Câu 1:
PT \(\Leftrightarrow x^2+3x+8=(x+5)\sqrt{x^2+x+2}\)
\(\Leftrightarrow (x^2+x+2)+2(x+5)-4=(x+5)\sqrt{x^2+x+2}\)
Đặt \(\sqrt{x^2+x+2}=a; x+5=b(a\geq 0)\)
\(PT\Leftrightarrow a^2+2b-4=ba\)
\(\Leftrightarrow (a^2-4)-b(a-2)=0\)
\(\Leftrightarrow (a-2)(a+2-b)=0\Rightarrow \left[\begin{matrix} a=2\\ a+2=b\end{matrix}\right.\)
Nếu \(a=2\Rightarrow x^2+x+2=a^2=4\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow (x-1)(x+2)=0\Rightarrow x=1; x=-2\) (đều thỏa mãn)
Nếu \(a+2=b\Leftrightarrow \sqrt{x^2+x+2}+2=x+5\)
\(\Leftrightarrow \sqrt{x^2+x+2}=x+3\)
\(\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ x^2+x+2=(x+3)^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+3\geq 0\\ 5x+7=0\end{matrix}\right.\Rightarrow x=\frac{-7}{5}\) (thỏa mãn)
Vậy..........
Câu 2:
ĐKXĐ: \(x\geq 1\) hoặc \(x\leq \frac{1}{2}\)
\(10x^2-9x-8x\sqrt{2x^2-3x+1}+3=0\)
\(\Leftrightarrow 3(2x^2-3x+1)-8x\sqrt{2x^2-3x+1}+4x^2=0\)
Đặt \(\sqrt{2x^2-3x+1}=a(a\geq 0)\)
Khi đó PT \(\Leftrightarrow 3a^2-8xa+4x^2=0\)
\(\Leftrightarrow (a-2x)(3a-2x)=0\) \(\Rightarrow \left[\begin{matrix} a=2x\\ 3a=2x\end{matrix}\right.\)
Nếu \(a=\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2-3x+1=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 2x^2+3x-1=0\end{matrix}\right.\Rightarrow x=\frac{-3+\sqrt{17}}{4}\) (t/m)
Nếu \(3a=3\sqrt{2x^2-3x+1}=2x\Rightarrow \left\{\begin{matrix} x\geq 0\\ 9(2x^2-3x+1)=4x^2\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} x\geq 0\\ 14x^2-27x+9=0\end{matrix}\right.\Rightarrow x=\frac{3}{2}; x=\frac{3}{7}\) (t/m)
Vậy...........
3:
ĐK: \(x\geq -\sqrt[3]{3}\)
Đặt \(\sqrt{x^3+3}=a(a\geq 0)\)
PT \(\Leftrightarrow (x^3+3)+6x^2-2x-(5x-1)\sqrt{x^3+3}=0\)
\(\Leftrightarrow a^2+6x^2-2x-(5x-1)a=0\)
\(\Leftrightarrow 6x^2-x(5a+2)+(a^2+a)=0\)
Coi đây là pt bậc 2 ẩn $x$.
Ta thấy \(\Delta=(5a+2)^2-24(a^2+a)=(a-2)^2\)
\(\Rightarrow x=\frac{(5a+2)\pm \sqrt{\Delta}}{12}\Rightarrow x=\frac{a}{2}\) hoặc \(x=\frac{a+1}{3}\)
Nếu \(x=\frac{a}{2}=\frac{\sqrt{x^3+3}}{2}\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2=\frac{x^3+3}{4}\end{matrix}\right.\)
\( \Rightarrow \left\{\begin{matrix} x\geq 0\\ x^3-4x^2+3=0\end{matrix}\right. \Rightarrow \left\{\begin{matrix} x\geq 0\\ (x-1)(x^2-3x-3)=0\end{matrix}\right.\)
\( \Rightarrow \left[\begin{matrix} x=1\\ x=\frac{3+\sqrt{21}}{2}\end{matrix}\right.\) (t.m)
Nếu \(x=\frac{a+1}{3}\Rightarrow 3x-1=a=\sqrt{x^3+3}\)
\( \Rightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ (3x-1)^2=x^3+3\end{matrix}\right. \Rightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ x^3-9x^2+6x+2=0\end{matrix}\right.\)
\( \Rightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ (x-1)(x^2-8x-2)=0\end{matrix}\right.\Rightarrow x=1; x=4+3\sqrt{2}\)
Vậy \(x\in\left\{1; 4+3\sqrt{2}; \frac{3+\sqrt{21}}{2}\right\}\)
Câu 4:
ĐK: \(-1\leq x\leq 1\)
Đặt \(\sqrt{x+1}=a; \sqrt{1-x}=b(a,b\geq 0)\)
\(\Rightarrow 2a^2-b^2=3x+1\)
PT \(\Leftrightarrow 2\sqrt{1-x}+\sqrt{1-x^2}+(3x+1)-4\sqrt{x+1}=0\)
\(\Leftrightarrow 2b+ab+2a^2-b^2-4a=0\)
\(\Leftrightarrow (4a^2-b^2)-(4a-2b)-(2a^2-ab)=0\)
\(\Leftrightarrow (2a-b)(2a+b)-2(2a-b)-a(2a-b)=0\)
\(\Leftrightarrow (2a-b)(a+b-2)=0\)
\(\Rightarrow \left[\begin{matrix} 2a=b\\ a+b=2\end{matrix}\right.\)
Nếu \(2a=b\Rightarrow 4a^2=b^2\Rightarrow 4(x+1)=1-x\Rightarrow x=\frac{-3}{5}\) (t.m)
Nếu \(a+b=2\Rightarrow a^2+b^2+2ab=4\)
\(\Leftrightarrow 2+2ab=4\Rightarrow ab=1\)
Theo đl Vi-et đảo thì $a,b$ là nghiệm của $X^2-2X+1=0$ hay $a=b=1$
\(\Rightarrow x=0\) (t/m)
Vậy..........
5.
ĐK: \(x\geq -3; x\neq 0\)
\(PT\Rightarrow 4x^2+x+2=4x\sqrt{x+3}\)
\(\Leftrightarrow (2x)^2-4x\sqrt{x+3}+(x+3)=1\)
\(\Leftrightarrow (2x-\sqrt{x+3})^2=1\)
\(\Rightarrow 2x-\sqrt{x+3}=\pm 1\)
Nếu \(2x-\sqrt{x+3}=1\Leftrightarrow 2x-1=\sqrt{x+3}\)
\(\Rightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ (2x-1)^2=x+3\end{matrix}\right.\Rightarrow x=\frac{5+\sqrt{57}}{8}\) (chọn)
Nếu \(2x-\sqrt{x+3}=-1\Leftrightarrow 2x+1=\sqrt{x+3}\)
\(\Rightarrow \left\{\begin{matrix} x\geq -\frac{1}{2}\\ (2x+1)^2=x+3\end{matrix}\right.\Rightarrow x=\frac{-3+\sqrt{41}}{8}\) (chọn)
Vậy............
