a: \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)\)

\(=\dfrac{37}{3}-3-4-\dfrac{3}{2}\)

\(=\dfrac{74-9}{6}-7=\dfrac{65}{6}-7=\dfrac{65-42}{7}=\dfrac{23}{7}\)

b: \(3\dfrac{5}{6}+2\dfrac{1}{6}\cdot6\)

\(=3+\dfrac{5}{6}+\dfrac{13}{6}\cdot6\)

\(=16+\dfrac{5}{6}=\dfrac{101}{6}\)

c: \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

\(=3+\dfrac{1}{2}+4+\dfrac{5}{7}-5-\dfrac{5}{14}\)

\(=2+\dfrac{7+10-5}{14}=2+\dfrac{12}{14}=2+\dfrac{6}{7}=\dfrac{20}{7}\)

d: \(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99+2}{22}=\dfrac{101}{22}\)

các bn giải mk nha

Làm như zậy bạn nhé ^_^"

Đặt :

\(A=\left(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\right)\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)( mik lười viết bạn thông cảm nhé !!! )

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)

\(A=\frac{1}{1}-\frac{1}{6}\)

\(A=\frac{5}{6}\)

Thay A vào biểu thức :

Ta có : \(\frac{5}{6}.10-x=0\)

=> \(\frac{25}{3}=x+0\)

=> \(x=\frac{25}{3}\)

(1/1*1/2+1/2*1/3+1/3*1/4+1/4*1/5+1/5*1/6)*10-x=0

=> (1/1.2+1/2.3+1/3.4+1/4.5+1/5.6)*10-x=0

=> (1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6)*10-x=0

=> (1-1/6)*10-x=0

=> 5/6*10-x=0

=> 25/3-x =0

=> x=0+25/3

=> x=25/3

Vậy x=25/3

a.\(\frac{1}{6}.6^x+6^x.36=6^{15}\left(1+6^3\right)\)

\(6^x.\frac{217}{6}=6^{15}.217\)

\(6^x=6^{16}\)

\(x=16\)

Huhu, giúp mk đi mừ!!!

B. 1/6 ; 2/3 ; 3/4 ; 5/4

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\)

\(\Leftrightarrow A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{4052169}-1\right)\left(\frac{1}{\text{​​}\text{​​}4056196}-1\right)\)

\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....\frac{-4056195}{\text{​​​​}4056196}\)

\(\Leftrightarrow A=\frac{\left(-1\right)3}{2^2}.\frac{\left(-2\right)4}{3^3}.\frac{\left(-3\right)5}{4^2}.....\frac{\left(-2013\right)2015}{\text{​​​​}2014^2}\)

\(\Leftrightarrow A=\frac{\left(-1\right)\left(-2\right)....\left(-2013\right)}{2.3...1014}.\frac{3.4......2015}{2.3......2014}\)

\(\Leftrightarrow A=\frac{-1}{1014}.\frac{2015}{2}=\frac{-2015}{4028}\)

VÌ \(\frac{-2015}{4028}< \frac{-1}{2}\)

\(\Rightarrow A< \frac{-1}{2}\Leftrightarrow A< B\)

Ta có \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)=\frac{-3}{2^2}.\frac{-8}{3^2}...\frac{-4056195}{2014^2}\)

\(=-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{2013.2015}{2014^2}\right)=-\left(\frac{1.3.2.4...2013.2015}{2.2.3.3...2014.2014}\right)\)

\(=-\left(\frac{\left(1.2.3...2013\right)\left(3.4.5...2015\right)}{\left(2.3.4...2014\right)\left(2.3.4...2014\right)}\right)=-\frac{2015}{2014.2}=-\frac{2015}{4028}< \frac{-2014}{4028}< \frac{1}{2}=B\)

=> A < B

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{2014^2}-1\right)=\frac{-3}{3^2}\cdot\frac{-8}{3^2}...\frac{-4056195}{2014^2}\)

\(A=-\left(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{2013.2015}{2014^2}\right)=-\left(\frac{1.3.2.4....2013.2015}{2.2.3.3...2014.2014}\right)\)

\(A=-\left(\frac{\left(1.2.3...2014\right)\left(3.4.5...2015\right)}{\left(2.3.4....2014\right)\left(2.3.4....2014\right)}\right)=-\frac{2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}< \frac{1}{2}=B\)

\(\Rightarrow A< B\)

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

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot\dfrac{5}{6}\)

\(=\dfrac{1}{6}\)

\(\cdot DuyNam\)

\(A=-\dfrac{7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=-\dfrac{7}{21}+\dfrac{4}{3}\)

\(A=1\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+-\dfrac{6}{9}\right)\)

\(B=\dfrac{2}{15}+-\dfrac{1}{9}\)

\(B=\dfrac{1}{45}\)

\(C=\left(-\dfrac{1}{5}+\dfrac{3}{12}\right)+-\dfrac{3}{4}\)

\(C=\dfrac{1}{20}+-\dfrac{3}{4}\)

\(C=-\dfrac{7}{10}\)