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

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

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

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

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

\(\Rightarrow10b=12a\)

\(\Rightarrow5b=6a\)

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

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:

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

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 \(B=3\)

\(a^5+b^2+ab+6\ge3a^2b+6\)

\(\Rightarrow P\le\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{\sqrt{a^2b+2}}+\dfrac{1}{\sqrt{b^2c+2}}+\dfrac{1}{\sqrt{c^2a+2}}\right)\le\sqrt{\dfrac{1}{a^2b+2}+\dfrac{1}{b^2c+2}+\dfrac{1}{c^2a+2}}=\sqrt{Q}\)

\(Q=\dfrac{c}{a+2c}+\dfrac{a}{b+2a}+\dfrac{b}{c+2b}=\dfrac{1}{2}\left(1-\dfrac{a}{a+2c}+1-\dfrac{b}{b+2a}+1-\dfrac{c}{c+2b}\right)\)

\(Q=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{a^2}{a^2+2ac}+\dfrac{b^2}{b^2+2ab}+\dfrac{c^2}{c^2+2bc}\right)\)

\(Q\le\dfrac{3}{2}-\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)

\(\Rightarrow P\le\sqrt{1}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

a) Ta có: \(a\left(-\dfrac{3}{2}\right)+a\cdot\dfrac{1}{4}-a\cdot\dfrac{5}{6}\)

\(=a\left(-\dfrac{3}{2}+\dfrac{1}{4}-\dfrac{5}{6}\right)\)

\(=a\left(\dfrac{-18}{12}+\dfrac{3}{12}-\dfrac{10}{12}\right)\)

\(=a\cdot\dfrac{-25}{12}\)(1)

Thay \(a=\dfrac{3}{5}\) vào biểu thức (1), ta được:

\(\dfrac{3}{5}\cdot\dfrac{-25}{12}=\dfrac{-75}{60}=\dfrac{-5}{4}\)

35⋅−2512=−7560=−54

a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)

b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)

c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)

d)\(\dfrac{a}{b}=3\times\dfrac{2}{7}=\dfrac{6}{7}\)

e)\(\dfrac{a}{b}=\dfrac{7}{5}-\left(\dfrac{2}{5}\times\dfrac{1}{2}\right)=\dfrac{7}{5}-\dfrac{1}{5}=\dfrac{6}{5}\)

a 22/5

b 59/42

c 9/10

d 6/7

e 6/5

bài 1

a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)

d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)

bài 2

a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)

b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)

bài 4 :

145  69 + 22 x  145 +145 x 8 + 145 

\(=145\times\left(69+22+8+1\right)=145\times100=14500\)

bài 1:

a, \(\dfrac{2}{5}+\dfrac{3}{8}=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b,\(\dfrac{7}{6}-\dfrac{2}{3}=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c,\(\dfrac{5}{9}x6=\dfrac{5}{9}x\dfrac{6}{1}=\dfrac{30}{9}\)

d,\(\dfrac{8}{5}:\dfrac{4}{7}=\dfrac{8}{5}x\dfrac{7}{4}=\dfrac{14}{5}\)

bài 2 :

\(a,\dfrac{4}{5}+x=\dfrac{5}{6}\)

            \(x=\dfrac{5}{6}-\dfrac{4}{5}\)

            \(x=\dfrac{1}{30}\)

b, \(x:\dfrac{7}{10}=5\)

    \(x\)         \(=5x\dfrac{7}{10}\)

    \(x\)         \(=\dfrac{35}{10}\)    

bài 3 :

đổi :16 tấn 8 tạ = 168 tạ

        2 tấn 6 tạ = 26 tạ 

xe ô tô thứ nhất chở số tạ hàng là:

      ( 168 + 26 ) : 2= 97 ( tạ)

xe ô tô thứ hai chở số tạ hàng là:

         97 - 26 = 71 ( tạ)

               đáp số :xe ô tô thứ nhất : 97 tạ thóc

                            xe ô tô thứ hai  : 71 tạ thóc