Đời cơ bản là cạn lời =)), quy đồng mãnh liệt thật :))
\(M=\sqrt{1+x^2+\dfrac{x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(=\sqrt{\left(x+1\right)^2-2x+\dfrac{x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(=\sqrt{\left(x+1\right)^2-2.\left(x+1\right).\dfrac{x}{x+1}+\dfrac{x^2}{\left(x+1\right)^2}}\)
\(=\sqrt{\left(x+1-\dfrac{x}{x+1}\right)^2}+\dfrac{x}{x+1}\)
\(=\left|x+1-\dfrac{x}{x+1}\right|+\dfrac{x}{x+1}\)
Tới đây thì tùy x mà khử tiếp
\(M=\sqrt{1+x^2+\dfrac{x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\sqrt{\dfrac{\left(x+1\right)^2+\left(x+1\right)^2.x^2+x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\sqrt{\dfrac{x^2+2x+1+\left(\left(x+1\right).x\right)^2+x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\sqrt{\dfrac{x^2+2x+1+\left(x^2+x\right)^2+x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\sqrt{\dfrac{x^2+2x+1+x^4+2x^3+x^2+x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\sqrt{\dfrac{3x^2+2x+1+x^4+2x^3}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\sqrt{\dfrac{x^4+2x^3+3x^2+2x+1}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\)
\(M=\dfrac{\sqrt{x^4+2x^3+3x^2+2x+1}}{x+1}+\dfrac{x}{x+1}\)
\(M=\dfrac{\sqrt{x^4+2x^3+3x^2+2x+1}+x}{x+1}\)
>> Có lẽ là như thek ..... <<
\( M=\sqrt{1+x^2+\dfrac{x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\\ =\sqrt{\dfrac{x^2+2x+1+x^4+2x^3+x^2+x^2}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\\ =\sqrt{\dfrac{x^4+2x^3+3x^2+2x+1}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\\ =\sqrt{\dfrac{\left(x^2+x+1\right)}{\left(x+1\right)^2}}+\dfrac{x}{x+1}\\ =\dfrac{\left|x^2+x+1\right|}{\left|x+1\right|}+\dfrac{x}{x+1}\)
Nếu còn rút gọn nữa thì chia làm 2 trường hợp rồi giải tiếp
