Cho \(\frac{a}{b}=\frac{c}{d}.\) Chứng minh \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
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 \(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}\)
Chứng minh a = b = c = d
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}==\frac{a+b+c+d}{3b+3c+3d+3a}=\frac{a+b+c+d}{3.\left(a+b+c+d\right)}=\frac{1}{3}\)
suy ra:
\(\frac{a}{3b}=\frac{1}{3}\Rightarrow a=\frac{1}{3}.3b=b\)
\(\frac{b}{3c}=\frac{1}{3}\Rightarrow b=\frac{1}{3}.3c=c\)
\(\frac{c}{3d}=\frac{1}{3}\Rightarrow c=\frac{1}{3}.3d=d\)
=>a=b=c=d
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3b+3c+3d+3a}=\frac{a+b+c+d}{3\left(b+c+d+a\right)}=\frac{1}{3}\)
\(\Rightarrow a=\frac{1}{3}.3b=b\) (1)
\(b=\frac{1}{3}.3c=c\) (2)
\(c=\frac{1}{3}.3d=d\) (3)
\(d=\frac{1}{3}.3a=a\) (4)
Từ (1), (2), (3), (4) suy ra: a = b = c = d (đpcm)
Cho \(\frac{a}{b}=\frac{c}{d}\ne\frac{-1}{3}\).
Chứng minh: \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
Xét \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(\Leftrightarrow a.\left(3c+d\right)=c.\left(3a+b\right)\)
\(\Leftrightarrow3ac+ad=3ca+cb\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\) ( Đúng theo giả thiết )
Vậy \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
Cho tỉ lệ thức \(\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\). Chứng minh rằng \(\frac{a}{b}=\frac{c}{d}\).
Ta có: \(\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\)
\(\Rightarrow\frac{3a+4b}{3a-4b}-1=\frac{3c+4d}{3c-4d}-1\)
\(\Leftrightarrow\frac{8b}{3a-4b}=\frac{8d}{3c-4d}\)
\(\Rightarrow b\left(3c-4d\right)=d\left(3a-4b\right)\)
\(\Leftrightarrow3bc=3ad\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
Cho các số a,b,c,d thõa mãn điều kiện:\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}\)và a+b+c+d khác 0.Chứng minh rằng a=b=c=d
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 \(\frac{a}{b}=\frac{c}{d}\).Chứng minh rằng :
a ) \(\frac{a+c}{c}=\frac{b+d}{d}\)
b ) \(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
c ) \(\frac{a^2+c^2}{b^2+d^2}=\frac{ab}{bd}\)
Lưu ý : spam + tl linh tinh,cop bài vớ vẩn = báo cáo
\(a,\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{c}+1=\frac{b}{d}+1\)
\(\Rightarrow\frac{a}{c}+\frac{c}{c}=\frac{b}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+c}{c}=\frac{b+d}{d}\)
Ta có :
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Leftrightarrow\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\)
\(\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3a+5d}{3c-5d}\)
1.Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)
Chứng minh rằng :
a) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
Lười suy nghĩ nên ta cứ dùng cách đặt k.
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)ĐK:...
\(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\) (1)
Lại có: \(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\) (2)
Từ (1) và (2) ta suy ra đpcm: \(\frac{a}{3a+b}=\frac{c}{3c+d}\left(=\frac{k}{3k+1}\right)\)
Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}.\)
Mà \(\frac{a}{c}=\frac{3a}{3c}\)
\(\Rightarrow\frac{3a}{3c}=\frac{b}{d}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\)
Ta có \(\frac{a}{c}=\frac{3a+b}{3c+d}\)
=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\left(đpcm\right).\)
Chúc bạn học tốt!
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: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)