Bài 5:
ĐKXĐ: \(x^2+5x+2>=0\)
=>\(\left[{}\begin{matrix}x>=\dfrac{-5+\sqrt{17}}{2}\\x< =\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)
\(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
=>\(x^2+5x+4-3\sqrt{x^2+5x+2}-6=0\)
=>\(x^2+5x+2-3\sqrt{x^2+5x+2}-4=0\)
=>\(\left(\sqrt{x^2+5x+2}-4\right)\left(\sqrt{x^2+5x+2}+1\right)=0\)
=>\(\sqrt{x^2+5x+2}-4=0\)
=>\(\sqrt{x^2+5x+2}=4\)
=>\(x^2+5x+2=16\)
=>\(x^2+5x-14=0\)
=>\(\left(x+7\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
Bài 1:
a: Khi x=16 thì \(A=\dfrac{\sqrt{16}+2}{\sqrt{16}+1}=\dfrac{4+2}{4+1}=\dfrac{6}{5}\)
b: \(B=\dfrac{x+12}{x-4}+\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{2-\sqrt{x}}\)
\(=\dfrac{x+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)
\(=\dfrac{x+12+\sqrt{x}-2-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(P-2=\dfrac{\sqrt{x}-1-2\sqrt{x}-2}{\sqrt{x}+1}\)
=>\(P-2=\dfrac{-\sqrt{x}-3}{\sqrt{x}+1}< 0\)
=>P<2
d: B<2/3
=>B-2/3<0
=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{2}{3}< 0\)
=>\(\dfrac{3\sqrt{x}-3-2\sqrt{x}-4}{3\left(\sqrt{x}+2\right)}< 0\)
=>\(\dfrac{\sqrt{x}-7}{3\left(\sqrt{x}+2\right)}< 0\)
=>\(\sqrt{x}-7< 0\)
=>\(\sqrt{x}< 7\)
=>0<=x<49
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< 49\\x< >4\end{matrix}\right.\)
