Tính giá trí của biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\dfrac{3}{5} \cdot \dfrac{6}{7}+\dfrac{3}{7}: \dfrac{5}{3}-\dfrac{2}{7}: 1 \dfrac{2}{3}$;
b) $\mathrm{B}=\left(-13 \cdot \dfrac{2}{5}+\dfrac{-2}{9}: 2 \dfrac{1}{2}+\dfrac{2}{5}. \dfrac{11}{9}\right) \cdot 2 \dfrac{1}{2}$;
c) $\mathrm{C}=\left(\dfrac{-4}{5}+\dfrac{5}{7}\right): \dfrac{2}{3}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right): \dfrac{2}{3}$;
d) $\mathrm{D}=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)+\dfrac{4}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)$.
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\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
Cách 1: Tính giá trị từng biểu thức trong ngoặc
A=
Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp
A =
= (6-5-3) -
= -2 -0 - = - (2 + ) = -2
Lời giải:
Cách 1: Tính giá trị từng biểu thức trong ngoặc
A=
Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp
A =
= (6-5-3) -
= -2 -0 - = - (2 + ) = -2
a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A = - 2 + (-1)\\A = - 3\end{array}\)
b,
Cách 1:
\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)
Cách 2:
\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B = \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
a) A=35.67+37.35−27.35A=35.67+37.35−27.35
=35⋅(67+37−27)=35=35⋅(67+37−27)=35
b) B=(−13⋅25+−29⋅25+25⋅119)⋅52B=(−13⋅25+−29⋅25+25⋅119)⋅52
=(−13−29+119)⋅25⋅52=−13+(119−29)=−12.=(−13−29+119)⋅25⋅52=−13+(119−29)=−12.
c) C=(−45+57)⋅32+(−15+27)⋅32=(−45+57+−15+27)⋅32=((−45+−15)+(57+27))⋅32=0.C=(−45+57)⋅32+(−15+27)⋅32=(−45+57+−15+27)⋅32=((−45+−15)+(57+27))⋅32=0.
d) D=49:(115−1015)+49:(222−522)D=49:(115−1015)+49:(222−522)
=49:−35+49:−322=49⋅−53+49.−223=49:−35+49:−322=49⋅−53+49.−223
=49⋅(−53+−223)=49.−273=−4.
a) \mathrm{A}=\dfrac{3}{5}. \dfrac{6}{7}+\dfrac{3}{7}. \dfrac{3}{5}-\dfrac{2}{7}. \dfrac{3}{5}A=53. 76+73. 53−72. 53
b) \mathrm{B} =\left(-13 \cdot \dfrac{2}{5}+\dfrac{-2}{9} \cdot \dfrac{2}{5}+\dfrac{2}{5} \cdot \dfrac{11}{9}\right) \cdot \dfrac{5}{2} B=(−13⋅52+9−2⋅52+52⋅911)⋅25
=\left(-13-\dfrac{2}{9}+\dfrac{11}{9}\right) \cdot \dfrac{2}{5} \cdot \dfrac{5}{2}=-13+\left(\dfrac{11}{9}-\dfrac{2}{9}\right)=-12 .=(−13−92+911)⋅52⋅25=−13+(911−92)=−12.
c) \mathrm{C} =\left(\dfrac{-4}{5}+\dfrac{5}{7}\right) \cdot \dfrac{3}{2}+\left(\dfrac{-1}{5}+\dfrac{2}{7}\right) \cdot \dfrac{3}{2} =\left(\dfrac{-4}{5}+\dfrac{5}{7}+\dfrac{-1}{5}+\dfrac{2}{7}\right) \cdot \dfrac{3}{2}=\left(\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)+\left(\dfrac{5}{7}+\dfrac{2}{7}\right)\right) \cdot \dfrac{3}{2}=0 .C=(5−4+75)⋅23+(5−1+72)⋅23=(5−4+75+5−1+72)⋅23=((5−4+5−1)+(75+72))⋅23=0.
d) \mathrm{D}=\dfrac{4}{9}:\left(\dfrac{1}{15}-\dfrac{10}{15}\right)+\dfrac{4}{9}:\left(\dfrac{2}{22}-\dfrac{5}{22}\right)D=94:(151−1510)+94:(222−225)