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 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 =\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
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 = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
= \(\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)-\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
= \(\left(-2\right)-0+\dfrac{1}{2}\)
= \(-\dfrac{5}{2}\)
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\) thực hiện phép tính theo 2 cách
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(\Rightarrow A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(\Rightarrow A=\left(7-6-5\right)-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}+\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\)
\(\Rightarrow A=-4-\dfrac{1}{4}+0\)
\(\Rightarrow A=-\dfrac{17}{4}\)
Cách 2 tính trong dấu ngoặc đơn của A, bạn tự tính nhé.
Tính bằng cách hợp lý:
\(a,\dfrac{2}{3}+\left(\dfrac{5}{7}+\dfrac{-2}{3}\right)\)
\(b,\left(\dfrac{-1}{4}+\dfrac{5}{8}\right)+\dfrac{-3}{8}\)
\(c,\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)
\(d,6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
\(e,\left(31,12-5,97\right)-\left(-68,88+4,03\right)\)
\(h,3,7.\left(-10,56\right)+3,7.110,56\)
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 một cách hợp lí
f) \(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\) g) \(\left(\dfrac{-3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)
g)\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right).\dfrac{7}{3}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right).\dfrac{7}{3}\)
\(=\left(-\dfrac{3}{4}+-\dfrac{1}{4}+\dfrac{2}{5}+\dfrac{3}{5}\right).\dfrac{7}{3}\)
\(=\left(-1+1\right).\dfrac{7}{3}=0.\dfrac{7}{3}=0\)
f) \(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)
\(=\dfrac{3^3.5^3+5.5^2.3^2-5^3}{3^3.6^3+6.6^2.3^2-6^3}\)
\(=\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)
\(=\dfrac{5^3}{6^3}\)
\(=\dfrac{125}{216}\)
Bài 1.( 2 điểm)Tính bằng cách hợp lí:
a) \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
b) \(\left(\dfrac{18}{23}+\dfrac{7}{12}\right)+\left(\dfrac{-13}{19}-\dfrac{3}{4}\right)+\left(\dfrac{-6}{19}+\dfrac{5}{23}\right)\)
c) \(\dfrac{4}{3}+\dfrac{-5}{6}+\dfrac{-1}{4}\)
d) \(\dfrac{5}{6}-\dfrac{7}{5}+\dfrac{17}{30}\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
Bài 1: Tính một cách hợp lí
d) (\(^{2^2}\) : \(\dfrac{4}{3}\) - \(^{\dfrac{1}{2}}\) ) x \(\dfrac{6}{5}\) - 17
h) \(\dfrac{\left(-1\right)^3}{15}\) + \(\left(-\dfrac{2}{3}\right)^2\) : \(2\dfrac{2}{3}\) - \(\left|-\dfrac{5}{6}\right|\)
k) \(\dfrac{2.6^9-2^5.18^4}{2^2.6^8}\)
n) 3 - \(\left(-\dfrac{7}{8}\right)^0\) + \(\left(\dfrac{1}{2}\right)^3\) . 16
Mg giải gấp giúp mình ạ
d: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{6}\cdot\dfrac{6}{5}-17=1-17=-16\)
h: \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)
\(=-\dfrac{1}{15}+\dfrac{-8}{27}:\dfrac{8}{3}-\dfrac{5}{6}\)
\(=-\dfrac{1}{15}-\dfrac{1}{9}-\dfrac{5}{6}\)
\(=\dfrac{-6-10-75}{90}=\dfrac{-91}{90}\)
k: \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^8\cdot3^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}\)
\(=\dfrac{5}{2}\)
n: \(3-\left(-\dfrac{7}{8}\right)^0+\left(\dfrac{1}{2}\right)^3\cdot16\)
\(=3-1+\dfrac{1}{8}\cdot16\)
=2+2
=4
Thực hiện phép tính (tính nhanh nếu có thể):
a) \(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(\dfrac{5}{3}-\dfrac{3}{2}\right)+\left(\dfrac{7}{3}-\dfrac{5}{2}\right)\)
b) \(\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}\)
\(a.=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}=\dfrac{1+3-5}{2}-\dfrac{2+5-7}{3}=\dfrac{-1}{2}\)
\(b.\left(\dfrac{3}{4}-1\dfrac{1}{6}\right)^2:\sqrt{\dfrac{25}{144}}=\left(-\dfrac{5}{12}\right)^2:\dfrac{5}{12}=\dfrac{5}{12}\)
