đây là toán đâu phải văn. bạn bị say rượu à

a) Ta có:

\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)

\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)

Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)

b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)

Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)

- Ta có:

\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)

- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)

a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)

=>Bằng nhau

b: a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)

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

a. \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)

\(=1.\frac{7}{12}:\frac{7}{12}\)

\(=1\)

b. 

\(\frac{5}{9}.\frac{8}{11}+\frac{5}{9}.\frac{9}{11}-\frac{5}{9}.\frac{6}{11}\)

\(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)

\(=\frac{5}{9}.1\)

\(=\frac{5}{9}\)

Tk mk nha!

b)  \(=\frac{5}{9}.\left(\frac{8}{11}+\frac{9}{11}-\frac{6}{11}\right)\)

\(=\frac{5}{9}.1\)

\(=\frac{5}{9}\)

a,\(=\frac{2}{3}+\left(\frac{-4}{27}+\frac{5}{18}\right):\frac{7}{12}\)=\(\frac{2}{3}+\frac{-16}{63}+\frac{10}{21}\)=\(\frac{8}{9}\)

\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72\left(not27\right)}=\frac{5.18-10.27+15.36}{4\left(5.18-10.27+15.36\right)}=\frac{1}{4}\)

\(\frac{\frac{-6}{7}+\frac{6}{19}-\frac{6}{31}}{\frac{9}{7}-\frac{9}{19}+\frac{9}{31}}=\frac{-6\left(\frac{1}{7}-\frac{1}{19}+\frac{1}{31}\right)}{9\left(\frac{1}{7}-\frac{1}{19}+\frac{1}{31}\right)}=\frac{-6}{9}=\frac{-2}{3}\)

a) \(\frac{a}{9}+\frac{3}{b}=\frac{1}{18}\Leftrightarrow\frac{ab+27}{9b}=\frac{1}{18}\left(1\right)\)

\(\Rightarrow9b=18\Rightarrow b=2\left(2\right)\)

Thay (2) vào (1) ta có 

\(\frac{2a+27}{18}=\frac{1}{18}\Rightarrow2a+27=1\Rightarrow2a=-26\Rightarrow a=-13\)

\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)

\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)

\(=-\frac{6}{9}=-\frac{2}{3}\)

a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875

b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067

c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333

                                     vậy a>b>c 

**************************l i k e***********************************8

A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2

B:  Tử âm ; mẫu dương => B < 0

C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)

= 25  \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)

Dễ có: B < A < C 

\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)

\(\frac{A}{B}=10\)

\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)

Tách 9=1+1+...+1 ( có 9 số 1)

\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)

\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)

\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )

Vậy \(A:B=10\)

a: \(A=\dfrac{1}{9}:\dfrac{1}{9}:\left(\dfrac{10+7}{15}:\dfrac{12-5}{30}\right)\)

\(=1:\left(\dfrac{17}{15}\cdot\dfrac{30}{7}\right)=1:\dfrac{34}{7}=\dfrac{7}{34}\)

b: \(=\left(5.6+0.64\right)\cdot1.25\cdot\dfrac{19}{3}+31.64\)

\(=\dfrac{39}{5}\cdot\dfrac{19}{3}+\dfrac{791}{25}=\dfrac{2026}{25}\)