\(\dfrac{a+4}{a-4}=\dfrac{b+5}{b-5}\)

=>\(\left(a+4\right)\left(b-5\right)=\left(a-4\right)\left(b+5\right)\)

\(\Leftrightarrow ab-5a+4b-20=ab+5a-4b-20\)

\(\Leftrightarrow-10a=-8b\)

=>a/b=4/5

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

f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)

l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)

m: \(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)

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

nhân ra ik ròi suy ra đpcm :D

\(đk:a;b\ne\dfrac{5}{3}\)

\(\dfrac{3b-28}{3a-5}-\dfrac{38-3a}{5-3b}=\dfrac{3b-28}{3\left(11+b\right)-5}-\dfrac{38-3\left(11+b\right)}{5-3b}=1-1=0\)

Bài 1:

Giải:

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\)

Vậy x = 6, y = 10

Bài 2:

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

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

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

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

\(\Rightarrow10b=12a\)

\(\Rightarrow6a=5b\)

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

\(\Rightarrowđpcm\)

B1 :

+ Theo bài ra :

\(\dfrac{x}{3}=\dfrac{y}{5}\left(1\right)\)và \(x+y=16\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)

+ Do đó :

\(\dfrac{x}{3}=2\Rightarrow x=2.3=6\)

\(\dfrac{y}{5}=2\Rightarrow y=2.5=10\)

Vậy x = 6 ; y = 10

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

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

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

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

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

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

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

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)