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=2.3+(2.3).4+(2.3).49
3.5+(3.5).4+(3.5).49
=2.3.(1+4+49)
3.5.(1+4+49)
=2.3
3.5
=6
15
\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.\left(3+2.6+7.21\right)}{5.\left(3+2.6+7.21\right)}=\frac{2}{5}\)
\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+8.3+98.3}{3.5+3.20+3.45}=\frac{3.\left(2+8+98\right)}{3.\left(5+20+45\right)}=\frac{108}{70}=\frac{54}{35}=1\frac{19}{35}\)
2.3 + 4.6 + 14.21 / 3.5 + 6.10 + 21.35
= 2.3.(1 + 2.2 + 7.7) / 3.5.(1 + 2.2 + 7.7)
= 2/5
\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.3+4.6+2.7.3.7}{3.5+2.3.5.2+3.7.5.7}\)
\(=\frac{6.1+4.6+6.49}{15.1+15.4+15.49}\)
\(=\frac{6\left(1+4+49\right)}{35\left(1+4+49\right)}=\frac{6}{35}\)
tren tu : dat 2.3 ra ngoai
trong ngoặc: 1+2.2+7.7
dưới mẫu : dat 3.5 ra ngoai
trong ngoặc: 1+2.2+7.7
rút gọn: 2.3/3.5=2/5
cho mk vs
\(\dfrac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}=\dfrac{2\left(3+2\cdot3+7\cdot21\right)}{3\left(5+2\cdot10+7\cdot35\right)}=\dfrac{2\cdot3\left(1+2\cdot3+7\cdot5\right)}{3\cdot5\left(1+2\cdot3+7\cdot5\right)}=\dfrac{2}{5}\)
\(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
\(=\dfrac{6+24+294}{15+60+735}\)
\(=\dfrac{324}{810}\)
\(=\dfrac{2}{5}\)
\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
=\(\frac{2\left(3+2.6+7.21\right)}{5\left(3+2.6+7.21\right)}\)
=\(\frac{2}{5}\)
ta có:
\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
bài trên có các số ở tử và mẫu giống nhau đó là: 21;6;3
=>ta cho chúng hết cho nhau
vậy ta còn phân số:
\(\frac{2+4+14}{3+10+35}\)
=\(\frac{20}{48}\)=\(\frac{5}{12}\)
a)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)
\(=\frac{1\cdot2\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}{3\cdot4\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}\)
\(=\frac{1\cdot2}{3\cdot4}=\frac{1}{6}\)
b) \(\frac{12\cdot13+24\cdot26+36\cdot39}{24\cdot26+48\cdot52+72\cdot78}\)
\(=\frac{12\cdot13\cdot\left(1+2\cdot2+3\cdot3\right)}{24\cdot26\cdot\left(1+2\cdot2+3\cdot3\right)}\)
\(=\frac{12\cdot13}{24\cdot26}=\frac{1}{4}\)
\(\frac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}=\frac{2\cdot3\cdot\left(1+2\cdot2+7\cdot7\right)}{3\cdot5\cdot\left(1+2\cdot2+7\cdot7\right)}=\frac{2\cdot3}{3\cdot5}=\frac{2}{5}\)
Giải:
a) \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\)
\(=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2.2^6}{5^4.3^2.2^{13}}\)
\(=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}\)
\(=\dfrac{1}{2^6}=\dfrac{1}{64}\)
b) \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\)
\(=\dfrac{15^2.16^3.16-15^2.16^3.15}{12^2.20^2.20-20^2.12^2.12}\)
\(=\dfrac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\)
\(=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3.1}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.8}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.\left(2^2\right)^2.\left(2^2\right)^2.5^2.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^4.2^4.2^3}\)
\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}\)
\(=2\)
c) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
\(=\dfrac{2.3+2.3.2.2+2.3.7.7}{3.5+3.5.2.2+3.5.7.7}\)
\(=\dfrac{2.3+2.3.4+2.3.49}{3.5+3.5.4+3.5.49}\)
\(=\dfrac{2.3.\left(1+4+49\right)}{3.5.\left(1+4+49\right)}\)
\(=\dfrac{2.3}{3.5}\)
\(=\dfrac{2}{5}\)
Chúc bạn học tốt!
\(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\)
= \(\dfrac{2.3+8.3+98.3}{3.5+3.20+245.3}\)
= \(\dfrac{3.\left(2+8+98\right)}{3.\left(5+20+245\right)}\)
= \(\dfrac{3.108}{3.270}\)
= \(\dfrac{108}{270}\) = \(\dfrac{2}{5}\)
cảm ơn nha