Câu a giải rồi
b/ ĐKXĐ: \(3x^2+5x+1\ge0\)
\(\Leftrightarrow\sqrt{3x^2+5x+8}=\sqrt{3x^2+5x+1}+1\)
\(\Leftrightarrow3x^2+5x+8=3x^2+5x+2+2\sqrt{3x^2+5x+1}\)
\(\Leftrightarrow\sqrt{3x^2+5x+1}=3\)
\(\Leftrightarrow3x^2+5x-8=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{8}{3}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ge3\)
\(\Leftrightarrow\sqrt{x-1}-2+2\left(\sqrt{x^2-9}-4\right)=0\)
\(\Leftrightarrow\frac{x-5}{\sqrt{x-1}+2}+\frac{2\left(x^2-25\right)}{\sqrt{x^2-9}+4}=0\)
\(\Leftrightarrow\frac{x-5}{\sqrt{x-1}+2}+\frac{2\left(x-5\right)\left(x+5\right)}{\sqrt{x^2-9}+4}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{x-1}+2}+\frac{2\left(x+5\right)}{\sqrt{x^2-9}+4}\right)=0\)
\(\Leftrightarrow x-5=0\) (ngoặc phía sau luôn dương \(\forall x\ge3\))
\(\Rightarrow x=5\)
Câu c:
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+34}=a\\\sqrt[3]{x-3}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^3-b^3=37\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\\left(a-b\right)\left(a^2+ab+b^2\right)=37\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\a^2+ab+b^2=37\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\a^2+ab+b^2=37\end{matrix}\right.\)
\(\Leftrightarrow\left(b+1\right)^2+b\left(b+1\right)+b^2-37=0\)
\(\Leftrightarrow3b^2+3b-36=0\Rightarrow\left[{}\begin{matrix}b=3\\b=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x-3}=3\\\sqrt[3]{x-3}=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-61\end{matrix}\right.\)
