12-12/7-12/289-12/85 trên 4-4/7-4/289-4/85 chia cho 3/13+3/169+3/91 trên 7+7/13+7/169+7/91 23/09/2021 Bởi aihong 12-12/7-12/289-12/85 trên 4-4/7-4/289-4/85 chia cho 3/13+3/169+3/91 trên 7+7/13+7/169+7/91
$\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}:\dfrac{3+\dfrac{3}{13}+\dfrac{3}{169}+\dfrac{3}{91}}{7+\dfrac{7}{13}+\dfrac{7}{169}+\dfrac{7}{91}}$ $=\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{3\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{7\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}$ `=12/4:3/7` `=3:3/7` `=7` Bình luận
Đáp án: `7` Giải thích các bước giải: Cách 1: `(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 ):(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)` Ta có : `(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 )` `=[12(1-1/7-1/289-1/85 )]/[4(1-1/7-1/289-1/85 )]` `=12/4=3(1)` Ta có : `(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)` `=[3(1/13+1/169+1/91 )]/[ 7(1+1/13+1/169+1/91)` `=3/7(2)` Vì `(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 ):(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)` nên `(1):(2)` `=3:3/7` `=3. 7/3` `=7` Cách 2 : `(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 ):(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)` `=[12(1-1/7-1/289-1/85 )]/[4(1-1/7-1/289-1/85 )]:[3(1/13+1/169+1/91 )]/[ 7(1+1/13+1/169+1/91)` `=12/4:3/7` `=3:3/7` `=3. 7/3` `=7` Bình luận
$\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}:\dfrac{3+\dfrac{3}{13}+\dfrac{3}{169}+\dfrac{3}{91}}{7+\dfrac{7}{13}+\dfrac{7}{169}+\dfrac{7}{91}}$
$=\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{3\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{7\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}$
`=12/4:3/7`
`=3:3/7`
`=7`
Đáp án: `7`
Giải thích các bước giải:
Cách 1:
`(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 ):(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)`
Ta có :
`(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 )`
`=[12(1-1/7-1/289-1/85 )]/[4(1-1/7-1/289-1/85 )]`
`=12/4=3(1)`
Ta có :
`(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)`
`=[3(1/13+1/169+1/91 )]/[ 7(1+1/13+1/169+1/91)`
`=3/7(2)`
Vì `(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 ):(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)`
nên `(1):(2)`
`=3:3/7`
`=3. 7/3`
`=7`
Cách 2 :
`(12-12/7-12/289-12/85 )/(4-4/7-4/289-4/85 ):(3/13+3/169+3/91 )/( 7+7/13+7/169+7/91)`
`=[12(1-1/7-1/289-1/85 )]/[4(1-1/7-1/289-1/85 )]:[3(1/13+1/169+1/91 )]/[ 7(1+1/13+1/169+1/91)`
`=12/4:3/7`
`=3:3/7`
`=3. 7/3`
`=7`