a)\(x^2+x+12\sqrt{x+1}=36\)
\(pt\Leftrightarrow x^2+x-12+12\sqrt{x+1}-24=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\frac{144\left(x+1\right)-576}{12\sqrt{x+1}+24}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\frac{144\left(x-3\right)}{12\sqrt{x+1}+24}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4+\frac{144}{12\sqrt{x+1}+24}\right)=0\)
Dễ thấy: \(x+4+\frac{144}{12\sqrt{x+1}+24}>0\forall x\ge-1\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
b)\(x+\sqrt{x-2}=2\sqrt{x-1}\)
\(pt\Leftrightarrow x-2+\sqrt{x-2}=2\sqrt{x-1}-2\)
\(\Leftrightarrow x-2+\frac{x-2}{\sqrt{x-2}}=2\left(\sqrt{x-1}-1\right)\)
\(\Leftrightarrow x-2+\frac{x-2}{\sqrt{x-2}}-2\cdot\frac{x-1-1}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow x-2+\frac{x-2}{\sqrt{x-2}}-2\cdot\frac{x-2}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(1+\frac{1}{\sqrt{x-2}}-\frac{2}{\sqrt{x-1}+1}\right)=0\)
Suy ra x-2=0=>x=2
c)Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:
\(VT=\sqrt{x+3}+\sqrt{1-x}\)
\(\ge\sqrt{x+3+1-x}=\sqrt{4}=2=VP\)
Xảy ra khi \(\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
1) ĐK: \(x\ge-1\)
\(PT\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12.\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
\(\Leftrightarrow x=3\text{ hoặc }\frac{12}{\sqrt{x+1}+2}+x+4=0\) (*)
VT của (*) luôn dương với \(x\ge-1\)
=> x = 3