Cho biểu thức M=1/(√x+√(x-1))-1/(√x-√(x-1))-(√(x^3)-x)/(1-√x) Rút gọn M 14/07/2021 Bởi Melanie Cho biểu thức M=1/(√x+√(x-1))-1/(√x-√(x-1))-(√(x^3)-x)/(1-√x) Rút gọn M
Giải thích các bước giải: ĐKXĐ: \(x > 1\) Ta có: \(\begin{array}{l}M = \dfrac{1}{{\sqrt x + \sqrt {x – 1} }} – \dfrac{1}{{\sqrt x – \sqrt {x – 1} }} – \dfrac{{\sqrt {{x^3}} – x}}{{1 – \sqrt x }}\\ = \dfrac{{\left( {\sqrt x – \sqrt {x – 1} } \right) – \left( {\sqrt x + \sqrt {x – 1} } \right)}}{{\left( {\sqrt x + \sqrt {x – 1} } \right)\left( {\sqrt x – \sqrt {x – 1} } \right)}} – \dfrac{{x\sqrt x – x}}{{1 – \sqrt x }}\\ = \dfrac{{ – 2\sqrt {x – 1} }}{{x – \left( {x – 1} \right)}} – \dfrac{{x\left( {\sqrt x – 1} \right)}}{{1 – \sqrt x }}\\ = \dfrac{{ – 2\sqrt {x – 1} }}{1} + x\\ = x – 2\sqrt {x – 1} \\ = \left( {x – 1} \right) – 2\sqrt {x – 1} + 1\\ = {\left( {\sqrt {x – 1} – 1} \right)^2}\end{array}\) Bình luận
Giải thích các bước giải:
ĐKXĐ: \(x > 1\)
Ta có:
\(\begin{array}{l}
M = \dfrac{1}{{\sqrt x + \sqrt {x – 1} }} – \dfrac{1}{{\sqrt x – \sqrt {x – 1} }} – \dfrac{{\sqrt {{x^3}} – x}}{{1 – \sqrt x }}\\
= \dfrac{{\left( {\sqrt x – \sqrt {x – 1} } \right) – \left( {\sqrt x + \sqrt {x – 1} } \right)}}{{\left( {\sqrt x + \sqrt {x – 1} } \right)\left( {\sqrt x – \sqrt {x – 1} } \right)}} – \dfrac{{x\sqrt x – x}}{{1 – \sqrt x }}\\
= \dfrac{{ – 2\sqrt {x – 1} }}{{x – \left( {x – 1} \right)}} – \dfrac{{x\left( {\sqrt x – 1} \right)}}{{1 – \sqrt x }}\\
= \dfrac{{ – 2\sqrt {x – 1} }}{1} + x\\
= x – 2\sqrt {x – 1} \\
= \left( {x – 1} \right) – 2\sqrt {x – 1} + 1\\
= {\left( {\sqrt {x – 1} – 1} \right)^2}
\end{array}\)