Hưng Gia 422 Questions 546 Answers 0 Best Answers 15 Points View Profile0 Hưng Gia Asked: Tháng Mười 28, 20202020-10-28T11:57:00+00:00 2020-10-28T11:57:00+00:00In: Môn Toánmấy anh chị giúp em vs ạ0mấy anh chị giúp em vs ạ ShareFacebookRelated Questions Một hình thang có đáy lớn là 52cm ; đáy bé kém đáy lớn 16cm ; chiều cao kém đáy ... Useful news and important articles APROTININ FROM BOVINE LUNG CELL CULTURE купить онлайн1 AnswerOldestVotedRecentDoris 845 Questions 2k Answers 0 Best Answers 19 Points View Profile Doris 2020-10-28T11:58:08+00:00Added an answer on Tháng Mười 28, 2020 at 11:58 sáng Đáp án:a) $\left\{ \begin{array}{l}a≥0\\a \neq \{1;4\}\end{array} \right.$b) $A=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}$c) $A=\dfrac{\sqrt[]{5}}{3\sqrt[]{5}+6}$Giải thích các bước giải:a) ĐKXĐ: $\left\{ \begin{array}{l}\sqrt[]{a}-1 \neq 0\\\sqrt[]{a} \neq 0\\\sqrt[]{a}-2 \neq 0\\a≥0\end{array} \right.$$↔ \left\{ \begin{array}{l}a≥0\\a \neq \{1;4\}\end{array} \right.$b) $A=\Bigg(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\Bigg):\Bigg(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\Bigg)$$=\dfrac{\sqrt[]{a}-(\sqrt[]{a}-1)}{\sqrt[]{a}(\sqrt[]{a}-1)}:\dfrac{(\sqrt[]{a}+1)(\sqrt[]{a}-1)-(\sqrt[]{a}+2)(\sqrt[]{a}-2)}{(\sqrt[]{a}-2)(\sqrt[]{a}-1)}$$=\dfrac{1}{\sqrt[]{a}(\sqrt[]{a}-1)}.\dfrac{(\sqrt[]{a}-2)(\sqrt[]{a}-1)}{a-1-a+4}$$=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}$c) $A=\dfrac{1}{9} ↔ \dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}=\dfrac{1}{9}$$↔ 3\sqrt[]{a}=9\sqrt[]{a}-18$$↔ 6\sqrt[]{a}=18$$↔ \sqrt[]{a}=3$$→ a=9$d) $a=9+4\sqrt[]{5}$$→ \sqrt[]{a}=\sqrt[]{9+4\sqrt[]{5}}$$↔ \sqrt[]{a}=\sqrt[]{(\sqrt[]{5})^2+2.\sqrt[]{5}.2+4}$$↔ \sqrt[]{a}=\sqrt[]{(\sqrt[]{5}+2)^2}$$↔ \sqrt[]{a}=|\sqrt[]{5}+2|$$↔ \sqrt[]{a}=\sqrt[]{5}+2$Vậy $A=\dfrac{\sqrt[]{5}+2-2}{3(\sqrt[]{5}+2)}$$=\dfrac{\sqrt[]{5}}{3\sqrt[]{5}+6}$0Reply Share ShareShare on FacebookLeave an answerLeave an answerHủy By answering, you agree to the Terms of Service and Privacy Policy .* Lưu tên của tôi, email, và trang web trong trình duyệt này cho lần bình luận kế tiếp của tôi.
Doris
Đáp án:
a) $\left\{ \begin{array}{l}a≥0\\a \neq \{1;4\}\end{array} \right.$
b) $A=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}$
c) $A=\dfrac{\sqrt[]{5}}{3\sqrt[]{5}+6}$
Giải thích các bước giải:
a) ĐKXĐ:
$\left\{ \begin{array}{l}\sqrt[]{a}-1 \neq 0\\\sqrt[]{a} \neq 0\\\sqrt[]{a}-2 \neq 0\\a≥0\end{array} \right.$
$↔ \left\{ \begin{array}{l}a≥0\\a \neq \{1;4\}\end{array} \right.$
b) $A=\Bigg(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\Bigg):\Bigg(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\Bigg)$
$=\dfrac{\sqrt[]{a}-(\sqrt[]{a}-1)}{\sqrt[]{a}(\sqrt[]{a}-1)}:\dfrac{(\sqrt[]{a}+1)(\sqrt[]{a}-1)-(\sqrt[]{a}+2)(\sqrt[]{a}-2)}{(\sqrt[]{a}-2)(\sqrt[]{a}-1)}$
$=\dfrac{1}{\sqrt[]{a}(\sqrt[]{a}-1)}.\dfrac{(\sqrt[]{a}-2)(\sqrt[]{a}-1)}{a-1-a+4}$
$=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}$
c) $A=\dfrac{1}{9} ↔ \dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}=\dfrac{1}{9}$
$↔ 3\sqrt[]{a}=9\sqrt[]{a}-18$
$↔ 6\sqrt[]{a}=18$
$↔ \sqrt[]{a}=3$
$→ a=9$
d) $a=9+4\sqrt[]{5}$
$→ \sqrt[]{a}=\sqrt[]{9+4\sqrt[]{5}}$
$↔ \sqrt[]{a}=\sqrt[]{(\sqrt[]{5})^2+2.\sqrt[]{5}.2+4}$
$↔ \sqrt[]{a}=\sqrt[]{(\sqrt[]{5}+2)^2}$
$↔ \sqrt[]{a}=|\sqrt[]{5}+2|$
$↔ \sqrt[]{a}=\sqrt[]{5}+2$
Vậy $A=\dfrac{\sqrt[]{5}+2-2}{3(\sqrt[]{5}+2)}$
$=\dfrac{\sqrt[]{5}}{3\sqrt[]{5}+6}$