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bài này bạn cứ đặt a=bk, c=dk là được dễ tính lắm sao đó thì thay vào rồi rút gọn là được khi đó bạn sẽ chứng minh được dễ dàng hihi
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{a+b+c}{b+c+d}\)
\(\dfrac{b}{c}=\dfrac{a+b+c}{b+c+d}\)
\(\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}.\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(1\right)\)
Từ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Ta xét tích: \(\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)=\dfrac{a}{d}\left(dpcm\right)\)
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{ac}{c^2}\)=\(\dfrac{bd}{d^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{d^2}{c^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)= \(\dfrac{2c^2-ac}{2c^2-bd}\)
=> \(\dfrac{a}{b}\)=\(\dfrac{2c^2-ac}{2c^2-bd}\)=>\(\dfrac{a^2}{b^2}\)=\(\dfrac{2c^2-ac}{2d^2-bd}\)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)= \(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)=\(\dfrac{ma+nb}{mc+nd}\)=\(\dfrac{ma-nb}{mc-nd}\)
=> \(\dfrac{ma+nb}{ma-nb}\)=\(\dfrac{mc+nd}{mc-nd}\)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^3}{c^3}\)=\(\dfrac{b^3}{d^3}\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^3\)(2)
Từ (1) và (2) suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^3\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)
2.
Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(1\right)\)
Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(dpcm\right)\)
Giải:
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\left(đpcm\right)\)
Vậy...
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=t\)
Ta có : \(\left\{{}\begin{matrix}\left(\dfrac{a+b+c}{b+c+d}\right)^3=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)
Ta có đpcm
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)
\(\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{b^3}{d^3}\)
Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
\(\Leftrightarrow\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
\(\Leftrightarrow\dfrac{a}{d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(đpcm\right)\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;b=ck;c=dk\) (1)
Thay (1) vào đề bài:
\(VT=\left(\frac{bk+ck+dk}{ck+dk+d}\right)^3=\left[\frac{k\left(c+d\right)+bk}{k\left(c+d\right)+d}\right]^3=\left(\frac{bk}{d}\right)^3=\frac{bk}{d}\)
\(VP=\frac{bk}{d}\)
\(\Rightarrow VT=VP\)
hay \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\rightarrowđpcm.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}=\dfrac{a^3}{c^3}=\dfrac{c^3}{b^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}\left(1\right)\)
Từ \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}\)
Ta xét tích: \(\left(\dfrac{a}{c}\right)^3=\dfrac{a}{c}.\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{c}{b}.\dfrac{b}{d}=\dfrac{a}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\left(dpcm\right)\)
hay