Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{bk+b}{b}=\frac{dk+d}{d}\)

Xét VT \(\frac{bk+b}{b}=\frac{b\left(k+1\right)}{b}=k+1\left(1\right)\)

Xét VP \(\frac{dk+d}{d}=\frac{d\left(k+1\right)}{d}=k+1\left(2\right)\)

Từ (1) và (2) -->Đpcm

b)Đặt tương tự ta có:

\(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\Leftrightarrow\frac{5bk+3b}{5bk-3b}=\frac{5dk+3d}{5dk-3d}\)

Xét VT \(\frac{5bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-2}\left(1\right)\)

Xét VP \(\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\left(2\right)\)

Từ (1) và (2) -->Đpcm

Bạn xem lại đề nhé :)

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

2) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{5}{3}.\frac{a}{b}=\frac{5}{3}.\frac{c}{d}\Rightarrow\frac{5a}{3b}-1=\frac{5c}{3d}-1\Rightarrow\frac{5a-3b}{3b}=\frac{5c-3d}{3d}\)

\(\Rightarrow\frac{3b}{5a-3b}=\frac{3d}{5c-3d}\Rightarrow\frac{6b}{5a-3b}=\frac{6d}{5c-3d}\Rightarrow\frac{6b}{5a-3b}+1=\frac{6d}{5c-3d}+1\)

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

1) Vì a/b = c/d

=> a/b + 1 = c/d + 1

=> a + b/b = c + d/d (đpcm)

2) Vì a/b = c/d

=> a/c = b/d

=> 5a/5c = 3b/3d = 5a + 3b/5c + 3d = 5a - 3b/5c - 3d ( theo tc DTSBN )

=> 5a + 3b/5a - 3b = 5c + 3d/5c - 3d

1,a/b=c/d

=>\(\frac{a}{b}+1=\frac{c}{d}+1\)

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

a) \(\dfrac{a}{b}=\dfrac{c}{d}\left(a;b;c;d\ne0\right)\)

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

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

\(\Rightarrow dpcm\)

b) \(\dfrac{a}{b}=\dfrac{c}{d}\)

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

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

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

\(\Rightarrow dpcm\)

Thanks

Lời giải:

Đặt $\frac{a}{b}=\frac{c}{d}=k$

$\Rightarrow a=bk; c=dk$. Khi đó:

1.

$\frac{a+b}{b}=\frac{bk+b}{b}=\frac{b(k+1)}{b}=k+1(1)$

$\frac{c+d}{d}=\frac{dk+d}{d}=\frac{d(k+1)}{d}=k+1(2)$
Từ $(1); (2)\Rightarrow \frac{a+b}{b}=\frac{c+d}{d}$

2.

$\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}(3)$

$\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}(4)$

Từ $(3); (4)\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}$ (đpcm)

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

a) \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)\(=\dfrac{\dfrac{a}{k}.b}{\dfrac{c}{k}.d}=\dfrac{ab}{cd}=VT\)

Vậy...

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

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

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

c) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(bk\right)^2+3\left(bk\right).b}{11\left(bk\right)^2-8b^2}\)\(=\dfrac{7k^2+3k}{11k^2-8}\)

\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3\left(dk\right).d}{11\left(dk\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)

Suy ra \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)

a) Có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

=> \(ad=bc\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\) => \(\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)

(theo tính chất dãy tỉ số bằng nhau)

=> (đpcm)

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

=> \(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)(theo tính chất dãy tỉ số bằng nhau)

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

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

=> \(\dfrac{a^2}{c^2}=\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)          => \(\dfrac{7a^2}{7c^2}=\dfrac{3ab}{3cd}=\dfrac{11a^2}{11c^2}=\dfrac{8b^2}{8d^2}\)

=> \(\dfrac{7a^2+3ab}{7c^2+3cd}=\dfrac{11a^2-8b^2}{11c^2-8d^2}\) (theo tính chất dãy tỉ số bằng nhau)

=> \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)(đpcm)

#Ayumu

tra mạng có bài giải chi tiết rồi đó bạn

cảm ơn

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

vãi cả loz sao lại sai ?