Mik có sửa lại đề

C1: \(\dfrac{6}{7}\times\dfrac{1}{2}+\dfrac{6}{7}\times\dfrac{1}{4}=\dfrac{3}{7}+\dfrac{3}{14}=\dfrac{6}{14}+\dfrac{3}{14}=\dfrac{9}{14}\)

C2:\(\dfrac{6}{7}\times\dfrac{1}{2}+\dfrac{6}{7}\times\dfrac{1}{4}=\dfrac{6}{7}\times\left(\dfrac{1}{2}+\dfrac{1}{4}\right)=\dfrac{6}{7}\times\left(\dfrac{2}{4}+\dfrac{1}{4}\right)=\dfrac{6}{7}\times\dfrac{3}{4}=\dfrac{9}{14}\)

  1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6+1/6x1/7+1/7x1/8+1/8x1/9

=7/18

\(1\frac{1}{3}.1\frac{1}{4}.1\frac{1}{5}.1\frac{1}{6}.1\frac{1}{7}.1\frac{1}{8}=\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}.\frac{8}{7}.\frac{9}{8}=\frac{9}{3}=3\)

=11/3x4x5x6x7x8

=11/20160

Bài 1 :

Gọi \(A=5+5^2+5^3+...+5^{98}+5^{99}\\ 5A=5^2+5^3+5^4+...+5^{99}+5^{100}\\ 5A-A=\left(5^2+5^3+5^4+...+5^{99}+5^{100}\right)-\left(5+5^2+5^3+...+5^{98}+5^{99}\right)\\ 4A=5^{100}-5\\ A=\dfrac{5^{100}-5}{4}\)

Bài 2:

\(\left(12x-4\right)\cdot8^{2022}=4\cdot8^{2023}\\ 12x-4=4\cdot8^{2023}:8^{2022}\\ 12x-4=4\cdot8\\ 12x-4=32\\ 12x=36\\ x=3\)

= -97/1176 ^{lớp 4 học số âm à???}

-97/1176 

-97/1176 

\(=\dfrac{3}{7}\times\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{1}{8}\right)\\ =\dfrac{3}{7}\times\left(\dfrac{24}{5\times24}+\dfrac{2\times40}{3\times40}-\dfrac{15}{8\times15}\right)\\ =\dfrac{3}{7}\times\left(\dfrac{24+80-15}{120}\right)\\ =\dfrac{3}{7}\times\dfrac{89}{120}\\ =\dfrac{89}{280}\)

\(\dfrac{3}{7}\times\dfrac{1}{5}+\dfrac{2}{3}\times\dfrac{3}{7}-\dfrac{3}{7}\times\dfrac{1}{8}\)
\(=\dfrac{3}{7}\times\left(\dfrac{1}{5}+\dfrac{3}{7}-\dfrac{1}{8}\right)\)
\(=\dfrac{3}{7}\times\dfrac{56+120-35}{280}\)
\(=\dfrac{3}{7}\times\dfrac{141}{280}\)
\(=\dfrac{423}{1960}\)