\(A=\dfrac{1}{7}\left(\dfrac{555}{777}+\dfrac{4444}{12221}+\dfrac{33333}{244442}+\dfrac{11}{30}+\dfrac{13}{30}\right)\)

\(=\dfrac{1}{7}\left(\dfrac{5}{7}+\dfrac{4}{11}+\dfrac{3}{22}+\dfrac{23}{30}\right)\\ =\dfrac{1}{7}\left(\dfrac{5}{7}+\dfrac{1}{2}+\dfrac{23}{30}\right)\\ =\dfrac{1}{7}\cdot\dfrac{208}{105}\\ =\dfrac{208}{735}\)

(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2.................999 x 2 x 2 x 2 ?

A.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 > 999 x 2 x 2 x 2 = 9990 > 7992

B.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 < 999 x 2 x 2 x 2 = 9990 < 7992

C.(111 + 222 + 333 + 444 + 555 + 666 + 777 + 888 + 999)x 2 = 999 x 2 x 2 x 2 = 9990 = 7992

Đáp án A nhé

1110 + 1110 + 1110 + 1110 + 1110 = 5550

ko rảnh nha thằng kia

Ta có: \(ac=b^2\)

=>\(\frac{a}{b}=\frac{b}{c}\)

Ta có: \(ab=c^2\)

=>\(\frac{b}{c}=\frac{c}{a}\)

=>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\)

=>\(\begin{cases}c=ak\\ b=ck=ak\cdot k=ak^2\\ a=bk=ak^2\cdot k=ak^3\end{cases}\Rightarrow\begin{cases}c=ak\\ b=ak^2\\ a\left(k^3-1\right)=0\end{cases}\)

=>\(\begin{cases}c=ak\\ b=ak^2\\ k^3-1=0\end{cases}\Rightarrow\begin{cases}k=1\\ c=a\cdot1=a\\ b=a\cdot1^2=a\end{cases}\)

=>a=b=c

\(P=\frac{a^{555}}{b^{222}\cdot c^{333}}+\frac{b^{555}}{c^{222}\cdot a^{333}}+\frac{c^{555}}{a^{222}\cdot b^{333}}\)

\(=\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}\)

=1+1+1

=3

 111 + 333 + 666 + 777 +999 +444 + 555 +222

=(111+999) + (333+777) + (666 + 444) +(555+ 222)

=1110 + 1110+1110 + 777

=1110 * 3 + 777

= 3330+ 777=4107

= 4885 nha tk mk nhé , xin bn lun đó

bằng 4995 nha

\(999+888+777+666+555+444+333+222+111\)

\(=\left(999+111\right)+\left(888+222\right)+\left(777+333\right)+\left(666+444\right)+555\)

\(=1110+1110+1110+1110+555\)

\(=\left(1110\times4\right)+555\)

\(=4440+555\)

\(=4995\)