ĐKXĐ : \(1\le x\le11\)
Ta có \(\sqrt{11-x}=\sqrt{3x+10}-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{11-x}+\sqrt{x-1}=\sqrt{3x+10}\)
\(\Leftrightarrow\left(\sqrt{11-x}+\sqrt{x-1}\right)^2=3x+10\)
\(\Leftrightarrow2\sqrt{\left(11-x\right).\left(x-1\right)}=3x\)
\(\Leftrightarrow4\left(11-x\right)\left(x-1\right)=9x^2\)
\(\Leftrightarrow13x^2-48x+44=0\)
\(\Leftrightarrow\left(x-2\right).\left(13x-22\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{13}\end{matrix}\right.\left(tm\right)\)
Tập nghiêm S = \(\left\{2;\dfrac{22}{13}\right\}\)
ĐKXĐ :1≤x≤11
ta có \(\sqrt{11-x}=\sqrt{3x+10}-\sqrt{x-1}\)
<=> \(\sqrt{11-x}+\sqrt{x-1}=\sqrt{3x+10}\)
<=>\((\sqrt{11-x}+\sqrt{x-1})^2\)=3x+10
<=>\(2\sqrt{\left(11-x\right).\left(x-1\right)}=3x\)
<=> 4(11−x)(x−1)=9\(x^2\)
<=>13\(x^2\)−48x+44=0
<=>(x−2).(13x−22)=0
\(< =>\left\{{}\begin{matrix}x=2\\x=\dfrac{22}{13}\end{matrix}\right.
\)(tm)
Tập nghiêm S = \(\left\{2;\dfrac{22}{13}\right\}\)
biến đổi pt ta có: √11-x=√3x+10-√x-1
⇔√11-x+√x-1=√3x+10
⇔√11-x+√x-1^2
⇔11-x+2.√(11-x)(√x-1)+x-1=3x+10
⇔2.√-11+12x-x^2=3x
⇔3x≥0 ⇔x≥0
4-11+12x-x^2=9x^2 13x^2-48x+44=0
⇔x=2
x=22/13
thử lại ta thấy pt đã cho có 2 nghiệm là :x=2
x=22/13
