\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

=>(a+5)(b-6)=(a-5)(b+6)

=>ab-6a+5b-30=ab+6a-5b-30

=>-6a+5b=6a-5b

=>-12a=-10b

=>6a=5b

hay a/b=5/6

1.Tìm x, biết:

=>x=6; y=10

2.Cho

CMR:

Lại có: (*)=a+5-a+5/b+6-b+6=10/12=5/6 (2)

Từ 1 và 2 suy ra a/b=5/6 (đpcm)

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\\ \Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\\ \Leftrightarrow12a=10b\\ \Leftrightarrow6a=5b\Leftrightarrow\dfrac{a}{b}=\dfrac{5}{6}\)

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)

nhân ra ik ròi suy ra đpcm :D

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Rightarrow ab+5b-6a-30=ab-5b+6a-30\)

\(\Rightarrow5b-6a=-5b+6a\)

\(\Rightarrow10b=12a\)

\(\Rightarrow5b=6a\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\left(đpcm\right)\)

Vậy \(\dfrac{a}{b}=\dfrac{5}{6}\)

\(\dfrac{a+5}{a-5}=\dfrac{a+6}{a-6}\)suy ra \(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(a+6\right)\)

suy ra: \(6a=5b\)

suy ra: \(\dfrac{a}{b}=\dfrac{5}{6}\)

Một cách giải khác:

TH1: b = -6

VP = 0 => VT = 0 => a = -5

=> \(\dfrac{a}{b}=\dfrac{-5}{-6}=\dfrac{5}{6}\)

TH2: b \(\ne\) -6 nên:

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)

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

\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{a}{b}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}\)

\(\Leftrightarrow7\cdot\left(x-1\right)=6\cdot\left(x+5\right)\)

\(\Leftrightarrow7x-7-6x-30=0\)

\(\Leftrightarrow x-37=0\)

\(\Leftrightarrow x=37\left(N\right)\)

Ta có: \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

\(\Leftrightarrow7x-7-6x-30=0\)

\(\Leftrightarrow x=37\)

\(a)\)\(\dfrac{a}{b}\times4+\dfrac{1}{6}=\dfrac{19}{6}\)

   \(\dfrac{a}{b}\times4=\dfrac{19}{6}-\dfrac{1}{6}\)

   \(\dfrac{a}{b}\times4=\dfrac{18}{6}\)

   \(\dfrac{a}{b}=\dfrac{18}{6}\div4\)

   \(\dfrac{a}{b}=\dfrac{18}{6}\times\dfrac{1}{4}\)

\(\dfrac{a}{b}=\dfrac{18}{24}\)

Sửa câu a:

(x - 2)² - 36 = 0

(x - 2 - 6)(x - 2 + 6) = 0

(x - 8)(x + 4)= 0

\(\Leftrightarrow \begin{bmatrix} x - 8= 0 & & \\ x + 4 = 0 & & \end{bmatrix}\)

\(\Leftrightarrow \begin{bmatrix} x = 8 & & \\ x = - 4 & & \end{bmatrix}\)

pn bỏ dấu ngoặc bên phải nhé

Vậy x = 8; x = - 4

2:

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)

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

\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{2a}{2b}=\dfrac{a}{b}\)

Từ đó suy ra \(\dfrac{a}{b}=\dfrac{5}{6}\)

\(\RightarrowĐPCM\)

1.

a) \(\frac{x - 3}{5}=\frac{7}{x - 1} \)

(x - 3)(x - 1) = 35

x² - x - 3x + 3 - 35 = 0

x² - 4x - 32 = 0

x² - 4x + 4 - 36 = 0

(x - 2)² - 36 = 0

(x - 2 - 36)(x - 2 + 36) = 0

(x - 38)(x + 34) = 0

\(\Leftrightarrow \begin{bmatrix} x - 38 = 0 & & \\ x + 34 = 0 & & \end{bmatrix}\)

\(\Leftrightarrow \begin{bmatrix} x = 38 & & \\ x = - 34 & & \end{bmatrix}\)

pn bỏ dấu ngoặc bên phải nhé

Vậy x = 38 ; x = - 34

b) (x - 5)^{x + 1} - (x - 5)^{x + 11} = 0

\((x - 5)^{x + 1}\left [ 1- (x - 5)^{10} \right ]= 0\)

tới đây pn giải giống câu a