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`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+4+...+2018}\right)\)
\(A=\frac{2}{1+2}\cdot\frac{2+3}{1+2+3}\cdot\frac{2+3+4}{1+2+3+4}\cdot...\cdot\frac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018}\)
Đến chỗ này đố ai tính được ?!!?!
gạch các số của tử số và các số của mẫu số giống nhau
ví dụ như bạn nói:
\(\dfrac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018} =1\)
Giải:
\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
`@` `\text {Ans}`
`\downarrow`
`a,`
\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{6}{5}-x=\dfrac{2}{3}\)
\(x=\dfrac{6}{5}-\dfrac{2}{3}\)
\(x=\dfrac{8}{15}\)
Vậy, `x = \dfrac{8}{15}`
`b,`
\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)
\(x\times3\dfrac{1}{3}=\dfrac{40}{51}\)
\(x=\dfrac{40}{51}\div3\dfrac{1}{3}\)
\(x=\dfrac{4}{17}\)
Vậy, `x=`\(\dfrac{4}{17}\)
`c,`
\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)
\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)
\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)
\(x=\dfrac{34}{7}\)
Vậy, `x=`\(\dfrac{34}{7}\)
a,\(\dfrac{3}{2}.\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{4}{5}-x=\dfrac{2}{3}:\dfrac{3}{2}\)
\(\dfrac{4}{5}-x=\dfrac{4}{9}\)
\(x=\dfrac{4}{5}-\dfrac{4}{9}\)
\(x=\dfrac{16}{45}\)
`4/3: x=1/2`
`x=4/3 :1/2`
`x= 4/3 xx 2`
`x=8/3`
__
`x-2/7 =1/3`
`x=1/3 +2/7`
`x= 7/21 + 6/21`
`x= 13/21`
__
`x:2/3 =1/3`
`x=1/3 xx 2/3`
`x= 2/9`
__
`2 xx x +1/5 =1`
`2 xx x =1-1/5`
`2xx x = 5/5 -1/5`
`2 xx x=4/5`
`x= 4/5 :2`
`x= 4/5 xx 1/2`
`x= 4/10`
`x=2/5`
\(\dfrac{4}{3}:x=\dfrac{1}{2}\) \(x-\dfrac{2}{7}=\dfrac{1}{3}\) \(x:\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{4}{3}\times2\) \(x=\dfrac{1}{3}+\dfrac{2}{7}\) \(x=\dfrac{1}{3}\times\dfrac{2}{3}\)
\(x=\dfrac{8}{3}\) \(x=\dfrac{13}{21}\) \(x=\dfrac{2}{9}\)
\(2\times x+\dfrac{1}{5}=1\)
\(2\times x=1-\dfrac{1}{5}\)
\(2\times x=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}\times\dfrac{1}{2}\)
\(x=\dfrac{2}{5}\)
a)\(\dfrac{2}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}+\dfrac{1}{3}\right).\dfrac{4}{5}=1.\dfrac{4}{5}=\dfrac{4}{5}\)
b)\(\dfrac{2}{3}.\dfrac{4}{5}-\dfrac{1}{3}.\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right).\dfrac{4}{5}=\dfrac{1}{3}.\dfrac{4}{5}=\dfrac{4}{15}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}+\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=\dfrac{4}{5}\times1=\dfrac{4}{5}\)
b) \(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{3}=\dfrac{4}{15}\)
c) \(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{2}{3}=\dfrac{8}{9}\)
d) \(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{4}{3}\times\dfrac{1}{3}=\dfrac{4}{9}\)
#)Giải :
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{1}{13}\right)x...x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1x4x7x...x94x100}{4x7x10x...x97x100}\)
\(=\frac{1}{100}\)
#~Will~be~Pens~#
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)\left(1-\frac{1}{13}\right)...\left(1-\frac{1}{97}\right)\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.\frac{10}{13}...\frac{94}{97}.\frac{97}{100}\)
\(=\frac{1}{100}\)
\(\dfrac{x-1}{3}=\dfrac{1}{3}\)
=>x-1=1
=>x=1+1=2
\(\dfrac{\left(x-1\right)}{3}=\dfrac{1}{3}\)
\(\Rightarrow x-1=1\)
\(x=1+1\)
\(x=2\)
Vậy \(x=2\)