Chứng minh: \(\frac{a}{b}=\frac{c}{d}\)
a) \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
b) \(\frac{2017a^2+ab}{2018a^2-8b^2}=\frac{2017c^2+cd}{2018c^2-8d^2}\)
Những câu hỏi liên quan
\(\frac{2017a^2+ab}{2018a^2-8b^2}=\frac{2017c^2+cd}{2018c^2+8d}\)
Cho\(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng:
a)\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
b)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
c)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
a)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)(T/C...)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
b)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)(T/C...)
\(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
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c)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}\)
\(\Rightarrow\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}=\frac{7a^2+3ab}{7c^2+3cd}=\frac{11a^2-8b^2}{11c^2-8d^2}\)
\(\Rightarrow\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)
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Cho \(\frac{a}{b}=\frac{c}{d}\), chứng minh rằng
a) \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
bạn ơi viết phân số như bạn thế nào rồi mk giải cho
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Nguyễn Thị Thu Huyền ơi , viết phân số = dấu /
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a) \(\frac{a}{b}=\frac{c}{d}=k\) => a=bk;c=dk
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\) (1)
(2)\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2}{d^2}\)
từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
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Xem thêm câu trả lời
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng
\(1.\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\) \(2.\frac{2a+b}{2a-b}=\frac{2c+d}{2c-d}\) \(3.\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\) \(4.\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng ta có tỉ lệ thức sau :
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Dăm ba mấy bài đặt k:v
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Ta có:
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018b^2k^2+2019b^2}{2018b^2k^2-2019b^2}=\frac{b^2\left(2018k^2+2019\right)}{b^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)
\(\frac{2018c^2+2019d^2}{2018c^2-2019d^2}=\frac{2018d^2k^2+2019d^2}{2018d^2k^2-2019d^2}=\frac{d^2\left(2018k^2+2019\right)}{d^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)
Từ đó \(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
B1: Cho tỷ lệ thức frac{a}{b} frac{c}{d}. Chứng minh rằng
a) frac{a-b}{b} frac{c-d}{d}
b) frac{2016a-2017b}{2018c+2019d} frac{2016c-2017d}{2018a+2019b}
c) frac{7a^2+3ab}{11a^2-8b^2} frac{7c^2+3cd}{11c^2-8d^2}
B2: Tìm GTNN của:
A ( x4 + 3)2 B | 0,5 + x | + (y - 1,3)4 + 20 C frac{5x-19}{x-4}( x thuộc Z)
Đọc tiếp
B1: Cho tỷ lệ thức \(\frac{a}{b}\) = \(\frac{c}{d}\). Chứng minh rằng
a) \(\frac{a-b}{b}\)= \(\frac{c-d}{d}\)
b) \(\frac{2016a-2017b}{2018c+2019d}\) = \(\frac{2016c-2017d}{2018a+2019b}\)
c) \(\frac{7a^2+3ab}{11a^2-8b^2}\) = \(\frac{7c^2+3cd}{11c^2-8d^2}\)
B2: Tìm GTNN của:
A = ( x4 + 3)2 B = | 0,5 + x | + (y - 1,3)4 + 20 C = \(\frac{5x-19}{x-4}\)( x thuộc Z)
Bài 1:
a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)
\(\Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}.\)
\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\left(đpcm\right).\)
Mình làm được thế thôi nhé.
Chúc bạn học tốt!
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1) Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh: \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
2) Cho\(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{a^2-d^2}{c^2-d2}=\frac{ab}{cd}\)
b) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
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Cho tỉ lệ thức a/b=c/d CMR :
a) \(\frac{7a+8b}{7a-8b}=\frac{7c+8d}{7c-8d}\)
b) \(\frac{11a-5b}{3a+4b}=\frac{11c-5d}{3c+4d}\)
c) \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
d) \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
e) \(\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
help me 3 l-i-k-e
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
b) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
GIúp 1 câu cx dc, ai giúp dc hai câu càng tốt
a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{a.a}{c.c}=\frac{b.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{a}{c}.\frac{b}{d}=\frac{a-b}{c-d}.\frac{a-b}{c-d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
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