Bài 171 trang 67 SGK Toán 6 tập 2

Đề bài

Tính giá trị các biểu thức sau:

$A = 27 + 46 + 79 + 34 + 53;$

$B = - 377 - (98 - 277)$

$C = - 1,7. 2,3 + 1,7. (- 3,7) - 1,7.3 - 0,17:0,1$

$\displaystyle D = 2{3 \over 4}.\left( { - 0,4} \right) - 1{3 \over 5}.2,75 + \left( { - 1,2} \right):{4 \over {11}}$

$E = \displaystyle {{\left( {{2^3}.5.7} \right)\left( {{5^2}{{.7}^3}} \right)} \over {{{\left( {{{2.5.7}^2}} \right)}^2}}}$

Lời giải chi tiết

$\displaystyle A = 27 + 46 + 79 + 34 + 53$$\displaystyle = (27 + 53)  +  (46 + 34) + 79$$\displaystyle = 80 + 80 + 79 = 239$

$\displaystyle B = - 377 - (98 - 277)$

   $\displaystyle = - 377 - 98 + 277$

   $\displaystyle =  (- 377+ 277)-98$

   $\displaystyle = -100-98$ 

   $\displaystyle =           -198$

$\displaystyle C = - 1,7. 2,3 + 1,7. (- 3,7)$$\displaystyle - 1,7.3 - 0,17:0,1$

  $\displaystyle = - 1,7. 2,3 + 1,7. (- 3,7)$$\displaystyle - 1,7.3 - 0,17.10$

 $\displaystyle = - 1,7. 2,3 - 1,7.  3,7$$\displaystyle - 1,7.3 - 1,7$

  $\displaystyle = -1,7. (2,3 + 3,7 + 3 + 1)$

  $\displaystyle = -1,7 . 10$

  $\displaystyle =  -17.$

$\displaystyle D=2{3 \over 4}.\left( { - 0,4} \right) - 1{3 \over 5}.2,75$$\displaystyle + \left( { - 1,2} \right):{4 \over {11}}$

$\displaystyle ={{11} \over 4}.{{ - 4} \over {10}} - {8 \over 5}.{{275} \over 100} + {{ - 12} \over 10}.{{11} \over 4}$

   $\displaystyle ={{11} \over 4}.{{ - 2} \over {5}} - {8 \over 5}.{{11} \over 4} + {{ - 6} \over 5}.{{11} \over 4}$

   $\displaystyle ={{11} \over 4}.\left( {{{ - 2} \over {5}} - {8 \over 5} + {{ - 6} \over 5}} \right)$

   $\displaystyle  = {{11} \over 4}.{{ - 2 - 8 - 6} \over 5}$

   $\displaystyle  = {{11} \over 4}.{{ - 16} \over 5}$

   $\displaystyle  = {{ - 44} \over 5}$ 

$\displaystyle E = {{\left( {{2^3}.5.7} \right).\left( {{5^2}{{.7}^3}} \right)} \over {{{\left( {{{2.5.7}^2}} \right)}^2}}}$

$\begin{array}{l} = \dfrac{{{2^3}{{.5.7.5}^2}{{.7}^3}}}{{{2^2}{{.5}^2}.{{\left( {{7^2}} \right)}^2}}} = \dfrac{{{2^3}{{.5}^{1 + 2}}{{.7}^{1 + 3}}}}{{{2^2}{{.5}^2}{{.7}^{2.2}}}}\\ = \dfrac{{{2^3}{{.5}^3}{{.7}^4}}}{{{2^2}{{.5}^2}{{.7}^4}}} = \dfrac{{2.5.1}}{{1.1.1}} = 10 \end{array}$

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