giải các phương trình
1) \(\sqrt{4x-20}\) +3\(\sqrt{\dfrac{x-5}{9}}\) \(-\dfrac{1}{3}\sqrt{9x-45}=6\)
2)\(\sqrt{x+1}+\sqrt{x+6}=5\)
3) \(x^2-6x+\sqrt{x^2-6x+7}=5\)
4)\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=4\)
5)\(\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}}=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)\)
6)\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
7)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
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 .............
3)
ĐK: \(x^2-6x+7\geq 0\)
Đặt \(\sqrt{x^2-6x+7}=a(a\geq 0)\) \(\Rightarrow x^2-6x=a^2-7\)
PT trở thành: \(a^2-7+a=5\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow (a-3)(a+4)=0\Rightarrow a=3\) (do \(a\geq 0)\)
\(\Rightarrow \sqrt{x^2-6x+7}=3\)
\(\Rightarrow x^2-6x+7=9\)
\(\Leftrightarrow x^2-6x-2=0\) \(\Rightarrow x=3\pm \sqrt{11}\) (đều thỏa mãn)
4)
ĐK: \(x\geq 1\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=4\)
\(\Leftrightarrow \sqrt{(x-1)-4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=6\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=4\)
\(\Leftrightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=4(*)\)
TH1: \(\sqrt{x-1}\geq 3\)
\((*)\Leftrightarrow \sqrt{x-1}-2+\sqrt{x-1}-3=4\)
\(\Leftrightarrow 2\sqrt{x-1}=9\Rightarrow \sqrt{x-1}=\frac{9}{2}\Rightarrow x=\frac{85}{4}\) (t/m)
TH2: \(\sqrt{x-1}< 2\)
\((*)\Leftrightarrow 2-\sqrt{x-1}+3-\sqrt{x-1}=4\)
\(\Leftrightarrow 2\sqrt{x-1}=1\Rightarrow \sqrt{x-1}=\frac{1}{2}\Rightarrow x=\frac{5}{4}\) (t/m)
TH3: \(2\leq \sqrt{x-1}< 3\)
\((*)\Leftrightarrow \sqrt{x-1}-2+3-\sqrt{x-1}=4\)
\(\Leftrightarrow 1=4\) (vô lý)
Vậy \(x=\frac{85}{4}\) hoặc \(x=\frac{5}{4}\)
5)
\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}(2x^3+x^2+2x+1)\)
\(\Leftrightarrow \sqrt{x^2-\frac{1}{4}+\sqrt{(x+\frac{1}{2})^2}}=\frac{1}{2}(x^2+1)(2x+1)\)
\(\Leftrightarrow \sqrt{x^2-\frac{1}{4}+|x+\frac{1}{2}|}=\frac{1}{2}(x^2+1)(2x+1)(*)\)
Vì vế trái của pt luôn không âm nên:
\(\frac{1}{2}(x^2+1)(2x+1)\geq 0\Rightarrow 2x+1\geq 0\Rightarrow x\geq \frac{-1}{2}\)
\(\Rightarrow |x+\frac{1}{2}|=x+\frac{1}{2}\)
Do đó:
\((*)\Leftrightarrow \sqrt{x^2-\frac{1}{4}+x+\frac{1}{2}}=\frac{1}{2}(x^2+1)(2x+1)\)
\(\Leftrightarrow \sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}(x^2+1)(2x+1)\)
\(\Leftrightarrow \sqrt{(x+\frac{1}{2})^2}=\frac{1}{2}(x^2+1)(2x+1)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}(x^2+1)(2x+1)\)
\(\Leftrightarrow 2x+1=(x^2+1)(2x+1)\)
\(\Leftrightarrow x^2(2x+1)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-\frac{1}{2}\end{matrix}\right.\) (đều thỏa mãn)
Vậy.............
6)
\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+30}=8\)
\(\Leftrightarrow \sqrt{3(x^2+2x+1)+9}+\sqrt{5(x^4-x^2+1)+25}=8\)
\(\Leftrightarrow \sqrt{3(x+1)^2+9}+\sqrt{5(x^2-1)^2+25}=8\)
Ta thấy:
\(\sqrt{3(x+1)^2+9}\geq \sqrt{3.0+9}=3\)
\(\sqrt{5(x^2-1)^2+25}\geq \sqrt{5.0+25}=5\)
\(\sqrt{3(x+1)^2+9}+\sqrt{5(x^2-1)^2+25}\geq 8\)
Dấu "=" xảy ra khi \((x+1)^2=(x^2-1)^2=0\Leftrightarrow x=-1\) (t/m)
Vậy..............
7)
Ta có:
\(\sqrt{3x^2+6x+7}=\sqrt{3(x^2+2x+1)+4}=\sqrt{3(x+1)^2+4}\)
\(\geq \sqrt{3.0+4}=2\)
\(\sqrt{5x^2+10x+14}=\sqrt{5(x^2+2x+1)+9}=\sqrt{5(x+1)^2+9}\)
\(\geq \sqrt{5.0+9}=3\)
\(\Rightarrow \sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}\geq 5\)
Mà \(4-2x-x^2=5-(x^2+2x+1)=5-(x+1)^2\leq 5-0=5\)
Do đó:
\(\sqrt{3x^2+6x+7}+\sqrt{5x2+10x+14}=4-2x-x^2=5\)
\(\Rightarrow x=-1\) (t/m)
