Kiệt Gia 913 Questions 2k Answers 0 Best Answers 27 Points View Profile0 Kiệt Gia Asked: Tháng Mười 25, 20202020-10-25T02:15:57+00:00 2020-10-25T02:15:57+00:00In: Môn ToánAi đó giúp em bài 5 với ạ???? em đang gấp lắm0Ai đó giúp em bài 5 với ạ???? em đang gấp lắm 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 AnswerOldestVotedRecentAdela 859 Questions 2k Answers 0 Best Answers 19 Points View Profile Adela 2020-10-25T02:17:24+00:00Added an answer on Tháng Mười 25, 2020 at 2:17 sáng Giải thích các bước giải:Ta có:\(\begin{array}{l}a,\\\dfrac{{3{x^2} – 2x – 5}}{A} = \dfrac{{3x – 5}}{{2x – 3}}\\ \Leftrightarrow \dfrac{{3{x^2} – 2x – 5}}{A} = \dfrac{{\left( {3x – 5} \right)\left( {x + 1} \right)}}{{\left( {2x – 3} \right)\left( {x + 1} \right)}}\\ \Leftrightarrow \dfrac{{3{x^2} – 2x – 5}}{A} = \dfrac{{3{x^2} – 2x – 5}}{{2{x^2} – x – 3}}\\ \Leftrightarrow A = 2{x^2} – x – 3\\b,\\\dfrac{{2{x^2} + 3x – 2}}{{{x^2} – 4}} = \dfrac{B}{{{x^2} – 4x + 4}}\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {2x – 1} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{B}{{{x^2} – 4x + 4}}\\ \Leftrightarrow \dfrac{{2x – 1}}{{x – 2}} = \dfrac{B}{{{x^2} – 4x + 4}}\\ \Leftrightarrow \dfrac{{\left( {2x – 1} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x – 2} \right)}} = \dfrac{B}{{{x^2} – 4x + 4}}\\ \Leftrightarrow \dfrac{{2{x^2} – 5x + 2}}{{{x^2} – 4x + 4}} = \dfrac{B}{{{x^2} – 4x + 4}}\\ \Leftrightarrow B = 2{x^2} – 5x + 2\\c,\\\dfrac{C}{{2x + 1}} = \dfrac{{10{x^2} – 5x}}{{4{x^2} – 1}}\\ \Leftrightarrow \dfrac{C}{{2x + 1}} = \dfrac{{5x.\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}\\ \Leftrightarrow \dfrac{C}{{2x + 1}} = \dfrac{{5x}}{{2x + 1}}\\ \Rightarrow C = 5x\\d,\\\dfrac{{4{x^2} – 16x + 16}}{{{x^2} – 4}} = \dfrac{D}{{x + 2}}\\ \Leftrightarrow \dfrac{{4.\left( {{x^2} – 4x + 4} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{D}{{x + 2}}\\ \Leftrightarrow \dfrac{{4.{{\left( {x – 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{D}{{x + 2}}\\ \Leftrightarrow \dfrac{{4.\left( {x – 2} \right)}}{{x + 2}} = \dfrac{D}{{x + 2}}\\ \Leftrightarrow D = 4\left( {x – 2} \right)\end{array}\)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.
Adela
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{3{x^2} – 2x – 5}}{A} = \dfrac{{3x – 5}}{{2x – 3}}\\
\Leftrightarrow \dfrac{{3{x^2} – 2x – 5}}{A} = \dfrac{{\left( {3x – 5} \right)\left( {x + 1} \right)}}{{\left( {2x – 3} \right)\left( {x + 1} \right)}}\\
\Leftrightarrow \dfrac{{3{x^2} – 2x – 5}}{A} = \dfrac{{3{x^2} – 2x – 5}}{{2{x^2} – x – 3}}\\
\Leftrightarrow A = 2{x^2} – x – 3\\
b,\\
\dfrac{{2{x^2} + 3x – 2}}{{{x^2} – 4}} = \dfrac{B}{{{x^2} – 4x + 4}}\\
\Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {2x – 1} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{B}{{{x^2} – 4x + 4}}\\
\Leftrightarrow \dfrac{{2x – 1}}{{x – 2}} = \dfrac{B}{{{x^2} – 4x + 4}}\\
\Leftrightarrow \dfrac{{\left( {2x – 1} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x – 2} \right)}} = \dfrac{B}{{{x^2} – 4x + 4}}\\
\Leftrightarrow \dfrac{{2{x^2} – 5x + 2}}{{{x^2} – 4x + 4}} = \dfrac{B}{{{x^2} – 4x + 4}}\\
\Leftrightarrow B = 2{x^2} – 5x + 2\\
c,\\
\dfrac{C}{{2x + 1}} = \dfrac{{10{x^2} – 5x}}{{4{x^2} – 1}}\\
\Leftrightarrow \dfrac{C}{{2x + 1}} = \dfrac{{5x.\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}\\
\Leftrightarrow \dfrac{C}{{2x + 1}} = \dfrac{{5x}}{{2x + 1}}\\
\Rightarrow C = 5x\\
d,\\
\dfrac{{4{x^2} – 16x + 16}}{{{x^2} – 4}} = \dfrac{D}{{x + 2}}\\
\Leftrightarrow \dfrac{{4.\left( {{x^2} – 4x + 4} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{D}{{x + 2}}\\
\Leftrightarrow \dfrac{{4.{{\left( {x – 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{D}{{x + 2}}\\
\Leftrightarrow \dfrac{{4.\left( {x – 2} \right)}}{{x + 2}} = \dfrac{D}{{x + 2}}\\
\Leftrightarrow D = 4\left( {x – 2} \right)
\end{array}\)