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

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

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

\(2A=1-\frac{1}{3^8}\)

\(A=\frac{1}{2}-\frac{1}{2.3^8}\)

lm hộ mik nhé mik cần gấp

1b) Ta có: \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}......\frac{101}{100}=\frac{3.4.5....101}{2.3.4....100}=\frac{101}{2}\)

AI K CHO MU TA THẾ

MK NÈ, BẠN ẤY LÀM TỐT MÀ!

\(1,2\sqrt{27}+5\sqrt{12}-3\sqrt{48}\\ =2.3\sqrt{3}+5.2\sqrt{3}-3.4\sqrt{3}\\ =6\sqrt{3}+10\sqrt{3}-12\sqrt{3}\\ =4\sqrt{3}\)

\(2,\sqrt{147}+\sqrt{75}-4\sqrt{27}\\ =7\sqrt{3}+5\sqrt{3}-4.3\sqrt{3}\\ =7\sqrt{3}+5\sqrt{3}-12\sqrt{3}\\ =\sqrt{3}\left(7+5-12\right)\\ =0\)

\(3,3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\\ =3\sqrt{2}.\left(4-\sqrt{2}\right)+3\left(1-4\sqrt{2}+8\right)\\ =12\sqrt{2}-6+3-12\sqrt{2}+24\\ =21\)

\(4,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\\ =\sqrt{5}\left(2-5-4+11\right)\\ =4\sqrt{5}\)

1: =6căn 3+10căn 3-12căn 3=4căn 3

2: =7căn 3+5căn 3-12căn 3=0

3: =12căn 2-6+3(9-4căn 2)

=12căn 2-6+27-12căn 2=21

4: =2căn 5-5căn 5+4căn 5+9 căn 5

=10căn 5

\(-\frac{1}{7}+\frac{5}{3}+\frac{5}{4}+\frac{1}{3}-\frac{3}{2}\)

\(=\left(-\frac{1}{7}+\frac{5}{3}-\frac{3}{2}\right)+\left(\frac{5}{3}+\frac{1}{3}\right)\)

\(=\frac{-6}{42}+\frac{70}{42}-\frac{63}{42}+\frac{6}{3}\)

\(=\frac{-6+70-63}{42}+2\)

\(=\frac{1}{42}+\frac{84}{42}\)

\(=\frac{85}{42}\)

21/4x32/5=168/5

21/5:11/3=63/55

a)\(2\frac{1}{4}\times3\frac{2}{5}\)

\(=\frac{9}{4}\times\frac{17}{5}\)

\(=\frac{153}{20}\)

b)\(2\frac{1}{5}:1\frac{1}{3}\)

\(=\frac{11}{5}:\frac{4}{3}\)

\(=\frac{11}{5}\times\frac{3}{4}\)

\(=\frac{33}{20}\)

Cứ thấy có vấn đề...

Linz

\(2\frac{1}{4}\times3\frac{2}{5}=\frac{9}{4}\times\frac{17}{5}=\frac{9\times17}{4\times5}=\frac{153}{20}\)

Hok Tốt !!!!!!!!!!!!!!!!!!!!

a) \(4\times\dfrac{1}{4^2}+25\times\left[\dfrac{3^3}{4^3}:\dfrac{5^3}{4^3}\right]:\dfrac{3^3}{2^3}\)

\(=\dfrac{1}{4}+5^2\times\dfrac{3^3}{4^3}\times\dfrac{4^3}{5^3}\times\dfrac{2^3}{3^3}\)

\(=\dfrac{1}{4}+\dfrac{2^3}{5}=\dfrac{1}{4}+\dfrac{8}{5}=\dfrac{37}{20}\)

b) \(2^3+3\times\left(\dfrac{1}{2}\right)^{0-1}+\left[\left(-2\right)^2:\dfrac{1}{2}\right]-8\)

\(=8+3\times\left(2^{-1}\right)^{-1}+2^2\times2-8\)

\(=3\times2+2^3=14\)

Muộn rùi ngủ thôi không mai lớn có một khúc à :v 

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn nhé. Viết thế này người đọc đề sẽ rất mệt.

Bổ sung câu `c,`

`c, (-1/3)^2 - (-3/5)^0 + (1/2)^2 : 2`

`= 1/9 - 1 + 1/4 xx 1/2`

`= 1/9 - 1 + 1/8`

`= 1/9 - 9/9 +1/8`

`= -8/9 +1/8`

`= -55/72`

Giải:

(1-1/2²).(1-1/3²).(1-1/4²).....(1-1/10²)

=3/2.2 . 8/3.3 . 15/4.4 . ... . 99/10.10

=1.3.2.4.3.5.....9.11/2.2.3.3.4.4.....10.10

=1.2.3.....9/2.3.4.....10 . 3.4.5....11/2.3.4.....10

=1/10.11/2

=11/20

Chúc bạn học tốt!