Tính giá trị biểu thức
21. [(1 245 + 987): 23 - 15. 12] + 21.
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=21 .[ 2232 : 8 - 180 ] + 21
=21.99 +21
=2100
\(21\cdot\left[\left(1245+987\right):2^3-15\cdot12\right]+21\)
\(=21\left(279-180+1\right)\)
\(=21\cdot100=2100\)
a) ( 12 + 21 - 23 ) - ( 23 - 21 + 10 )
= 12 + 21 - 23 - 23 - 21 + 10
= ( 12 + 10 ) - ( 21 - 21 ) - ( 23 - 23 )
= 22 - 0 - 0
= 22
b) ( 55 + 45 + 15 ) - ( 15 - 55 + 45 )
= 55 + 45 + 15 - 15 - 55 + 45
= ( 55 + 45 ) + ( 55 + 45 ) + ( 15 - 15)
= 100 + 100 + 0
= 200
\(a)\)\(-\left(12+21-23\right)-\left(23-21+10\right)\)
\(=-12-21+23-23+21-10\)
\(=\left(-21+21\right)+23-23-12-10\)
\(=0+0-12-10\)
\(=-22\)
\(b)\)\(\left(55+45+15\right)-\left(15-55+45\right)\)
\(=55+45+15-15+55-45\)
\(=\left(15-15\right)+\left(45-45\right)+55+55\)
\(=0+0+55+55\)
\(=110\)
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}\) \(\left(a\ne0\right)\)
Tại a = 12 biểu thức \(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}:\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}.\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{55}{96}=\dfrac{65}{96}\)
Để \(A=\dfrac{15}{23}< =>\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)
\(\Leftrightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{55}{92}< =>\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}< =>\dfrac{1}{a}=\dfrac{19}{184}< =>a=\dfrac{184}{19}\)
Thay \(a=12\) vào A ta có:
\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{65}{96}\)
Vậy:
____________________
Ta có:
\(A=\dfrac{15}{23}\) khi \(\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)
\(\Rightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right)\cdot\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{15}{23}\)
\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{55}{92}:\dfrac{5}{4}\)
\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{11}{184}\)
\(\Rightarrow a=\dfrac{1\cdot184}{11}=\dfrac{184}{11}\)
A = - \(\dfrac{1}{21}\) - \(\dfrac{1}{28}\)
A =- \(\dfrac{4}{84}\)- \(\dfrac{3}{84}\)
A = -\(\dfrac{7}{84}\)
A = - \(\dfrac{1}{12}\)
B = - \(\dfrac{8}{18}\) - \(\dfrac{15}{27}\)
B = - \(\dfrac{4}{9}\) - \(\dfrac{5}{9}\)
B = -1
C = - \(\dfrac{5}{12}\) + 0,75
C = -\(\dfrac{5}{12}\) + \(\dfrac{3}{4}\)
C = - \(\dfrac{5}{12}\) + \(\dfrac{9}{12}\)
C = \(\dfrac{4}{12}\)
C= \(\dfrac{1}{3}\)
D = 3,5 - { - \(\dfrac{2}{7}\))
D = \(\dfrac{7}{2}\) + \(\dfrac{2}{7}\)
D = \(\dfrac{49}{14}\) + \(\dfrac{4}{14}\)
D = \(\dfrac{53}{14}\)
\(21.[(1 245 + 987) : 2^3 - 15.12] + 21\)
= 21.(2232 : 8 - 180) + 21
= 21.(279 - 180) + 21
= 21. 99 + 21
= 21. (99+1)
= 21. 100
= 2 100.