Bài 76 trang 39 SGK Toán 6 tập 2

Đề bài

Tính giá trị biểu thức sau một cách hợp lí: 

$A=\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}$ ;

$B= \dfrac{5}{9}.\dfrac{7}{13}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}.\dfrac{3}{13}$ ;

$C=\left (\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117} \right ).\left (\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12} \right )$. 

Lời giải chi tiết

$A=\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}$

$A= \dfrac{7}{19}.\left (\dfrac{8}{11}+\dfrac{3}{11} \right )+\dfrac{12}{19}$

$A = \dfrac{7}{{19}}.\dfrac{{11}}{{11}} + \dfrac{{12}}{{19}}$

$A=\dfrac{7}{19}.1 +\dfrac{12}{19}$

$A=\dfrac{7+12}{19}$

$A=\dfrac{19}{19}=1$

$B= \dfrac{5}{9}.\dfrac{7}{13}+\dfrac{5}{9}.\dfrac{9}{13}-\dfrac{5}{9}.\dfrac{3}{13}$

$B=\dfrac{5}{9}.\left (\dfrac{7}{13}+\dfrac{9}{13}-\dfrac{3}{13} \right )$

$B=\dfrac{5}{9}.\dfrac{7+9-3}{13}$

$B=\dfrac{5}{9}.\dfrac{13}{13}=\dfrac{5}{9}.1=\dfrac{5}{9}$

$C=\left (\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117} \right ).\left (\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12} \right )$

$C=\left (\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117} \right ).\dfrac{4-3-1}{12}$

$C=\left (\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117} \right ).0=0$

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