Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{2}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b+c}{2+4+3}=\dfrac{180}{9}=20\)
=>a=20; b=80; c=60
Bài 3:
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\left(\dfrac{b}{d}\right)^2\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2-b^2}{c^2-d^2}\)
c: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
b: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
c: \(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{k-1}{k+1}\)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{k-1}{k+1}\)
Do đó: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
a. Câu hỏi của Nguyễn Ngọc Quế Anh - Toán lớp 7 - Học toán với OnlineMath
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
➩\(\dfrac{a}{c}=\dfrac{b}{d}\)➩\(\dfrac{2008a}{2009c}=\dfrac{2009b}{2010d}=\dfrac{2008a+2009b}{2009c+2010d}=\dfrac{2008a-2009b}{2009c-2010}\)
➩\(\dfrac{2008a-2009b}{2009c+2009c}=\dfrac{2008c-2009d}{2009a+2010d}\left(đpcm\right)\)
* đpcm : điều phải chứng minh
Chúc bạn học tốt !!!
Nếu thấy đúng thì tick cho mình nhé !!!
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
b, \(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
c, \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)