Bài 1
\(a,x-1>0\\ x>1;x-1\ne1=>x\ne2;b,\sqrt{\left(x-1\right)^2}>0\\ x-1\ne0\\ x\ne1\\ 2,\\ a,M=\left(4+\sqrt{3}\right).\sqrt{16-2.4.\sqrt{3}+3}\\ =\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}\\ =\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)\\ =16-3\\ =13\\ b,=\dfrac{\sqrt{2}.\sqrt{8-\sqrt{15}}}{2\sqrt{15}-2}\\ =\dfrac{\sqrt{15-2.15.1+1}}{2\left(\sqrt{15}-1\right)}\\ =\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\dfrac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}\\ =\dfrac{1}{2}\)
Câu 5:
\(x^2-2x+5=\left(x-1\right)^2+4>=4\)
=>P>=2
Dấu '=' xảy ra khi x=1
