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bạn Kudo Shinichi sai rồi: muốn tìm số chia mà thương chia cho số bị chia à
a) 7,2 * x = 6,49
x = 6,49 : 7,2
x = 649/720
b) \(\frac{15}{77}:x=\frac{3}{11}\)
\(x=\frac{3}{11}:\frac{15}{77}=\frac{7}{5}\)
c) 2,4 * x + 1,1 *x = 0,7
x * ( 2,4 + 1,1 ) = 0,7
x * 3,5 = 0,7
=> x = 0,7 : 3,5 = 0,2
phần d sai đề đó bạn, 25/31 chia rồi lại công là sai, sửa đề đi nha.
Ai thấy đúng thì ủng hộ nha !!!
\(x(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\) \(5\frac{1}{2}\)
\(x\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(x\cdot\frac{11}{12}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{11}{12}\)
\(x=6\)
Vậy x = 6
\(\frac{x}{2}+\frac{x}{6}+\frac{x}{12}+\frac{x}{20}+...+\frac{x}{132}=5\frac{1}{2}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)=\frac{11}{2}\)
\(\Rightarrow x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}=\frac{11}{2}\right)\)
\(\Rightarrow x\left(1-\frac{1}{12}\right)=\frac{11}{2}\)
\(\Rightarrow x.\frac{11}{12}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{11}{12}=1\)
Vậy x = 1
\(\left(a\right)\frac{34-x}{30}=\frac{5}{6}\)
\(\frac{34-x}{30}=\frac{25}{30}\)
34 - x = 25
x = 34 - 25 = 9
\(\left(b\right)\frac{x+13}{34}=\frac{12}{17}\)
\(\frac{x+13}{34}=\frac{24}{34}\)
x + 13 = 24
x = 24 - 13 = 11
\(\left(c\right)\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{56}{81}\)
\(4x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Ta có : \(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)
\(2A=1-\frac{1}{81}=\frac{80}{81}\)
\(A=\frac{80}{81}\div2=\frac{40}{81}\)
\(\Rightarrow4x+\frac{40}{81}=\frac{56}{81}\)
\(4x=\frac{56}{81}-\frac{40}{81}\)
\(4x=\frac{16}{81}\)
\(x=\frac{16}{81}\div4=\frac{4}{81}\)
a) \(\frac{1}{3}-\frac{2}{x}=\frac{1}{6}\)
=> \(\frac{2}{x}=\frac{1}{3}-\frac{1}{6}\)
=> \(\frac{2}{x}=\frac{1}{6}\)
\(\Rightarrow x=12\)
Vậy x = 12
b) \(\frac{x-1}{2}=\frac{3}{4}\)
\(\Rightarrow\left(x-1\right).4=2.3\)
=> 4x - 4 = 6
=> 4.(x-1) = 6
=> x-1 = 6:4 = 1,5
=> x = 1,5 + 1
=> x = 2,5
Vậy x = 2,5
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+...+\left(x+\frac{1}{512}\right)=1\)
\(9x+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)=1\)
\(9x+\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{256}-\frac{1}{512}\right)\right]=1\)
\(9x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}\right)=1\)
\(9x+\left(1-\frac{1}{512}\right)=1\)
\(9x+\frac{511}{512}=1\)
\(9x=1-\frac{511}{512}\)
\(9x=\frac{1}{512}\)
\(\Rightarrow x=\frac{1}{512}\div9=\frac{1}{4608}\)
3/4 * x + 1/2 * x -15 = 35
3/4 * x +1/2 * x = 35 + 15
3/4 * x +1/2 * x = 50
x * ( 3/4 + 1/2 ) = 50
x * 5/4 = 50
x = 50 : 5/4
x = 40
phan b mik ko nhap dc nen bn tu lm nha
a, \(\frac{3}{4}\times x+\frac{1}{2}\times x-15=35\)
\(x+x-15=\frac{3}{4}-\frac{1}{2}\)
\(x+x-15=\frac{2}{8}=\frac{1}{4}\)
\(x=35-15\)
\(x=20\)
Vậy \(x=\frac{1}{4}\)và \(x=20\)
b, Chịu