Cho \(\frac{a}{b}=\frac{c}{d}\).Chứng minh rằng :
a ) \(\frac{a+c}{c}=\frac{b+d}{d}\)
b ) \(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
c ) \(\frac{a^2+c^2}{b^2+d^2}=\frac{ab}{bd}\)
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Cho \(\frac{a}{b}=\frac{c}{d}.\)Chứng minh.
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a/ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b/ \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\left(\frac{a+b}{c+d}\right)^2\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{b^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}cmr\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{3a+5b}{3a-5b}=\frac{3bk+5b}{3bk-5b}=\frac{b\left(3k+5\right)}{b\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
\(\frac{3c+5d}{3c-5d}=\frac{3dk+5d}{3dk-5d}=\frac{d\left(3k+5\right)}{d\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
Vậy từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) chứng minh \(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Ta có:
a/b=c/d => a/c=b/d=2a/2c=3b/3d
= 2a+3b/2c+3d=2a-3b/2c-3d
=> 2a+3b/2a-3b=2c+3d/2c-3d (ĐPCM)
Cho tỉ lệ thúc a/b=c/d .C/minh rang ta có các tlt sau
a)\(\frac{3a+5b}{3a-5b}\)=\(\frac{3c+5d}{3c-5d}\)
b)\(\left(\frac{a+b}{c+d}\right)\) =\(\frac{a^2+b^2}{c^2+d^2}\)
c)\(\frac{a-b}{a+b}\)=\(\frac{c-d}{c+d}\)
d)\(\frac{ab}{cd}\)=\(\left(\frac{a-b}{c-d}\right)^2\)
Đặt Bằng a = bk
c = dk Rồi thay vào biểu thức nha bạn
cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh:\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Cho tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh
\(\frac{3a+5b}{3a-5b}\)= \(\frac{3c+5d}{3c-5d}\)
ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{d}{b}=\frac{c}{a}\Rightarrow\frac{c+d}{a+b}\Rightarrow\frac{3c+3d}{3a+3b}=\frac{3c-3d}{3a-3b}\)
\(\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)\(\left(điềuphảichứngminh\right)\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). Chứng minh
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh
a, \(\frac{a}{b}=\frac{c}{d}=\frac{3a-3c}{5b-5d}=\frac{2b-3c}{2b-3d}\)
b, \(\frac{a+b}{c+a}=\frac{a-b}{c-d}\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Khi đó:
a) Đề bài sai. Bạn xem lại đề.
b) Cần thêm điều kiện $a\neq \pm b; c\neq \pm d$
Khi đó \(t=\frac{a}{b}=\frac{c}{d}\neq \pm 1\)
\(\frac{a+b}{c+d}=\frac{bt+b}{dt+d}=\frac{b(t+1)}{d(t+1)}=\frac{b}{d}\)
\(\frac{a-b}{c-d}=\frac{bt-b}{dt-d}=\frac{b(t-1)}{d(t-1)}=\frac{b}{d}\)
\(\Rightarrow \frac{a+b}{c+d}=\frac{a-b}{c-d}\) (đpcm)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\).chứng minh
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b) \(\frac{2a+3b}{2a-3b}\)=\(\frac{2c+3d}{2c-3d}\)