Question

Giúp mình bài 2,4,5 với

Answer 1

2 dễ rồi, suy nghĩ đy nhé

4) \(15-\left\{2\left[x-\left(2x-4\right)\cdot5\right]-3\left(x+1\right)\right\}=12-x\)

\(\Rightarrow15-\left[2\left(x-10+20\right)-3x-3\right]=12-x\)

\(\Rightarrow15-\left(2x-20+40-3x-3\right)=12-x\)

\(\Rightarrow15+x-17=12-x\)

\(\Rightarrow x+x=12-15+17\)

\(\Rightarrow2x=14\Rightarrow x=\dfrac{14}{2}=7\)

5) \(\left(\dfrac{3}{1\cdot2}+\dfrac{3}{2\cdot3}+\dfrac{3}{3\cdot4}+...+\dfrac{3}{89\cdot90}\right)-2x=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\)

\(\Rightarrow3\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{89\cdot90}\right)-2x=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{97}-\dfrac{1}{99}\)

\(\Rightarrow3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{89}-\dfrac{1}{90}\right)-2x=1-\dfrac{1}{99}\)

\(\Rightarrow3\left(1-\dfrac{1}{90}\right)-2x=\dfrac{98}{99}\)

\(\Rightarrow-2x=\dfrac{98}{99}-3\left(1-\dfrac{1}{90}\right)=\dfrac{98}{99}-\dfrac{89}{30}=-\dfrac{1957}{990}\)

\(\Rightarrow x=-\dfrac{1957}{990}:\left(-2\right)=\dfrac{1957}{1980}\)

Vậy.............