## Cho A= (x√x+1)/(x-1)-(x-1)/(√x+1) 1/ Tìm ĐKXĐ 2/ rút gọn A 3/ Tính A khi x=1 , x=9/4 4/ Tìm x để A= (-1)/2 5/ Tìm x để A<1 6/ Tìm x để A ∈R #Giúp m

Question

Cho A= (x√x+1)/(x-1)-(x-1)/(√x+1)
1/ Tìm ĐKXĐ
2/ rút gọn A
3/ Tính A khi x=1 , x=9/4
4/ Tìm x để A= (-1)/2
5/ Tìm x để A<1 6/ Tìm x để A ∈R #Giúp mik vs :((

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4 years 2020-11-21T05:10:17+00:00 1 Answers 64 views 0

4) $$x = \dfrac{1}{9}$$
$$\begin{array}{l} 1)DK:x \ge 0;x \ne 1\\ 2)A = \dfrac{{x\sqrt x + 1}}{{x – 1}} – \dfrac{{x – 1}}{{\sqrt x + 1}}\\ = \dfrac{{\left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}} – \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{\sqrt x + 1}}\\ = \dfrac{{x – \sqrt x + 1}}{{\sqrt x – 1}} – \left( {\sqrt x – 1} \right)\\ = \dfrac{{x – \sqrt x + 1 – x + 2\sqrt x – 1}}{{\sqrt x – 1}}\\ = \dfrac{{\sqrt x }}{{\sqrt x – 1}}\\ 3)Thay:x = 1\left( l \right)\\ Thay:x = \dfrac{9}{4}\\ \to A = \dfrac{{\sqrt {\dfrac{9}{4}} }}{{\sqrt {\dfrac{9}{4}} – 1}} = \dfrac{3}{2}:\left( {\dfrac{3}{2} – 1} \right) = 3\\ 4)A = – \dfrac{1}{2}\\ \to \dfrac{{\sqrt x }}{{\sqrt x – 1}} = – \dfrac{1}{2}\\ \to 2\sqrt x = – \sqrt x + 1\\ \to 3\sqrt x = 1\\ \to \sqrt x = \dfrac{1}{3}\\ \to x = \dfrac{1}{9}\left( {TM} \right)\\ 5)A < 1\\ \to \dfrac{{\sqrt x }}{{\sqrt x – 1}} < 1\\ \to \dfrac{{\sqrt x – \sqrt x + 1}}{{\sqrt x – 1}} < 0\\ \to \dfrac{1}{{\sqrt x – 1}} < 0\\ \to \sqrt x – 1 < 0\\ \to x < 1;x \ge 0\\ 6)A \in R\\ \Leftrightarrow x \ge 0;x \ne 1 \end{array}$$