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a) \(4^8\cdot4^4=\left(2^2\right)^8\cdot\left(2^2\right)^4=2^{16}\cdot2^8=2^{16+8}=2^{24}\)
b) \(5^{12}\cdot7-5^{11}\cdot10\)
\(=5^{11}\cdot\left(5\cdot7-10\right)=5^{11}\cdot\left(35-10\right)=5^{11}\cdot25\)
\(=5^{11}\cdot5^2=5^{11+2}=5^{13}\)
d) \(27^{16}:9^{10}\)
\(=\left(3^3\right)^{16}:\left(3^2\right)^{10}=3^{48}:3^{20}=3^{48-20}=3^{28}\)
e) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5^{9-8}=5\)
f) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:4\right)^4-\left(32:16\right)^{12}\)
\(=6^4-2^{12}\)
\(=2^4\cdot\left(3^4-2^8\right)=2^4\cdot-175=-2800\)
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
a: =-1+17/20=-3/20
b: =(28/60-33/60)*(-25/3)
=(-1/12)*(-25/3)=1/12*25/3=25/36
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)
Câu 3:
a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)
\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)
b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)
\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)
Câu 4:
a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)
b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)
\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)
\(=-27\cdot100-23\cdot100\)
\(=100\left(-27-23\right)\)
\(=-50\cdot100=-5000\)
c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)
d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)
\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)
\(=-1+1+\dfrac{5}{6}\)
\(=\dfrac{5}{6}\)
1) \(\left(+15\right)+\left(+17\right)=15+17=32\)
2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)
3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)
4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)
5) \(\left(-15\right)+20=20-15=5\)
6) \(\left(-5\right)+8+7+5\)
\(=\left(-5+5\right)+\left(8+7\right)\)
\(=15\)
7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)
\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)
\(=\left(-10\right)+\left(-11\right)\)
\(=-21\)
8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)
\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)
\(=10+\left(-30\right)\)
\(=-20\)
9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)
\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)
\(=\left(-17\right)+\left(-100\right)\)
\(=-117\)
10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)
\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)
\(=\left(-400\right)+40+240\)
\(=\left(-360\right)+240\)
\(=-120\)
11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)
\(=\left(-596\right)+2001+1999-404+189\)
\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)
\(=\left(-1000\right)+2190+1999\)
\(=1190+1999\)
\(=3189\)
12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
\(\dfrac{15^2\cdot16^4-15^3\cdot16^3}{12^2\cdot20^3-20^2\cdot12^3}\)
\(=\dfrac{15^2\cdot16^3\left(16-15\right)}{12^2\cdot20^2\left(20-12\right)}\)
\(=\dfrac{3^2\cdot5^2\cdot2^{12}}{2^4\cdot3^2\cdot2^4\cdot5^2\cdot2^3}\)
\(=\dfrac{2^{12}}{2^{11}}=2^1=2\)
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