Question

[Nguyễn Ngọc Gia Linh]

C1: Cho M=( 1- \(\frac{4\sqrt{x}}{x-1}\) + \(\frac{1}{\sqrt{x-1}}\) ) : \(\frac{x-2\sqrt{x}}{x-1}\)

a, rút gọn M

b, tìm x để M = \(\frac{1}{2}\)

C2: giải phương trình

a, \(\sqrt{49x-98}-14\sqrt{\frac{x-2}{49}}=3\sqrt{x-2}+8\)

b, \(\sqrt{x+1}-\sqrt{x-2}=1\)

c, \(\sqrt{x^2+1}+\sqrt{4x^2-4x+5}=0\)

Answer 1

[Nguyễn Thành Trương]

\(\text{Câu 1: Sửa đề}\)

\( a)M = \left( {1 - \dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{x - 2\sqrt x }}{{x - 1}}\\ M = \left[ {1 - \dfrac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \left[ {1 + \dfrac{{ - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \left[ {1 + \dfrac{{ - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) - 3\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{x - 2\sqrt x }}\\ M = \sqrt x \left( {\sqrt x - 3} \right).\dfrac{1}{{x - 2\sqrt x }}\\ M = \dfrac{{x - 3\sqrt x }}{{x - 2\sqrt x }} \)

\( b)M = \dfrac{1}{2} \Rightarrow \dfrac{{x - 3\sqrt x }}{{x - 2\sqrt x }} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( {x - 3\sqrt x } \right) = x - 2\sqrt x \\ \Leftrightarrow 2x - 6\sqrt x = x - 2\sqrt x \\ \Leftrightarrow - 4\sqrt x = - x\\ \Leftrightarrow 16x = {x^2}\\ \Leftrightarrow 16x - {x^2} = 0\\ \Leftrightarrow x\left( {16 - x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 16 - x = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 16 \end{array} \right. \)

Answer 2

[Nguyễn Thành Trương]

\(\text{Câu 2}:\)

\( a)\sqrt {49x - 98} - 14\sqrt {\dfrac{{x - 2}}{{49}}} = 3\sqrt {x - 2} + 8\left( {x \ge 2} \right)\\ \Leftrightarrow 7\sqrt {x - 2} - 3\sqrt {x - 2} = 8 + 14\sqrt {\dfrac{{x - 2}}{{49}}} \\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 14\sqrt {\dfrac{{x - 2}}{{49}}} \\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 14\dfrac{{\sqrt {x - 2} }}{7}\\ \Leftrightarrow 4\sqrt {x - 2} = 8 + 2\sqrt {x - 2} \\ \Leftrightarrow 4\sqrt {x - 2} - 2\sqrt {x - 2} = 8\\ \Leftrightarrow 2\sqrt {x - 2} = 8\\ \Leftrightarrow \sqrt {x - 2} = 4\\ \Leftrightarrow x - 2 = 16\\ \Leftrightarrow x = 16 + 2 = 18 \text{(thỏa mãn điều kiện)} \)