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\(\left(1\right)\frac{121.75.130.169}{39.60.11.198}\)
\(=\frac{121.15.5.13.10.13.13}{13.3.15.4.11.18.11}\)
\(=\frac{121.5.13^2.10}{3.4.18.121}\)
\(=\frac{50.196}{12.18}\)
\(=\frac{9800}{216}=\frac{1225}{27}\)
\(\left(2\right)\frac{135.350+135.550}{900.100+35.900}\)
\(=\frac{135.\left(350+550\right)}{900.\left(100+35\right)}\)
\(=\frac{135.900}{900.135}=1\)
HOK TOT
1)=11^2.15.5.13.10.169/13.3.15.4.11.198
=5.10.169/3.4.18
=8450/216=4225/108
2)=135.(350+550)/900.(100+35)
=135.900/900.135=1
\(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11^2\cdot3\cdot5^2\cdot13\cdot2\cdot5\cdot13^2}{13\cdot3\cdot2^2\cdot3\cdot5\cdot11\cdot2\cdot11\cdot3^2}\)
\(=\frac{11^2\cdot3\cdot5^3\cdot13^3\cdot2}{13\cdot3^4\cdot2^3\cdot5\cdot11^2}=\frac{13\cdot5^2}{3^3\cdot2^2}=\frac{325}{36}\)
P/S:Không bt có đúng ko nx nhưng hướng làm là như vậy
\(a,\frac{121.75.130.169}{39.60.11.198}=\frac{11^2.5^2.3.13.2.5.13^2}{13.3.5.2^2.3.11.3^2.2.11}\)\(=\frac{11^2.5^3.3.13^3.2}{13.3^4.5.2^3.11^2}=\frac{5^2.13^2}{3^3.2^2}=\frac{4425}{108}\)
\(b,\frac{1989.1990+3978}{1992.1991-3984}=\frac{1989.1990+1989.2}{1992.1991-1992.2}\)\(=\frac{1989\left(1990+2\right)}{1992\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=\frac{1}{1}=1\)
\(c.\frac{135.350+135.550}{900.100+35.900}=\frac{135\left(350+550\right)}{900\left(100+35\right)}=\)\(\frac{135.900}{900.135}=\frac{1}{1}=1\)
\(d.\frac{243.650-243.350}{600.200+600.43}=\frac{243\left(650-350\right)}{600\left(200+43\right)}\)\(=\frac{243.300}{600.243}=\frac{300}{600}=\frac{1}{2}\)
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
b) \(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{2\cdot9}{5}=\frac{18}{5}\)
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot1\cdot11\cdot1}{1\cdot5\cdot7\cdot1}=\frac{22}{35}\)
c) \(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11\cdot11\cdot13\cdot10\cdot169}{13\cdot3\cdot6\cdot10\cdot11\cdot11\cdot6\cdot3}\)
\(=\frac{169}{3\cdot6\cdot6\cdot3}=\frac{169}{324}\)
d) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)
Ta có: \(\frac{x}{42}=\frac{15}{21}=\frac{5}{7}\Rightarrow7x=42.5\)
\(\Rightarrow7x=210\)
\(\Rightarrow x=30\)
Tương tự: \(\frac{45}{y}=\frac{5}{7}\Rightarrow5y=45.7\)
\(\Rightarrow5y=315\)
\(\Rightarrow y=63\)
\(\frac{120}{z}=\frac{5}{7}\Rightarrow5z=120.7\)
\(\Rightarrow5z=840\)
\(\Rightarrow z=168\)
Vậy x = 30; y = 63 và z = 168
Ta có : \(\frac{15}{21}=\frac{5}{7}\rightarrow\frac{x}{42}=\frac{45}{y}=\frac{120}{z}=\frac{5}{7}\)
Mà : \(\frac{x}{42}=\frac{5}{7}\rightarrow x=\frac{42\cdot5}{7}=30\)
\(\frac{45}{y}=\frac{5}{7}\rightarrow y=\frac{45\cdot7}{5}=63\)
\(\frac{120}{z}=\frac{5}{7}\rightarrow z=\frac{120.7}{5}=168\)
Ta có \(\frac{1}{4}=\frac{3942}{15768}\)
Vậy số đó là \(\frac{3942}{15768}\).
Đáp án thì biết rồi, mong các bạn trình bày chi tiết dùm
+) \(\frac{-180}{270}=\frac{-2}{3}\) +) \(\frac{22}{-165}=\frac{-2}{15}\)
+) \(\frac{-39}{261}=\frac{-13}{87}\) +) \(\frac{4.26}{52.20}=\frac{4.13.2}{4.13.2.10}=\frac{1}{10}\)
+) \(\frac{7.\left(-32\right)}{16.35}=\frac{7.\left(-2\right).16}{16.7.5}=\frac{-2}{5}\)
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