a,\(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
\(\Leftrightarrow\sqrt{x-1+4\sqrt{x-1+4}}+\sqrt{x-1-6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1+2}\right)^2}+\sqrt{\left(\sqrt{x-1-3}\right)^2}=5\)
\(\Leftrightarrow\sqrt{x-1}+2+|\sqrt{x-1}-3|=5\Leftrightarrow|\sqrt{x-1}-3|=3-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{x-1}-3\le0\left(|A|=-A\Leftrightarrow A\le0\right)\)
\(\Leftrightarrow\sqrt{x-1}\le3\Leftrightarrow0\le x-1\le3^2\Leftrightarrow1\le x\le10\)
Nghiệm của phương trình đã cho là : \(1\le x\le10\)
b, \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)=4\)
\(\Leftrightarrow\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]=4\)
\(\Leftrightarrow\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)=4\)
\(\Leftrightarrow\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)=4\)
\(\Leftrightarrow\left(12x^2+11x+\frac{1}{2}+\frac{3}{2}\right)\left(12x^2+11x+\frac{1}{2}-\frac{3}{2}\right)=4\)
\(\Leftrightarrow\left(12x^2+11x+\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2=4\Leftrightarrow\left(12x^2+11x+\frac{1}{2}\right)^2=4+\frac{9}{4}\)
\(\Leftrightarrow\left(12x^2+11x+\frac{1}{2}\right)^2=\left(\frac{5}{2}\right)^2\Leftrightarrow\orbr{\begin{cases}12x^2+11x+\frac{1}{2}=\frac{5}{2}\\12x^2+11x+\frac{1}{2}=-\frac{5}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}12x^2+11x-2=0\left(1\right)\\12x^2+11x+3=0\left(2\right)\end{cases}}\)
Giải (1) \(\Delta=121+96=217\)
\(x_1=\frac{-11+\sqrt{217}}{24};x_2=\frac{-11-\sqrt{217}}{24}\)
Giải (2) \(\Delta=121-144=-23< 0\).Phương trình vô nghiệm.
Phương trình có 2 nghiệm phân biệt :
\(x_1=\frac{-11+\sqrt{217}}{24};x_2=\frac{-11-\sqrt{217}}{24}\)
a) \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
\(\Leftrightarrow\sqrt{x+3+4\sqrt{x-1}}=5-\sqrt{x+8-6\sqrt{x-1}}\)
\(\Leftrightarrow\left(\sqrt{x+3+4\sqrt{x-1}}\right)^2=\left(5-\sqrt{x+8-6\sqrt{x-1}}\right)^2\)
\(\Leftrightarrow x+3+4\sqrt{x-1}=x+33-10\sqrt{x+8-6\sqrt{x-1}}-6\sqrt{x-1}\)
\(\Leftrightarrow4\sqrt{x-1}+3-10=-10\sqrt{x+8-6\sqrt{x-1}}+6\sqrt{x-1}+33\)
\(\Leftrightarrow4\sqrt{x-1}-30=-10\sqrt{x+8-6\sqrt{x-1}}-6\sqrt{x-1}\)
\(\Leftrightarrow\left(4\sqrt{x-1}-30\right)^2=\left(-10\sqrt{x+8-6\sqrt{x-1}}-6\sqrt{x-1}\right)^2\)
\(\Leftrightarrow-120x+120-240\sqrt{x-1}=120\sqrt{x-1}.\sqrt{x-6\sqrt{x-1}+8}-600\sqrt{x-1}\)
\(\Leftrightarrow\left(-120x+120-240\sqrt{x-1}\right)^2=\left(120\sqrt{x-1}.\sqrt{x-6\sqrt{x-1}+8}\right)^2\)
\(\Leftrightarrow x\le3\Rightarrow\sqrt{x-1}\le3\Rightarrow1\le x\le10\)
Vậy: Nghiệm của pt là: \(1\le x\le10\)
Câu b xíu làm thử =)
