\(\frac{16}{27}\) tích mình nha !

\(A=\frac{9}{7}\times\frac{13}{5}\times\frac{9}{4}=\frac{1053}{140}\)

\(B=\frac{27}{5}\times\frac{13}{7}\times\frac{1}{4}=\frac{351}{140}\)

Vậy \(\frac{A}{B}=A:B=\frac{1053}{140}:\frac{351}{140}=\frac{1053}{351}=3\)

=81/81+27/81+9/81+3/81+1/81=121/81

1. Tinh:

a) 1 + 1 phần 3 + 1 phần 9 + 1 phần 27 + 1 phần 81 = 81/81 + 27/81 + 9/81 + 3/81 + 1/81 = 121/81

đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

\(\Rightarrow3S=3+1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}\)

\(\Rightarrow3S-S=3-\frac{1}{3^4}\)

\(\Rightarrow2S=\frac{80}{81}\)

\(\Rightarrow S=\frac{\frac{80}{81}}{2}\)

\(=\frac{80}{162}\)

\(=\frac{40}{81}\)

tham khảo:

\(a) 2+5+8+...+(3n−1)=n(3n+1)2 (1) Đặt Sn=2+5+8+...+(3n−1) Với n=1 ta có: S1=2=1(3.1+1)2 Giả sử (1) đúng với n=k(k≥1), tức là Sk=2+5+8+...+(3k−1)=k(3k+1)2 Ta chứng minh (1) đúng với n=k+1 hay Sk+1=(k+1)(3k+4)2 Thật vậy ta có: Sk+1=2+5+8+...+(3k−1)+[3(k+1)−1]=Sk+3k+2=k(3k+1)2+3k+2=3k2+k+6k+42=3k2+7k+42=(k+1)(3k+4)2 Vậy (1) đúng với mọi k≥1 hay (1) đúng với mọi n∈N∗ b) 3+9+27+...+3n=12(3n+1−3) (2) Đặt Sn=3+9+27+...+3n=12(3n+1−3) Với n=1, ta có: S1=3=12(32−3) (hệ thức đúng) Giả sử (2) đúng với n=k(k≥1) tức là Sk=3+9+27+...+3k=12(3k+1−3) Ta chứng minh (2) đúng với n=k+1, tức là chứng minh Sk+1=12(3k+2−3) Thật vậy, ta có: Sk+1=3+9+27+...+3k+1=Sk+3k+1=12(3k+1−3)+3k+1=32.3k+1−32=12(3k+2−3)(đpcm) Vậy (2) đúng với mọi k≥1 hay đúng với mọi n∈N∗\)

\(=1\frac{364}{729}\)\(=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}=1+\frac{243}{729}+\frac{81}{729}+\frac{27}{729}+\frac{9}{729}+\frac{3}{729}+\frac{1}{729}=1\frac{ }{ }\)

\(=1\frac{364}{729}\)

\(\frac{-x}{4}=\frac{-9}{x}\)

\(\Rightarrow-x\cdot x=-9\cdot4\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow-x^2=-6^2\)

\(\Rightarrow-x=-6\)

\(\Rightarrow\) \(x=6\)

\(\frac{x-1}{9}=\frac{8}{3}\)

\(\Rightarrow\left(x-1\right)\cdot3=9\cdot8\)

\(\Rightarrow\) \(3x-3=72\)

\(\Rightarrow\) \(3x=72+3\)

\(\Rightarrow\) \(3x=75\)

\(\Rightarrow\) \(x=75\div3\)

\(\Rightarrow\) \(x=25\)

a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)

\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)

hay \(x=-\dfrac{41}{12}\)

Vậy: \(x=-\dfrac{41}{12}\)

\(\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}=\frac{\frac{9}{27}+\frac{3}{27}-\frac{1}{27}}{\frac{45}{27}+\frac{15}{27}-\frac{5}{27}}=\frac{\frac{9+3-1}{27}}{\frac{45+15-5}{27}}=\frac{\frac{11}{27}}{\frac{55}{27}}=\frac{11}{55}=\frac{1}{5}.\)

\(\dfrac{3}{9}=\dfrac{3:3}{9:3}=\dfrac{1}{3}\)

\(\dfrac{27}{45}=\dfrac{27:9}{45:9}=\dfrac{3}{5}\)

\(\dfrac{18}{36}=\dfrac{18:18}{36:18}=\dfrac{1}{2}\)

\(\dfrac{9}{27}=\dfrac{9:9}{27:9}=\dfrac{1}{3}\)

\(\dfrac{8}{24}=\dfrac{8:8}{24:8}=\dfrac{1}{3}\)

\(\dfrac{110}{220}=\dfrac{110:110}{220:110}=\dfrac{1}{2}\)

\(\dfrac{3}{9}=\dfrac{3:3}{9:3}=\dfrac{1}{3}\)

\(\dfrac{27}{45}=\dfrac{27:9}{45:9}=\dfrac{3}{5}\)

\(\dfrac{18}{36}=\dfrac{18:18}{36:18}=\dfrac{1}{2}\)

\(\dfrac{9}{27}=\dfrac{9:9}{27:9}=\dfrac{1}{3}\)

\(\dfrac{8}{24}=\dfrac{8:8}{24:8}=\dfrac{1}{3}\)

\(\dfrac{110}{220}=\dfrac{110:110}{220:110}=\dfrac{1}{2}\)