Bài 2 ; Tính a, $\frac{24.47-23}{24+47.23}$ b, $\frac{3+\frac{3}{7} -\frac{3}{11}+ \frac{3}{1001}-\frac{3}{13} }{\frac{9}{1001}-\frac{9}{13}+\frac{

Bài 2 ; Tính
a, $\frac{24.47-23}{24+47.23}$
b, $\frac{3+\frac{3}{7} -\frac{3}{11}+ \frac{3}{1001}-\frac{3}{13} }{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7} -\frac{9}{11}+9 }$
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  1. a, Ta có: $\frac{24.47-23}{24+47.23}$

    = $\frac{24.47-23}{47.23+47-23}$ 

    = $\frac{24.47-23}{47.24-23}$ 

    =1

    b, Ta có:

    $\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{1}{11}+9}$

    = $\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{3(\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}+3)}$

    = $\frac{1}{3}$ 

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  2. `a, {24.47-23}/{24+47.23}`

    `={(23+1).47-23}/{24+47.23}`

    `={23.47+47-23}/{24+47.23}`

    `={23.47+24}/{24+47.23}`

    `={24+23.47}/{24+23.47}`

    `=1`

    `b,{3+3/7-3/11+3/1001-3/13}/{9/1001-9/13+9/7-9/11+9}`

    `={3+3. 1/7-3. 1/11+3. 1/1001-3. 1/13}/{9. 1/1001-9. 1/13+9. 1/7-9. 1/11+9}`

    `={3.(1/7-1/11+1/1001-1/13)}/{9.(1/1001-1/13+1/7-1/11)}`

    `={3.(1/7-1/11+1/1001-1/13)}/{9.(1/7-1/11+1/1001-1/13)}`

    `=3/9`

    `=1/3`

     

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