Tính giá trị cưa biểu thức có nhiều dấu ngoặc:
A= \(\left(6-\dfrac{2}{3}+\dfrac{1}{3}\right)\) - \(\left(5+\dfrac{5}{7}-\dfrac{3}{2}\right)\) - \(\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
Cho biểu thức:
\(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
Hãy tính giá trị biểu thức A theo hai cách:
Cách 1: Trước hêt, tính giá trị của từng biểu thức trong ngoặc.
Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
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
cách 1:
A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
= \(\left(\dfrac{6}{1}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{5}{1}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{3}{1}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
= \(\left(\dfrac{18}{3}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{15}{3}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{9}{3}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
= \(\left(\dfrac{16}{3}+\dfrac{1}{2}\right)-\left(\dfrac{20}{3}-\dfrac{3}{2}\right)-\left(\dfrac{2}{3}+\dfrac{5}{2}\right)\)
= \(\left(\dfrac{32}{6}+\dfrac{3}{6}\right)-\left(\dfrac{40}{6}-\dfrac{9}{6}\right)-\left(\dfrac{4}{6}+\dfrac{15}{6}\right)\)
= \(\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
= \(-\dfrac{15}{6}=-\dfrac{5}{2}\)
cách 2:
A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
= \(\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)-\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
= \(\left(-2\right)-0+\dfrac{1}{2}\)
= \(-\dfrac{5}{2}\)
Tính giá trị của biểu thức:
\(A=\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(-2022\right)^0\)
\(B=0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2.\left(-\dfrac{1}{9}\right)\)
\(C=2\dfrac{6}{7}.\left[\left(\dfrac{-7}{5}-\dfrac{3}{2}:\dfrac{-5}{-4}\right)+\left(\dfrac{3}{2}\right)^2\right]\)
\(D=\dfrac{2}{7}+\dfrac{5}{7}.\left(\dfrac{3}{5}-0,25\right).\left(-2\right)^2+35\%\)
\(E=1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):1\dfrac{2}{5}\)
\(F=\dfrac{\dfrac{5}{3}-\dfrac{5}{7}+\dfrac{5}{9}}{\dfrac{10}{3}-\dfrac{10}{7}+\dfrac{10}{9}}\)
Tính bằng cách hợp lí giá trị của biểu thức.
A = \(\left(3-\dfrac{1}{4} +\dfrac{3}{2}\right)\)- \(\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)\)-\(\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
B =\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
Tính bằng hai cách:
A=\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
Cách 1: Trước hết, tính giá trị của từng biểu thức trong ngoặc.
Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
tính giá trị biểu thức sau
a) \(A=\dfrac{9^4}{3^2}\)
b) \(B=81.\left(\dfrac{5}{3}\right)^4\)
c) \(C=\left(\dfrac{4}{7}\right)^{-4}.\left(\dfrac{2}{7}\right)^3\)
d) \(D=7^{-6}.\left(\dfrac{2}{3}\right)^0.\left(\dfrac{7}{5}\right)^6\)
e) \(E=8^3:\left(\dfrac{2}{3}\right)^5.\left(\dfrac{1}{3}\right)^2\)
f) \(F=\left(\dfrac{7}{9}\right)^{-2}.\left(\dfrac{1}{\sqrt{3}}\right)^8\)
g) \(G=\left(\dfrac{-4}{5}\right)^{-2}.\left(\dfrac{2}{5}\right)^2.\left(\sqrt{2}\right)^3\)
a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729
b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)
c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)
\(=\dfrac{2^3}{4^4}\cdot7\)
\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)
d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)
\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)
\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)
e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)
\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)
\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)
f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)
\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)
g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)
\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)
\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)
Tính giá trị của biểu thức sau:
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
Áp dụng tính chất các phép tính và quy tắc dấu ngoặc để tính giá trị các biểu thức sau :
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
Tính giá trị của biểu thức:
a) (-7,05 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\left(-\dfrac{1}{5}\right)\) : (-3)
b) \(\left(\dfrac{3}{25}-1,12\right)\) : \(\dfrac{3}{7}\) : [\(\left(3\dfrac{1}{2}\right)\) - \(\left(3\dfrac{2}{3}\right)\) : \(\dfrac{1}{14}\)]
a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)
= \(-\dfrac{73}{10}\) : - 5
= \(\dfrac{73}{50}\)
b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)
= \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)
= \(\dfrac{2}{41}\)
Tính giá trị của các biểu thức sau :
a)\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)\)+(0,4 - 5) - \(\left(4\dfrac{1}{4}-1\right)\)
b)\(\dfrac{2}{3}\) - \(\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
c)\(\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right)\):\(\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
d)3 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
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\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)
\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)
\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)
\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)
\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)
\(=7-\dfrac{26}{5}\)
\(=\dfrac{9}{5}\)
\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)
\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)
\(=\dfrac{2}{3}+\dfrac{21}{8}\)
\(=\dfrac{79}{24}\)
\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)
\(=\dfrac{31}{4}:\dfrac{49}{8}\)
\(=\dfrac{62}{49}\)
\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)