-5/9 x 7/13 + 5/9 x -6/13 + 2 5/9

= -5/9 x 7/13 + 5/9 x -6/13 + 23/9

= 5/9 x -7/13 + 5/9 x -6/13 + 23/9

= 5/9 x (-7/13 - 6/13) + 23/9

= 5/9 x -1 + 23/9

= -5/9 + 23/9

= 2

Câu 1: A

Câu 2: D

Câu 3: B

Câu 4: C

Câu 5: B

Câu 6: A

Câu 7: A

Câu 8: D

Ta có công thức: \(\dfrac{n\left(n-1\right)}{2}\)

Thay vào bài, ta được:

\(\dfrac{n\left(n-1\right)}{2}=91\\ n\left(n-1\right)=91.2\\ n\left(n-1\right)=182\\ 14\left(14-1\right)=182\)

Vậy \(n=14\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{20^2}< \dfrac{1}{19\cdot20}=\dfrac{1}{19}-\dfrac{1}{20}\)

Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

=>\(A< 1-\dfrac{1}{20}\)

=>A<1

=>0<A<1

=>A không là số tự nhiên

=>a<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)

=>a<1-\(\dfrac{1}{100}\)<\(\dfrac{3}{4}\)

=>a<\(\dfrac{3}{4}\)

\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\\ =\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}\)

Ta có: \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}-\dfrac{1}{100}\\ \dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{49}{100}< \dfrac{50}{100}=\dfrac{1}{2}\)

Hay \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}< \dfrac{1}{2}\) 

Vì \(\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+\dfrac{1}{6.6}+...+\dfrac{1}{100.100}=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

Vậy biểu thức \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

1/3² + 1/4² + 1/5² + 1/6² + ... + 1/100²

< 1/(2.3) + 1/(3.4) + 1/(4.5) + 1/(5.6) + ... + 1/(99.100)

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/99 - 1/100

= 1/2 - 1/100 < 1/2

\(\dfrac{1}{3^2}< \dfrac{1}{1\cdot3}=1-\dfrac{1}{3}\)

\(\dfrac{1}{5^2}< \dfrac{1}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}\)

...

\(\dfrac{1}{99^2}< \dfrac{1}{97\cdot99}=\dfrac{1}{97}-\dfrac{1}{99}\)

Do đó: \(\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}< 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\)

=>\(B=1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{9^2}< 1+1=2\)

=>1<B<2

=>B không là số tự nhiên

a, 2/3 + 1/3 : x = 3/5

1/3 : x = -1/15

x = -1/45

b, x - 5/9 = -2/3

x = -1/9