{4x+1}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{4x+1}\le7\\25\left(x+2\right)=49-14\sqrt{4x+1}+4x+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le4x+1\le49\\21x=-14\sqrt{4x+1}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\441x^2=196\left(4x+1\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\441x^2-784x-196=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\49\left(9x+2\right)\left(x-2\right)=0\end{cases}}\)
\(\Leftrightarrow x=-\frac{2}{9}\left(TmĐKXĐ\right)\)
Vậy
Incursion_03 em thử nha, sai thì thôi ạ, em hơi nghiện liên hợp r.
ĐK: x>=-1/4
PT \(\Leftrightarrow2x+\frac{31}{9}+\sqrt{4x^2+9x+2}-\frac{4}{9}=2\sqrt{x+2}-\frac{8}{3}+\sqrt{4x+1}-\frac{1}{3}+3\)
\(\Leftrightarrow2\left(x+\frac{2}{9}\right)+\frac{\left(x+\frac{2}{9}\right)\left(4x+\frac{73}{9}\right)}{\sqrt{4x^2+9x+2}+\frac{4}{9}}=\frac{4\left(x+\frac{2}{9}\right)}{2\sqrt{x+2}+\frac{8}{3}}+\frac{4\left(x+\frac{2}{9}\right)}{\sqrt{4x+1}+\frac{1}{3}}\)
\(\Leftrightarrow\left(x+\frac{2}{9}\right)\left[2+\frac{4x+\frac{73}{9}}{\sqrt{4x^2+9x+2}+\frac{4}{9}}-4\left(\frac{1}{2\sqrt{x+2}+\frac{8}{3}}+\frac{1}{\sqrt{4x+1}+\frac{1}{3}}\right)\right]=0\)
Cái ngoặc to em chịu:( đang suy nghĩ
ĐK \(x\ge-\frac{1}{4}\)
Cách liên hợp
\(\left(x+\frac{26}{9}-2\sqrt{x+2}\right)+\left(3x+1-\sqrt{4x+1}\right)+\left(\sqrt{4x^2+9x+2}-2x-\frac{8}{9}\right)=0\)
<=>\(\frac{x^2+\frac{16}{9}x+\frac{28}{81}}{x+\frac{26}{9}+2\sqrt{x+2}}+\frac{9x^2+2x}{3x+1+\sqrt{4x+1}}+\frac{\frac{49}{9}x+\frac{98}{81}}{\sqrt{4x^2+9x+2+2x+\frac{8}{9}}}=0\)
<=>\(\frac{\left(x+\frac{2}{9}\right)\left(x+\frac{14}{9}\right)}{x+\frac{26}{9}+2\sqrt{x+2}}+\frac{9x\left(x+\frac{2}{9}\right)}{3x+1+\sqrt{4x+1}}+\frac{\frac{49}{9}\left(x+\frac{2}{9}\right)}{\sqrt{4x^2+9x+2}+2x+\frac{8}{9}}=0\)
=>\(\orbr{\begin{cases}x=-\frac{2}{9}\left(tmĐK\right)\\\frac{x+\frac{14}{9}}{x+\frac{26}{9}+2\sqrt{x+2}}+\frac{9x}{3x+1+\sqrt{4x+1}}+\frac{\frac{49}{9}}{\sqrt{4x^2+9x+2}+2x+\frac{8}{9}}0\left(2\right)\end{cases}}\)
(2) Quy đồng phân số thứ 2 và thứ 3 ta được \(\frac{18x^2+\frac{73}{3}x+\frac{49}{9}+BT}{MS}>0\)
ta sẽ Cm được \(VT>0\forall x\ge-\frac{1}{4}\)=> PT (2) vô nghiệm
Vậy x=-2/9
