C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!

a) \(a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}\cdot a^{\dfrac{7}{6}}=a^{\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{7}{6}}=a^2\)

b) \(a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{4}}:a^{\dfrac{1}{6}}=a^{\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}}=a^{\dfrac{3}{4}}\)

c) \(\left(\dfrac{3}{2}a^{-\dfrac{3}{2}}\cdot b^{-\dfrac{1}{2}}\right)\left(-\dfrac{1}{3}a^{\dfrac{1}{2}}b^{\dfrac{2}{3}}\right)=\left(\dfrac{3}{2}\cdot-\dfrac{1}{3}\right)\left(a^{-\dfrac{3}{2}}\cdot a^{\dfrac{1}{2}}\right)\left(b^{-\dfrac{1}{2}}\cdot b^{\dfrac{2}{3}}\right)\)

\(=-\dfrac{1}{2}a^{-1}b^{-\dfrac{1}{3}}\)

dài thế ai mà tính đc

\(\frac{1}{3}\)

\(A=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(A=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4.(\frac{1}{9}-\frac{1}{7}-\frac{1}{11})}+\frac{3.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}{4.(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625})}\)

\(A=\frac{1}{4}+\frac{3}{4}\)(Vì\(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\ne0\)và\(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\ne0\))

\(A=1\)

Vậy A = 1

\(B=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-...-\frac{1}{6}-\frac{1}{2}\)

\(B=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-B=-\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+...+\frac{1}{2.1}\)

\(-B=-\frac{1}{9}-\frac{1}{10}+\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+...+1-\frac{1}{2}\)

\(-B=-\frac{1}{9}.2-\frac{1}{10}+1\)

\(-B=-\frac{2}{9}-\frac{1}{10}+1\)

\(-B=\frac{-20}{90}-\frac{9}{90}+\frac{90}{90}\)

\(-B=\frac{61}{90}\)

\(B=\frac{-61}{90}\)

Vậy\(B=\frac{-61}{90}\)

Linz

a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)

       \(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)

       \(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\) 

       \(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)

       \(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)

       \(=3.\frac{196}{597}\)

       \(=\frac{196}{199}\)

\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có : 

\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)

\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)

\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)

Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)

PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))

nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm

a)|-10|:(-2):(-5)+(-3)²

    =1+9

     =10

b)1+(-2)+3+(-4)+5+(-6)+...+21+(-22)

   =[1+(-2)]+[3+(-4)]+[5+(-6)]+...+[21+(-22]

   =(-1)+(-1)+(-1)+...+(-1)

Mà từ 1 đến 22 có:(22-1):1+1:2=11(cặp)

        Suy ra:1+(-2)+3+(-4)+5+(-6)+...+21+(-22)=(-11)

c)\(\frac{3}{4}.\frac{5}{9}+\frac{3}{4}.\frac{4}{9}\)

\(=\frac{3}{4}.\left(\frac{5}{9}+\frac{4}{9}\right)\)

\(=\frac{3}{4}\)

d)\(-\frac{4}{17}+\frac{5}{19}+-\frac{13}{17}+\frac{14}{19}+\frac{3}{115}\)

\(=\left[\left(-\frac{4}{17}\right)+\left(-\frac{13}{17}\right)\right]+\left(\frac{5}{19}+\frac{4}{19}\right)+\frac{3}{115}\)

\(=\left(-\frac{27}{17}\right)+1+\frac{3}{115}\)

\(=-\frac{1099}{1955}\)

e)\(\left(\frac{3}{4}+-\frac{7}{2}\right).\left(\frac{10}{11}+\frac{2}{22}\right)\)

\(=\left(\frac{3}{4}-\frac{14}{4}\right).\left(\frac{20}{22}+\frac{2}{22}\right)\)

\(=\left(-\frac{11}{4}\right).\left(\frac{22}{22}\right)\)

\(=-\frac{11}{4}\)