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a; A = |-101| + |21| + |-99| - |25|
A = 101 + 21 + 99 - 25
A = (101 + 99) - (25 - 21)
A = 200 - 4
A = 196
b; B = ||17 - 42| - 64|
B = ||-25| - 64|
B = |25 - 64|
B = |-39|
B = 39
c, C = |27 - 72| + |33 - 34| + |103 - 35|
C = |128 - 49| + |27 - 81| + |1000 - 243|
C = |79| + |-54| + | 757|
C = 79 + 54 + 757
C = 133 + 757
C = 890
\(A=\dfrac{7}{3}+\dfrac{5}{7}+\dfrac{2}{3}-\dfrac{7}{12}+\dfrac{5}{2}=3+\dfrac{221}{84}=\dfrac{473}{84}\)
f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)
\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)
g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)
Bài giải
\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)
\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)
\(=2\)
\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)
Bài làm
\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=\frac{84}{12}+\left(\frac{7}{12}-\frac{6}{12}+\frac{36}{12}\right)-\left(\frac{1}{12}+\frac{60}{12}\right)\)
\(=\frac{84}{12}+\frac{37}{12}-\frac{61}{12}\)
\(=\frac{60}{12}\)
\(=5\)
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