2000/2010 x 2010/2000 x 2020/2030 x 2030/2040 x 2040/50

=1/1 x 1/1 x 202/203

=202/203

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

Mình làm tiếp câu b nha !

b,                                               Bài giải

\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)

\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)

\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)

\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)

\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)

\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)

\(x-2016=-2\text{ : }\frac{31}{56}\)

\(x-2016=-\frac{112}{31}\)

\(x=-\frac{112}{31}+2016\)

\(x=\frac{62384}{31}\)

\(Có: x=2010.2010\implies x=2010^2\)

        \(y=2000.2020\implies y=(2010-10).(2010+10)\implies 2010^2-10^2\)

Mà 2010² và 10² lớn hơn 0.

\(\implies 2010^2>2010^2-10^2\)

\(\implies x>y\)

Vậy x>y.

_Học tốt_

a,

104,4 . x -14,1 . x + 9,6 . x =25 

x(104,4 - 14,1 + 9,6 ) = 25

x.99,9 = 25

x = \(\frac{250}{999}\)

b)

\(\frac{2009.2010+2000}{2011.2010-2020}=\frac{2009.2010+2000}{\left(2009+2\right).2010-2020}\)

                                    \(=\frac{2009.2010+2000}{2009.2010+2.2010-2020}\)

                                    \(=\frac{2009.2010+2000}{2009.2010+4020-2020}\)

                                    \(=\frac{2009.2010+2000}{2009.2010+2000}\)         

                                     \(=1\)

Thanh you 

B nha bn

Lý thuyết : Những số nào có chữ số tạn cùng là 0 hoặc 5 thì chia hết cho 5.

⇒ Đáp án B. 2020 + 2000 + 2030

helps me

^-^