Câu 1: C

Câu 2: A

Câu 3: C

giúp mk nha hiện giờ mk đg cần gấp

\(VP=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot5}+\dfrac{1}{5\cdot13}+\dfrac{1}{13\cdot8}+\dfrac{1}{8\cdot19}+\dfrac{1}{19\cdot11}+\dfrac{1}{11\cdot25}\\ =\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}+\dfrac{2}{19\cdot22}+\dfrac{2}{22\cdot25}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}+\dfrac{3}{19\cdot22}+\dfrac{3}{22\cdot25}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\left(1-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\dfrac{24}{25}\\ =\dfrac{16}{25}\)\(a^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\\ a^2-\dfrac{9}{25}=\dfrac{16}{25}\\ a^2=\dfrac{16}{25}+\dfrac{9}{25}\\ a^2=1\\ \Rightarrow\left[{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-a=-1\\-a=1\end{matrix}\right.\)

Vậy \(-a=-1\) hoặc \(-a=1\)

D

tk đi bạn 

D

a=3

b=2

c=0

BÀI 1:

a)    \(\frac{4}{5}+\frac{3}{8}+\frac{1}{4}\)

\(=\frac{32}{40}+\frac{15}{40}+\frac{10}{40}\)

\(=\frac{32+15+10}{40}\)\(=\frac{57}{40}\)

b)    \(\frac{5}{9}+\frac{2}{3}+\frac{1}{2}\)

\(=\frac{10}{18}+\frac{12}{18}+\frac{9}{18}\)

\(=\frac{10+12+9}{18}=\frac{31}{18}\)

bài 23,, 5 nữa

1. 

a.57/40

b.31/18

2) Nếu a + 4b chia hết cho 13 => 10a + 40b chia hết cho 13 (1).

Lấy (1) - 39b (luôn chia hết cho 13) được 10a +b

=> 10a + b chia hết cho 13.

Ngược lại cũng tương tự.