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Câu 1:
a/ \(x\ge-11\)
Đặt \(\sqrt{x+11}=a\ge0\Rightarrow11=a^2-x\), pt đã cho trở thành:
\(x^2+a=a^2-x\Leftrightarrow x^2-a^2+x+a=0\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)
TH1: \(x+a=0\Leftrightarrow x+\sqrt{x+11}=0\Leftrightarrow-x=\sqrt{x+11}\)
\(\Leftrightarrow\left[{}\begin{matrix}-x\ge0\\x^2=x+11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2-x-11=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1-3\sqrt{5}}{2}\)
TH2: \(x-a+1=0\Leftrightarrow x+1=\sqrt{x+11}\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\\left(x+1\right)^2=x+11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-10=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{41}}{2}\)
b/ \(\sqrt{9+x}=x-9\Leftrightarrow\left\{{}\begin{matrix}x-9\ge0\\9+x=\left(x-9\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge9\\x^2-19x+72=0\end{matrix}\right.\) \(\Rightarrow x=\frac{19+\sqrt{73}}{2}\)
Câu 2:
a/
\(f\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-1\right)\left(x-4\right)}=\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x-4\right)}\)
Lập bảng xét dấu ta được:
\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -1\\x>4\\1< x< 3\end{matrix}\right.\)
\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-1< x< 1\\3< x< 4\end{matrix}\right.\)
\(f\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
\(f\left(x\right)\) ko xác định tại \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b/ \(h\left(x\right)=\frac{-x^2+3x-1}{\left(x^2-2x+3\right)\left(x+2\right)}\)
Lập bảng xét dấu ta được:
\(f\left(x\right)>0\) khi \(\left[{}\begin{matrix}x< -2\\\frac{3-\sqrt{5}}{2}< x< \frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)< 0\) khi \(\left[{}\begin{matrix}-2< x< \frac{3-\sqrt{5}}{2}\\x>\frac{3+\sqrt{5}}{2}\end{matrix}\right.\)
\(f\left(x\right)=0\) tại \(x=\frac{3\pm\sqrt{5}}{2}\)
\(f\left(x\right)\) ko xác định tại \(x=-2\)
Đệ biết là có người làm câu c,d nên xin xí câu e :3
ĐK: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)
\(PT\Leftrightarrow5+\sqrt{x+1}=7\left(x-2\right)\)
\(\Leftrightarrow\sqrt{x+1}=7x-19\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\x+1=49x^2-266x+361\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\49x^2-267x+360=0\end{matrix}\right.\)
\(\Rightarrow x=3\left(tm\right)\)
a/ \(\Leftrightarrow\left\{{}\begin{matrix}9-2x\ge0\\x^2-4x-12=\left(9-2x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{9}{2}\\3x^2-32x+93=0\end{matrix}\right.\)
Phương trình vô nghiệm
b/ \(\Leftrightarrow\left(x+1\right)\sqrt[3]{15x^2-x-1}-\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\sqrt[3]{15x^2-x-1}-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\\sqrt[3]{15x^2-x-1}-x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt[3]{15x^2-x-1}=x-1\)
\(\Leftrightarrow15x^2-x-1=x^3-3x^2+3x-1\)
\(\Leftrightarrow x^3-18x^2+4x=0\)
\(\Leftrightarrow x\left(x^2-18x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=9\pm\sqrt{77}\\\end{matrix}\right.\)