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\(\left(-6\right).3+6.\left(-11\right)+\left(-6\right).\left(-2\right)\)
\(=\left(-6\right).3+\left(-6\right).11+\left(-6\right).\left(-2\right)\)
\(=\left(-6\right).\left[3+11+\left(-2\right)\right]\)
\(=\left(-6\right).12\)
\(=-72\)
52x - 3 - 2 . 52 = 52 : 3
52x - 3 - 2 = 52 : 3 : 52
52x - 3 - 2 = 3
52x - 3 = 3 + 2
52x -3 = 5
52x -3 = 51
⇒ 2x -3 = 1
2x = 1 + 3
2x = 4
x = 4 : 2
x = 2
Vậy x = 2
1,
\(\begin{array}{l}2,5.\left( {4,1 - 3 - 2,5 + 2.7,2} \right) + 4,2:2\\ = 2,5.\left( {4,1 - 3 - 2,5 + 14,4} \right) + 4,2:2\\ = 2,5.\left( {1,1 - 2,5 + 14,4} \right) + 2,1\\ = 2,5.\left( { - 1,4 + 14,4} \right) + 2,1\\ = 2,5.13 + 2,1\\ = 32,5 + 2,1\\ = 34,6\end{array}\)
2,
Cách 1:
\(\begin{array}{l}2,86.4 + 3,14.4 - 6,01.5 + {3^2}\\ = 11,44 + 12,56 - 30,05 + 9\\ = \left( {11,44 + 12,56} \right) + \left( { - 30,05 + 9} \right)\\ = 24 + \left( { - 21,05} \right)\\ = 24 - 21,05\\ = 2,95\end{array}\)
Cách 2:
\(\begin{array}{l}2,86.4 + 3,14.4 - 6,01.5 + {3^2}\\ = 4.(2,86+3,14) - 30,05 + 9\\ = 4.6 + \left( { - 30,05 + 9} \right)\\ = 24 + \left( { - 21,05} \right)\\ = 24 - 21,05\\ = 2,95\end{array}\)
\( (-11).77 - 23.11 + (-29) \)
\(=11.(-77)-23.11+(-29)\)
\(=11.[(-77)-23]+(-29)\)
\(=11.(-100)+(-29)\)
= -1100 + (-29)
= -(1100+29)=-1071
\(2\left(x-\frac{1}{2}\right)-5\left(\frac{3}{10}-1\right)=7\)
\(\Rightarrow\left(2x-1\right)-\left(\frac{3}{2}-5\right)=7\)
\(\Rightarrow\left(2x-1\right)-\frac{-7}{2}=7\)
\(\Rightarrow\left(2x-1\right)=7+\frac{-7}{2}=\frac{7}{2}\)
\(\Rightarrow2x=\frac{7}{2}+1\)
\(\Rightarrow x=\frac{9}{2}:2\)
\(\Rightarrow x=\frac{9}{4}\)
2(x-1/2)-5(3/10-1)=7
2x-1-3/2+5=7
2x=7+1+3/2-5
2x=9/2
x=9/4
Vậy x=9/4.
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}\)
\(A-\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(A=\dfrac{1}{2^{2022}}-2\)
12A=12+(12)2+(12)3+(12)4+...+(12)202312A=12+(12)2+(12)3+(12)4+...+(12)2023
A−12A=(12)2023−1A−12A=(12)2023−1
12A=(12)2023−112A=(12)2023−1
A=122022−2
68,25
68,25