a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)

\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)

\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)

a) Ta co: a/b = c/d= k

=> a=bk

c=dk

Ta co: a-b/a+b = bk-b/bk+b = b(k-1)/b(k+1) = k-1/k+1 (1)

Ta co: c-d/c+d = dk-d/dk+d = d(k-1)/d(k+1) = k-1/k+1 (2)

Tu (1) va (2)

=> a-b/a+b=c-d/c+d

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)

a) Từ (*) ta có:

\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\) (1)

\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\) (2)

Từ (1) và (2) suy ra \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

b) Từ (*) ta có:

\(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{b\left(7k-4\right)}{b\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (3)

\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{d\left(7k-4\right)}{d\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (4)

Từ (3) và (4) suy ra \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)

c) Từ (*) ta có:

\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (5)

\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (6)

\(\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}=\dfrac{\left[\left(dk\right)-\left(bk\right)\right]^2}{\left(d-b\right)^2}=\dfrac{\left[k\left(d-b\right)\right]^2}{\left(d-b\right)^2}=k^2\) (7)

Từ (5), (6) và (7) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)

Ta có:

Nếu:

\(\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\Leftrightarrow\left(2a+c\right)\left(b-d\right)=\left(a-c\right)\left(2b+d\right)\)

\(\Leftrightarrow2a\left(b-d\right)+c\left(b-d\right)=a\left(2b+d\right)-c\left(2b+d\right)\)

\(\Leftrightarrow2ab-2ad+bc-cd=2ab+ad-2bc+cd\)

\(\Leftrightarrow ad=bc\)

\(\Leftrightarrow\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\left(đpcm\right)\)

Các bạn chỉ cần giúp mk câu b, c, e, f,

bạn cứ đặt công thức gốc là k sau đó thay vào các câu là được thui

Ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)

<=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{3a}{3c}=\dfrac{2b}{2d}\)

<=>\(\dfrac{5a-3b}{5c-3d}=\dfrac{3a-2b}{3c-2d}\)(đpcm)

Các câu sau tương tự

Nguyễn Thị Hồng Nhung chị làm bài f đc ko ạ ???

c/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
d/ tương tự câu b/
e/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2} \)(1)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)(đpcm)
f/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(1\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2-c^2}{b^2-d^2}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(a+c\right)^2}{a^2-c^2}=\dfrac{\left(b+d\right)^2}{b^2-d^2}\)

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: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

b,

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

c,

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

ta có: \(a=bk;c=dk\)

\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

d,

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

ta có:\(a=bk;c=dk\)

\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)

e,

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

Ta có:\(a=bk;c=dk\)

\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)

f,

(để hôm sau lm nha, mỏi tay quá)

a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)

\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)

Còn các phần còn lại làm giống thế

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

Ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

a) \(\dfrac{3a+5c}{3b+5d}=\dfrac{3\cdot bk+5\cdot dk}{3b+5d}=\dfrac{k\left(3b+5d\right)}{3b+5d}=k\) (1)

\(\dfrac{a-2c}{b-2d}=\dfrac{bk-2dk}{b-2d}=\dfrac{k\left(b-2d\right)}{b-2d}=k\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{3a+5c}{3b+5d}=\dfrac{a-2c}{b-2d}\left(dpcm\right)\)

b) \(\dfrac{a^2-b^2}{ab}=\dfrac{\left(bk\right)^2-b^2}{bk\cdot b}=\dfrac{b^2k^2-b^2}{b^2k}=\dfrac{b^2\left(k-1\right)}{b^2k}=\dfrac{k-1}{k}\)(1)

\(\dfrac{c^2-d^2}{cd}=\dfrac{\left(dk\right)^2-d^2}{dk\cdot d}=\dfrac{d^2k^2-d^2}{d^2k}=\dfrac{d^2\left(k-1\right)}{d^2k}=\dfrac{k-1}{k}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\left(dpcm\right)\)

c) \(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\dfrac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\dfrac{b^3}{d^3}\) (1)

\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\) (2)

Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{a^3+b^3}{c^3+d^3}\left(dpcm\right)\)

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