A

A

Câu A

\(\left(\dfrac{2x}{5}+3\right):\left(-5\right)=\dfrac{-1}{25}\\ \dfrac{2x}{5}+3=\dfrac{-1}{25}\cdot\left(-5\right)\\ \dfrac{2x}{5}+3=\dfrac{1}{5}\\ \dfrac{2x+15}{5}=\dfrac{1}{5}\\ 2x+15=1\\ 2x=-14\\ x=-7\)

Vậy x = -7

\(\left(\dfrac{2x}{5}+3\right)\):\(\left(-5\right)=\dfrac{-1}{25}\)

\(\dfrac{2x}{5}+3=\dfrac{-1}{25}.\left(-5\right)\)

\(\dfrac{2x}{5}+3=\dfrac{1}{5}\)

\(\dfrac{2x}{5}=\dfrac{1}{5}+3\)

\(\dfrac{2x}{5}=\dfrac{16}{5}\)

2x=16

x=16:2

x=8

Vậy x= 8

7

để \(\left|8-x\right|=8-x< =>8-x\ge0< =>x\le8\)

\(=>8-x=x^2+x< =>x^2+2x-8=0\)

\(< =>\left(x+1\right)^2-3^2=0< =>\left(x-2\right)\left(x+4\right)=0\)

\(=>\left[{}\begin{matrix}x=2\left(TM\right)\\x=-4\left(TM\right)\end{matrix}\right.\)

*để\(\left|8-x\right|=x-8< =>8-x< 0< =>x>8\)

\(=>x-8=x^2+x< =>x^2=-8\)(vô lí)

vậy x=2 hoặc x=-4

bài gì

đề bài đâu r

a) \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(\dfrac{21}{21}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\dfrac{20}{21}\)

\(A=\dfrac{10}{21}\)

b) \(B=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(B=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{96.97}+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{96}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\dfrac{98}{99}\)

\(B=-\dfrac{97}{99}\)