Lời giải:
ĐKXĐ: \(x\geq \frac{-1}{16}\)
PT \(\Leftrightarrow x^2-x-2\sqrt{16x+1}-2=0\)
\(\Leftrightarrow (x^2-x-20)-2(\sqrt{16x+1}-9)=0\)
\(\Leftrightarrow (x-5)(x+4)-2.\frac{16x+1-81}{\sqrt{16x+1}+9}=0\)
\(\Leftrightarrow (x-5)(x+4)-\frac{32(x-5)}{\sqrt{16x+1}+9}=0\)
\(\Leftrightarrow (x-5)\left[x+4-\frac{32}{\sqrt{16x+1}+9}\right]=0(1)\)
Ta thấy:
Với mọi \(x\geq \frac{-1}{16}\Rightarrow \left\{\begin{matrix} x+4\geq \frac{63}{16}>3,6\\ \frac{32}{\sqrt{16x+1}+9}\leq \frac{32}{9}<3,6\end{matrix}\right.\)
\(\Rightarrow x+4>\frac{32}{\sqrt{16x+1}+9}\Rightarrow x+4-\frac{32}{\sqrt{16x+1}+9}>0(2)\)
Từ \((1);(2)\Rightarrow x-5=0\Rightarrow x=5\) là nghiệm duy nhất.
Lời giải:
ĐKXĐ: \(x\geq \frac{-1}{16}\)
PT \(\Leftrightarrow x^2-x-2\sqrt{16x+1}-2=0\)
\(\Leftrightarrow (x^2-x-20)-2(\sqrt{16x+1}-9)=0\)
\(\Leftrightarrow (x-5)(x+4)-2.\frac{16x+1-81}{\sqrt{16x+1}+9}=0\)
\(\Leftrightarrow (x-5)(x+4)-\frac{32(x-5)}{\sqrt{16x+1}+9}=0\)
\(\Leftrightarrow (x-5)\left[x+4-\frac{32}{\sqrt{16x+1}+9}\right]=0(1)\)
Ta thấy:
Với mọi \(x\geq \frac{-1}{16}\Rightarrow \left\{\begin{matrix} x+4\geq \frac{63}{16}>3,6\\ \frac{32}{\sqrt{16x+1}+9}\leq \frac{32}{9}<3,6\end{matrix}\right.\)
\(\Rightarrow x+4>\frac{32}{\sqrt{16x+1}+9}\Rightarrow x+4-\frac{32}{\sqrt{16x+1}+9}>0(2)\)
Từ \((1);(2)\Rightarrow x-5=0\Rightarrow x=5\) là nghiệm duy nhất.
Truy ngược dấu liên hợp:)
ĐK: \(x\ge-\frac{1}{16}\)
\(PT\Leftrightarrow x^2-\frac{41}{9}x-\frac{20}{9}+\frac{2\sqrt{1+16x}}{9}\left(\sqrt{1+16x}-9\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+\frac{4}{9}\right)+\frac{2\sqrt{1+16x}}{9}\left(\frac{16\left(x-5\right)}{\sqrt{1+16x}+9}\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[x+\frac{4}{9}+\frac{32\sqrt{1+16x}}{9\left(\sqrt{1+16x}+9\right)}\right]=0\)
Cái ngoặc to hiển nhiên vô nghiệm.
Vậy x = 5
Is that true?