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

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

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

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

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

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

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

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

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

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

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

b) \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\frac{2a}{2c}=\frac{5b}{5d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\)

\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

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

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.

Bài 3:

a) Sửa điều kiện: \(\frac{a}{b}=\frac{c}{d}\neq -1\)

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

Theo đkđb thì \(k\neq -1\) nên \(k^3+1\neq 0\); \(k+1\neq 0\)

Ta có: \(\frac{a^3+b^3}{c^3+d^3}=\frac{(bk)^3+b^3}{(dk)^3+d^3}=\frac{b^3(k^3+1)}{d^3(k^3+1)}=\frac{b^3}{d^3}\)

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

\(\Rightarrow \frac{a^3+b^3}{c^3+d^3}=\frac{(a+b)^3}{(c+d)^3}\) (đpcm)

b)

Đặt \(\frac{a}{b}=k; \frac{c}{d}=t\Rightarrow a=bk; c=dt\)

Ta cần cm \(k=t\)

Khi đó:

\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b(2k+13)}{b(3k-7)}=\frac{2k+13}{3k-7}\)

\(\frac{2c+13d}{3c-7d}=\frac{2dt+13d}{3dt-7d}=\frac{d(2t+13)}{d(3t-7)}=\frac{2t+13}{3t-7}\)

Vì \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\Rightarrow \frac{2k+13}{3k-7}=\frac{2t+13}{3t-7}\)

\(\Rightarrow (2k+13)(3t-7)=(2t+13)(3k-7)\)

\(-14k+39t=-14t+39k\Rightarrow k=t\)

Ta có đpcm.

\(\dfrac{4a^2}{a-1}=\dfrac{a\left(a^2-1\right)+4}{a-1}=4\left(a+1\right)+\dfrac{4}{a-1}+8\ge8+8=16\)

\(\dfrac{5b^2}{b-1}=5\left(b-1\right)+\dfrac{5}{b-1}+10\ge20\)

\(\dfrac{3c^2}{c-1}=3\left(c-1\right)+\dfrac{3}{c-1}+6=12\)

\(\Rightarrow dpcm\)

Bằng 12

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