[toán 10] giải phương trình

[snowangel1103]

\[\sqrt{7x^2+25x+19}=7\sqrt{x+2}+\sqrt{x^2-2x-35}\]

\[\sqrt{x^2+x-6x}+3\sqrt{x-1}-\sqrt{3x^2-6x+19}=0\]

 

[ngonduoctrogdem]

\[\sqrt{7x^2+25x+19}=7\sqrt{x+2}+\sqrt{x^2-2x-35}\]

ĐK:\[\left\{\begin{matrix} x\geq -2 & \\ x^{2}-2x-35\geq 0 & \\ 7x^{2}-25x+19\geq 0 & \end{matrix}\right.\]
PT \[\Leftrightarrow 7x^{2}-25x+19=x^{2}-2x-35+49x+98+14\sqrt{(x+2)(x+5)(x-7)}\]
\[\Leftrightarrow 6x^{2}-22x-44=14\sqrt{(x+2)(x-7)(x+5)}\]
\[\Leftrightarrow 6(x^{2}-5x-14)+8(x+5)=14\sqrt{(x^{2}-5x-14)(x+5)}\]
Đặt \[a=\sqrt{x^2-5x-14},b=\sqrt{x+5}\]
\[=>6a^2-14ab+8b^2=0=>(a-b)(6a-8b)=0=>...\]