Cho : \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)
Chứng minh : \(\dfrac{a-2b}{c-2d}\)=\(\dfrac{b}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh \(\dfrac{a+2b}{c+2d}=\dfrac{a-2b}{c-2d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{bk+2b}{dk+2d}=\dfrac{b\left(k+2\right)}{d\left(k+2\right)}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{a-2b}{c-2d}=\dfrac{bk-2b}{dk-2d}=\dfrac{b\left(k-2\right)}{d\left(k-2\right)}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{a-2b}{c-2d}\rightarrowđpcm\)
ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow4ad=4bc\Leftrightarrow2ad+2ad=2bc+2bc\)
\(\Leftrightarrow2ad-2bc=2bc-2ad\Leftrightarrow ac+2ad-2bc-4bd=ac+2bc-2ad-4bd\)
\(\Leftrightarrow\left(c+2d\right)\left(a-2b\right)=\left(a+2b\right)\left(c-2d\right)\Leftrightarrow\dfrac{a+2b}{c+2d}=\dfrac{a-2b}{c-2d}\left(đpcm\right)\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+4c}{b+4d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a+2c}{3b+2d}\)
c) \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-2b}{c-2d}\)
d) \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a-2b}{5c-2d}\)
a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\left(đpcm\right)\)
b;c;d tương tự hết
b: a/b=c/d
nên 3a/3b=2c/2d
=>a/b=c/d=(3a+2c)/(3b+2d)
c: a/c=b/d nên a/c=2b/2d=(a-2b)/(c-2d)
d: a/c=b/d
nên 5a/5c=2b/2d
=>a/c=b/d=(5a-2b)/(5c-2d)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\left(b,d\ne0\right)\).Chứng minh rằng
\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Hãy chứng minh rằng :
\(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(\dfrac{a+2c}{b+2d}=\dfrac{a-2c}{b-2d}\)
\(\dfrac{a^2+2b^2}{c^2+2d^2}=\dfrac{a^2-2b^2}{c^2-2d^2}\)
\(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Mình hướng dẫn thôi nhé:
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\) . Sau đó thế vào biểu thức tính rồi suy ra đpcm
Ví dụ bài đầu tiên: Thế a = kb; c=kd vào biểu thức,ta có:
\(\dfrac{a}{a+b}=\dfrac{kb}{kb+b}=\dfrac{kb}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{kd}{kd+d}=\dfrac{kd}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) ,ta có đpcm: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Các bài sau làm tương tự:Thế a=kb ; c=kd vào biểu thức rồi tính từng vế . Sau đó so sánh hai vế. Thấy hai vế = nhau => đpcm
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (Giả thiết các tỉ số đều có nghĩa). Chứng minh:
a) \(\dfrac{5a+2b}{5a-2b}=\dfrac{5c+2d}{5a-2d}\) b)\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
a: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5bk+2b}{5bk-2b}=\dfrac{5k+2}{5k-2}\)
\(\dfrac{5c+2d}{5c-2d}=\dfrac{5dk+2d}{5dk-2d}=\dfrac{5k+2}{5k-2}\)
Do đó: \(\dfrac{5a+2b}{5a-2b}=\dfrac{5c+2d}{5c-2d}\)
Cho các số \(a,b,c,d\) nguyên dương đôi một khác nhau và thỏa mãn: \(\dfrac{2a+b}{a+b}+\dfrac{2b+c}{b+c}+\dfrac{2c+d}{c+d}+\dfrac{2d+a}{d+a}=6\). Chứng minh \(A=abcd\) là số chính phương.
Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)
\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))
\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)
\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac=bd\) (do \(b\ne d\))
Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)
Cho tỉ lệ thức : \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh :
a ) \(\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\)
b ) \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(k+2\right)}{b}=k+2\)
\(\dfrac{c+2d}{d}=\dfrac{dk+2d}{d}=\dfrac{d\left(k+2\right)}{d}=k+2\)
Vậy \(\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\Rightarrowđpcm\)
\(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
Vậy \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\Rightarrowđpcm\)
a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow ad+2bd=bc+2bd\)
\(\Leftrightarrow d\left(a+2b\right)=b\left(c+2d\right)\Leftrightarrow\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\left(đpcm\right)\)
b) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow2ad=2bc\Leftrightarrow ad+ad=bc+bc\)
\(\Leftrightarrow ad-bc=bc-ad\Leftrightarrow ac+ad-bc-bd=ac+bc-ad-bd\)
\(\Leftrightarrow a\left(c+d\right)-b\left(c+d\right)=c\left(a+b\right)-d\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\Leftrightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
a)\(\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=b.k\\c=dk\end{matrix}\right.\)
\(\dfrac{a+2b}{b}=\dfrac{bk+2b}{b}=\dfrac{b\left(2+k\right)}{b}=k+2\left(1\right)\)
\(\dfrac{c+2d}{d}=\dfrac{dk+2d}{d}=\dfrac{d\left(k+2\right)}{d}=k+2\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{a+2b}{b}=\dfrac{c+2d}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh :
a) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a+2c}{3b+2d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a+7c}{-2b+7d}\)
Ta có:
a/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)
PS: Xong
Cho \(\dfrac{a}{b}\)= \(\dfrac{c}{d}\) . Chứng minh :
\(\dfrac{3a+2b}{3a-2b}=\dfrac{3c+2d}{3c-2d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=b.k;b=d.k\)
Thay :
(1) : \(\dfrac{3a+2b}{3a-2b}=\dfrac{3bk+2b}{3bk-2b}=\dfrac{b.\left(3.k+2\right)}{b.\left(3.k-2\right)}=\dfrac{3.k+2}{3.k-2}\)
(2) : \(\dfrac{3c+2d}{3c-2d}=\dfrac{3dk+2d}{3dk-2d}=\dfrac{d.\left(3.k+2\right)}{d.\left(3.k-2\right)}=\dfrac{3.k+2}{3.k-2}\)
Do đó : \(\dfrac{3a+2b}{3a-2b}=\dfrac{3c+2d}{3c-2d}\)
Câu 1 : cho tỉ lệ thức a/b =c/d .Chứng minh : \(\dfrac{a+2b}{a-2b}\) = \(\dfrac{c+2d}{c-2d}\)
Câu 2 : Tìm x,y,z biết : (áp dụng công thức dãy tỉ số bằng nhau)
a) 2x=3y , 5y =7z và 3x+5y-7z =30.
b) \(\dfrac{x-1}{2}\)=\(\dfrac{y+3}{4}\)=\(\dfrac{z-5}{6}\)và 5z-3x-4y=50.
c) \(\dfrac{1}{2}\)x =\(\dfrac{2}{3}\)y=\(\dfrac{3}{4}\)z và x-y=15.