\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)

\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)

\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)

\(\Rightarrow x=3\)

\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)

\(\Leftrightarrow25-5x=6x-8\)

\(\Leftrightarrow-5x-6x=-8-25\)

\(\Leftrightarrow-11x=-33\)

\(\Leftrightarrow x=3\)

Vậy x = 3

:v lớp 10

D

a) 4126

b) 615

c) 927

b) 615

c) 927

sao thi muộn vậy bn ! mk thi xong lâu rùi

shinichi

shinichi

Theo mk được biết thì Shinichi và Kid là hai anh em nên mk thích cả hai

Thik cả 2 phương án trên

mk chưa học đến thì mk bảo chưa học có làm sao đâu, hay để mk đi hỏi bạn bè cho

?2 x=8,16,24,32,40

?3Ư(12)=1,2,3,4,6,12

?4Ước của 1 là 1

Bội của 1 là 1,2,3,4,5,6,7,8,9,........... vân vân và vân vân

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-\left(2+1+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{200-2-1-\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-...-\dfrac{2}{100}}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =2\)

Đề nhỏ quá!! mà t 4 mắt. cẩn thận

Đặt :

\(A=\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+.............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....................+\dfrac{99}{100}}\)

\(A=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+..............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+.................+1-\dfrac{1}{100}}\)

\(A=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+..................+\dfrac{2}{100}\right)}{\left(1+1+.....+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+...........+\dfrac{1}{100}\right)}\)

\(A=\dfrac{2\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.............+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..............+\dfrac{1}{100}\right)}\)

\(A=2\)

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...............+\dfrac{99}{100}}=2\rightarrowđpcm\)

Ta có: \(\dfrac{1}{2}\cdot y+\dfrac{2}{3}\cdot y=\dfrac{7}{6}\Rightarrow y\left(\dfrac{1}{2}+\dfrac{2}{3}\right)=\dfrac{7}{6}\Rightarrow\dfrac{7}{6}y=\dfrac{7}{6}\Rightarrow y=\dfrac{7}{6}:\dfrac{7}{6}=1\)

Vậy \(D=\left\{1\right\}\)

1

B. Sai

sai