$\frac{3}{1.4}$+$\frac{3}{4.7}$+$\frac{3}{7.10}$+…+$\frac{3}{40.43}$ 29/10/2021 Bởi Josephine $\frac{3}{1.4}$+$\frac{3}{4.7}$+$\frac{3}{7.10}$+…+$\frac{3}{40.43}$
Gọi tổng đó là `A`. Ta có: `A=`$\frac{3}{1.4}$`+`$\frac{3}{4.7}$`+` $\frac{3}{7.10}$`+…+`$\frac{3}{40.43}$ `A=`$\frac{1}{1}$`-`$\frac{1}{4}$`+`$\frac{1}{4}$`-`$\frac{1}{7}$`+…+`$\frac{1}{40}$`-`$\frac{1}{43}$ `A=1-`$\frac{1}{43}$`=`$\frac{42}{43}$ Vậy $\frac{3}{1.4}$`+`$\frac{3}{4.7}$`+` $\frac{3}{7.10}$`+…+`$\frac{3}{40.43}$`=`$\frac{42}{43}$ Bình luận
`3/(1.4)+3/(4.7)+3/(7.10)+…+3/(40.43)` `=1/1-1/4+1/4-1/7+1/7-1/10+…+1/40-1/43` `=1/1-1/43` `=43/43-1/43` `=42/43` Bình luận
Gọi tổng đó là `A`. Ta có:
`A=`$\frac{3}{1.4}$`+`$\frac{3}{4.7}$`+` $\frac{3}{7.10}$`+…+`$\frac{3}{40.43}$
`A=`$\frac{1}{1}$`-`$\frac{1}{4}$`+`$\frac{1}{4}$`-`$\frac{1}{7}$`+…+`$\frac{1}{40}$`-`$\frac{1}{43}$
`A=1-`$\frac{1}{43}$`=`$\frac{42}{43}$
Vậy $\frac{3}{1.4}$`+`$\frac{3}{4.7}$`+` $\frac{3}{7.10}$`+…+`$\frac{3}{40.43}$`=`$\frac{42}{43}$
`3/(1.4)+3/(4.7)+3/(7.10)+…+3/(40.43)`
`=1/1-1/4+1/4-1/7+1/7-1/10+…+1/40-1/43`
`=1/1-1/43`
`=43/43-1/43`
`=42/43`