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ok em có ngay nhé em
Muốn đưa về tỉ số % em chỉ cần lấy tử số chia mẫu số rồi nhân với 100 là được
\(\dfrac{1}{2}\) = 0,5 = 50%
\(\dfrac{3}{5}\) = 0,6 = 60%
\(\dfrac{2}{4}\)= 0,5 = 50%
\(\dfrac{4}{100}\) = 4%
\(\frac{1}{2}\div\frac{1}{3}\div\frac{1}{4}\div\frac{1}{5}\div...\div\frac{1}{98}\div\frac{1}{99}\div\frac{1}{100}\)
\(=\frac{1}{2}\cdot3\cdot4\cdot5\cdot...\cdot98\cdot99\cdot100\)
Phép tính như thế này tạm thời chuyển sang :
\(\frac{1}{2}\cdot\frac{100!}{2}=\frac{100!}{4}\)
\(x:3\dfrac{1}{15}\) - \(\dfrac{3}{4}\) = 2\(\dfrac{1}{4}\)
\(x\): \(\dfrac{46}{15}\) - \(\dfrac{3}{4}\) = \(\dfrac{9}{4}\)
\(x\) : \(\dfrac{46}{15}\) = \(\dfrac{9}{4}\) + \(\dfrac{3}{4}\)
\(x\) : \(\dfrac{46}{15}\) = \(\dfrac{12}{4}\)
\(x\) : \(\dfrac{46}{15}\) = \(3\)
\(x\) = 3 \(\times\) \(\dfrac{46}{15}\)
\(x\) = \(\dfrac{46}{5}\)
\(x\) \(\times\) 3\(\dfrac{2}{3}\) - 1\(\dfrac{2}{3}\) = 2\(\dfrac{1}{3}\)
\(x\) \(\times\) \(\dfrac{11}{3}\) - \(\dfrac{5}{3}\) = \(\dfrac{7}{3}\)
\(x\) \(\times\) \(\dfrac{11}{3}\) = \(\dfrac{7}{3}\) + \(\dfrac{5}{3}\)
\(x\) \(\times\) \(\dfrac{11}{3}\) = \(\dfrac{12}{3}\)
\(x\times\dfrac{11}{3}\) = 4
\(x\) = 4 : \(\dfrac{11}{3}\)
\(x\) = \(\dfrac{12}{11}\)
Bài 1 :
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\dfrac{1}{2}xA=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(A-\dfrac{1}{2}xA=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(\dfrac{1}{2}xA=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}-\dfrac{1}{64}-\dfrac{1}{128}\)
\(\dfrac{1}{2}xA=\dfrac{1}{2}-\dfrac{1}{128}\)
\(\dfrac{1}{2}xA=\dfrac{64}{128}-\dfrac{1}{128}\)
\(\dfrac{1}{2}xA=\dfrac{63}{128}\)
\(A=\dfrac{63}{128}:\dfrac{1}{2}=\dfrac{63}{128}x\dfrac{2}{1}=\dfrac{63}{64}\)
Bài 2 :
Số tự nhiên có 2 chữ số là \(\overline{ab}\left(a;b\inℕ\right)\)
Thêm số 12 vào bên trái số đó :
\(\overline{12ab}=26x\overline{ab}\)
\(1200+10xa+b=26x\left(10xa+b\right)\)
\(1200+10xa+b=260xa+26xb\)
\(260xa-10xa+26xb-b=1200\)
\(250xa+25xb=1200\)
\(25x\left(10xa+b\right)=1200\)
\(10xa+b=48\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=8\end{matrix}\right.\)
Vậy số đó là \(48\)
\(1\frac{3}{5}+2\frac{2}{3}-1=\frac{8}{5}+\frac{8}{3}-1=1,6+2,\left(6\right)-1=0,6+2\left(6\right)=3,2\left(6\right)\)
\(2\frac{1}{3}\cdot1\frac{1}{2}=\frac{7}{3}\cdot\frac{3}{2}=2,\left(3\right)\cdot1,5=3.5\)
\(4\frac{4}{5}:\frac{3}{100}=\frac{24}{5}\cdot\frac{100}{3}=\frac{8}{5}\cdot100=1,6\cdot100=160\)
\(2\frac{1}{2}:1\frac{1}{4}=\frac{5}{4}\cdot\frac{4}{5}=1,25\cdot0,8=1\)
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)