a, Tìm giá trị lớn nhất của B= √(3x-x) +x b, Tìm MaxA= √(x ²+x+1) + √(x ² -x+1) c, Tìm Max M=( √a+ √b) ² với a,b >0 và a+b=1 Em cảm ơn

$\begin{array}{l} a)\,\,E\,\,viet\,\,ro\,\,de\,\,bai\,\,\,nhe.\\ b)A = \sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} \\ \Rightarrow {A^2} = {\left( {\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} } \right)^2}\\ = {x^2} + x + 1 + {x^2} - x + 1 + 2\sqrt {\left( {{x^2} + 1 + x} \right)\left( {{x^2} + 1 - x} \right)} \\ = 2{x^2} + 2 + 2\sqrt {{{\left( {{x^2} + 1} \right)}^2} - {x^2}} \\ = 2{x^2} + 2 + 2\sqrt {{x^4} + {x^2} + 1} \\ Ta\,\,co:{x^2} \ge 0\,\,\forall x\\ \Rightarrow {A^2} = 2{x^2} + 2 + 2\sqrt {{x^4} + {x^2} + 1} \ge 2.0 + 2 + 2\sqrt {0 + 0 + 1} = 4\\ \Rightarrow A \ge 2.\\ Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow {x^2} = 0 \Leftrightarrow x = 0.\\ c)\,\,\,M = {\left( {\sqrt a + \sqrt b } \right)^2}\,\,\,\forall a,\,\,b > 0,\,\,a + b = 1\\ a + b = 1 \Rightarrow b = 1 - a.\\ \Rightarrow M = {\left( {\sqrt a + \sqrt b } \right)^2} = {\left( {\sqrt a + \sqrt {1 - a} } \right)^2} = a + 1 - a + 2\sqrt {a\left( {1 - a} \right)} \,\,\,\left( {Dk:\,\,\,0 < a \le 1} \right)\\ = 1 + 2\sqrt {a - {a^2}} = 1 + 2\sqrt { - \left( {{a^2} - a} \right)} = 1 + 2\sqrt { - \left( {{a^2} - 2.\frac{1}{2}.a + \frac{1}{4} - \frac{1}{4}} \right)} \\ = 1 + 2\sqrt {\frac{1}{4} - {{\left( {a - \frac{1}{2}} \right)}^2}} \\ Ta\,\,co:\,\,{\left( {a - \frac{1}{2}} \right)^2} \ge 0 \Rightarrow - {\left( {a - \frac{1}{2}} \right)^2} \le 0\\ \Rightarrow 0 < \frac{1}{4} - {\left( {a - \frac{1}{2}} \right)^2} \le \frac{1}{4}\\ \Rightarrow M \le 1 + 2.\frac{1}{4} = \frac{3}{2}.\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow a - \frac{1}{2} = 0 \Leftrightarrow a = \frac{1}{2} \Rightarrow b = 1 - a = \frac{1}{2}.\\ Vay\,\,MaxM = \frac{3}{2}\,\,\,khi\,\,a = b = \frac{1}{2}. \end{array}$