c1 (3/11+3/29-3/170+3/2015):(7/11+7/29-7/170+7/2015) c2 (1-1/2)x (1-1/3)-(1-1/4)….(1-1/2003) 26/08/2021 Bởi Claire c1 (3/11+3/29-3/170+3/2015):(7/11+7/29-7/170+7/2015) c2 (1-1/2)x (1-1/3)-(1-1/4)….(1-1/2003)
c1 Ta có : 311+329−3170+32015711+729−7170+72015311+329−3170+32015711+729−7170+72015 = 3.(111+129−1170+12015)7.(111+129−1170+12015)3.(111+129−1170+12015)7.(111+129−1170+12015) = 3737 c, (1-1212 )x (1-1313 )-(1-1414 )….(1-1200312003 ) = 1212 x 2323 x 3434 x …. x 2002200320022003 = 1x2x3x....x20022x3x4x...x20031x2x3x….x20022x3x4x…x2003 Bình luận
Đáp án: c1 Ta có : $\frac{\frac{3}{11}+\frac{3}{29}-\frac{3}{170}+\frac{3}{2015}}{\frac{7}{11}+\frac{7}{29}-\frac{7}{170}+\frac{7}{2015}}$ = $\frac{3.(\frac{1}{11}+\frac{1}{29}-\frac{1}{170}+\frac{1}{2015})}{7.(\frac{1}{11}+\frac{1}{29}-\frac{1}{170}+\frac{1}{2015})}$ = $\frac{3}{7}$ c, (1-$\frac{1}{2}$ )x (1-$\frac{1}{3}$ )-(1-$\frac{1}{4}$ )….(1-$\frac{1}{2003}$ ) = $\frac{1}{2}$ x $\frac{2}{3}$ x $\frac{3}{4}$ x …. x $\frac{2002}{2003}$ = $\frac{1x2x3x….x2002}{2x3x4x…x2003}$ = $\frac{1}{2003}$ Giải thích các bước giải: Bình luận
c1 Ta có :
311+329−3170+32015711+729−7170+72015311+329−3170+32015711+729−7170+72015
= 3.(111+129−1170+12015)7.(111+129−1170+12015)3.(111+129−1170+12015)7.(111+129−1170+12015)
= 3737
c, (1-1212 )x (1-1313 )-(1-1414 )….(1-1200312003 )
= 1212 x 2323 x 3434 x …. x 2002200320022003
= 1x2x3x....x20022x3x4x...x20031x2x3x….x20022x3x4x…x2003
Đáp án:
c1 Ta có :
$\frac{\frac{3}{11}+\frac{3}{29}-\frac{3}{170}+\frac{3}{2015}}{\frac{7}{11}+\frac{7}{29}-\frac{7}{170}+\frac{7}{2015}}$
= $\frac{3.(\frac{1}{11}+\frac{1}{29}-\frac{1}{170}+\frac{1}{2015})}{7.(\frac{1}{11}+\frac{1}{29}-\frac{1}{170}+\frac{1}{2015})}$
= $\frac{3}{7}$
c, (1-$\frac{1}{2}$ )x (1-$\frac{1}{3}$ )-(1-$\frac{1}{4}$ )….(1-$\frac{1}{2003}$ )
= $\frac{1}{2}$ x $\frac{2}{3}$ x $\frac{3}{4}$ x …. x $\frac{2002}{2003}$
= $\frac{1x2x3x….x2002}{2x3x4x…x2003}$
= $\frac{1}{2003}$
Giải thích các bước giải: