\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\Rightarrow\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{c\left(a+b+c\right)}\Rightarrow c\left(a+b\right)\left(a+b+c\right)=ab\left(-a-b\right)\)

\(\Rightarrow\left(a+b\right)\left(ca+cb+c^2\right)+ab\left(a+b\right)=0\Rightarrow\left(a+b\right)\left(ca+cb+c^2+ab\right)=0\)

\(\Rightarrow\left(a+b\right)\left(c+a\right)\left(b+c\right)=0\)

=> Trong 3 số a,b,c có 2 số đối nhau.Giả sử a = -b thì a⁹ + b⁹ = 0.

Tương tự giả sử b = -c hay a = -c thì b⁹⁹ + c⁹⁹ = 0 hay c⁹⁹⁹ + a⁹⁹⁹ = 0

Vậy biểu thức cần tính bằng 0.

bằng 0 quá dễ Hi Hi !!!

biểu thức trên = 0 đó

\(\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{1}{5}-\frac{1}{7}-\frac{2}{35}\right)\)

\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{7}{35}-\frac{5}{35}-\frac{2}{35}\right)\)

\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right).0\)

\(=0\)

bài dễ thế không ai làm được hay thật

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\text{ }\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)

\(Q=0\)

Q=(1/99+12/999+123/999).(1/2-1/3-1/6) =(1/99+12/999+123/999).0 Q=0

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(\Leftrightarrow Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\times0\)

\(\Leftrightarrow Q=0\)

Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\) 

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).0\) 

\(Q=0\)

A=(9/1999+99/999+999/9999).(1/5-1/4+1/20)

A=(9/1999+99/999+999/9999).(-1/20+1/20)

A=(9/1999+99/999+999/9999).0

A=0

Vì mọi số nhân vs 0 thì đều = 0 kể cả phân số

mk nhanh nhất ủng hộ nha

\(A=\left(\frac{9}{1999}+\frac{99}{999}+\frac{999}{9999}\right)\cdot0\)

A=0

\(\left( { - 1} \right) + \left( { - 3} \right) =  - \left( {1 + 3} \right) =  - 4\)

\(\left( { - 3} \right) + \left( { - 1} \right) =  - \left( {3 + 1} \right) =  - 4\)

\( \Rightarrow \left( { - 1} \right) + \left( { - 3} \right) = \left( { - 3} \right) + \left( { - 1} \right)\)

\(\left( { - 7} \right) + \left( { + 6} \right) =  - \left( {7 - 6} \right) =  - 1\)

\(\left( { + 6} \right) + \left( { - 7} \right) =  - \left( {7 - 6} \right) =  - 1\)

\( \Rightarrow \left( { - 7} \right) + \left( { + 6} \right) = \left( { + 6} \right) + \left( { - 7} \right)\)

\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)

\(=0\)

\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot\dfrac{0}{6}\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot0\)

\(Q=0\)