\(\dfrac{3a-b}{a+b}=\dfrac{3}{4}\)

\(\Rightarrow3\left(a+b\right)=4\left(3a-b\right)\)

\(\Rightarrow3\left(a+b\right)-4\left(3a-b\right)=0\)

\(\Rightarrow\left(3a+3b\right)-\left(12a-4b\right)=0\)

\(\Rightarrow3a+3b-12a+4b=0\)

\(\Rightarrow-9a+7b=0\)

\(\Rightarrow-9a=-7b\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{-7}{-9}\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{7}{9}\)

Vậy giá trị tỉ số \(\dfrac{a}{b}=\dfrac{7}{9}\)

Bài 2: 

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

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

b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)

\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)

a.Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

b.Vì\(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

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

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

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

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

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

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

*a/b=c/d=k=>a=bk;c=dk

Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3

Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3

=>2a+3b/2a-3b=2c+3d/2c-3d

*a/b=c/d=>a/c=b/d=k

=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)

k^2=a/c.b/d=ab/cd (2)

Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2

*a/b=c/d=>a/c=b/d=k=a+b/c+d

=>k^2=(a+b/c+d)^2

k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2

=>(a+b/c+d)^2=a^2+b^2/c^2+d^2

Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)

a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)

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

Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)

b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)

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

Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)

c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)

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

Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\)

\(\Rightarrow ab\left(b+c\right)=bc\left(a+b\right)\)

\(\Rightarrow ab^2+abc=abc+b^2c\)

\(\Rightarrow ab^2=b^2c\)

\(\Rightarrow a=c\)

Đến đây ko còn manh mối :v

Chọn D

1)

a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)

\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)

\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)

\(\Rightarrow x=-\dfrac{55}{6}\)

b) 4,25 : 8 = -3,5 : x

\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)

\(x=\dfrac{-3,5.8}{4,25}\)

\(x=\dfrac{-28}{4,25}\)

2.

\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)

\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)

Ta có thể lặp đc các tỉ lệ thức sau:

\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)

\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)

\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)

\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)

Từ TLT: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{\left(a+c\right)}{\left(b+d\right)}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{ac}{bd}\left(dpcm\right)\)

a, ta có :

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

áp dụng tính chất dă y tỉ số bằng nhau ta có:

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

\(\Rightarrow\dfrac{a+2b}{c+2d}=\dfrac{2a-b}{2c-d}\Rightarrow\dfrac{a+2b}{2a-b}=\dfrac{c+2d}{2c-d}\) (ĐPCM)

b, ta có:

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

áp dụng tính chất dă tỉ số bằng nhau ta có :

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

\(\Rightarrow\dfrac{a+3c}{b+3d}=\dfrac{a-c}{b-d}\)

\(\Rightarrow\left(a+3c\right)\left(b-d\right)=\left(b+3d\right)\left(a-c\right)\) (ĐPCM)