Tìm x
1, \(4x^4+12x^3+12x-47x^2+4=0\)\(4x^4+12x^3+12x-47x^2+4=0\)
2, \(x^2+\sqrt{x+1}=1\)
3.\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
4.\(x-2\sqrt{x-1}-\left(x-1\right)\sqrt{x}+\sqrt{x^2-x}=0\)
5.\(x\sqrt[3]{35-x^3}-\left(x+\sqrt[3]{35-x^3}\right)=30\)
6. \(x^3-1=2\sqrt[3]{2x-1}\)
1/\(4x^4+12x^3-47x^2+12x+4=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^3+20x^2-7x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(2x^2+11x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=\frac{-11\pm\sqrt{105}}{4}\end{matrix}\right.\)
Vậy ....
1, 4x^4+12x^3+12x−47x^2+4=0 nhé