\(Cách\)\(1:\)

\(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow\text{a=bk;c=dk (1)}\)

Ta có:\(\frac{a}{3a+b}=\frac{c}{3c+d}\)(thay(1) vào)

Ta dc:\(\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)(tiếp tục thay 1 vào)

\(\frac{dk}{3dk+1}=\frac{k}{3k+1}\)

\(Từ\)\(\left(1\right);\left(2\right)\RightarrowĐPCM\)

\(Cách\)\(2:\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow3ac+ad=3ac+bc\)

\(\Rightarrow\text{a(3c+d)=c(3a+b)}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\left(ĐPCM\right)\)

Chúc bn hok tốt!!!

Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)

\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)

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

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

a.d = b.c ⇒ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}\) = \(\dfrac{3a}{3c}=\dfrac{2b}{2d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}\) (1)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

Từ (1) và(2) ta có:

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

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a.b}{c.d}=\dfrac{a^2}{c^2}\) = \(\dfrac{b^2}{d^2}\) = \(\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)

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

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

\(\Rightarrow dpcm\)

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

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

\(\Rightarrow dpcm\)

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

\(\Rightarrow\left(b-3d\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)

\(\Rightarrow dpcm\)

Đính chính câu c

\(\Rightarrow\left(3a+5c\right)\left(b-3d\right)=\left(3b+5d\right)\left(a-3c\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)

Ta có : 

\(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)

\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)

Từ 1 và 2 

=> \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b.k,c=d.k\)

Ta có:

\(\frac{a}{3a+b}=\frac{b.k}{3.b.k+b}=\frac{b.k}{b.\left(3.k+1\right)}=\frac{k}{3.k+1}\) (1)

\(\frac{c}{3c+d}=\frac{d.k}{3.d.k+d}=\frac{d.k}{d.\left(3.k+1\right)}=\frac{k}{3.k+1}\) (2)

Từ (1) và (2) suy ra \(\frac{a}{3a+b}=\frac{b}{3c+d}\)

a, Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}\)

Áp dụng tính chất của day tỉ số bằng nhau ta được:

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

\(=>\dfrac{a}{c}=\dfrac{3a+b}{3c+d}=>\dfrac{a}{3a+b}=\dfrac{c}{3c+d}=>\left(đpcm\right)\)

Bài 1:

Ta có:\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta được:

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

(ĐPCM)

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

Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=x\)

Xét \(x^2=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{ab}{cd}\)

=>(đpcm)