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\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\left(8+4\right)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=5\)
Vậy : \(\frac{15\cdot8+15\cdot4}{12\cdot3}=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\frac{18.\left(-130\right)}{\left(-12\right).\left(-130\right)}=\frac{-3}{2}\)
b) \(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{15.12}{12.3}=5\)
a) Ta có: \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\frac{2.\left(-13\right).3^2.2.5}{3.2^2.5.2.13}\)
\(=\frac{\left(-13\right).2^2.3^2.5}{2^3.3.5.13}\)
\(=-\frac{3}{2}\)
b) Ta có: \(\frac{15.8+15.4}{12.3}=\frac{3.5.\left(8+4\right)}{2^2.3.3}\)
\(=\frac{3.5.2^2.3}{2^2.3.3}\)
\(=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{15.12}{12.3}=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}=\)\(\frac{2\cdot\left(-1\right)\cdot13\cdot3\cdot3\cdot2\cdot5}{3\cdot\left(-1\right)2\cdot2\cdot\left(-1\right)\cdot5\cdot2\cdot13}\)=\(-3.\)
b)Ta có: \(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\left(8+4\right)}{12\cdot3}\)=\(\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ta có \(\frac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}\)=\(\frac{2.\left(-13\right).\left(-3\right).\left(-3\right).2.\left(-5\right)}{\left(-3\right).2.2.\left(-5\right).\left(-2\right).\left(-13\right)}\)
rút gọn đi còn: \(\frac{-3}{-2}\)=\(\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có cách làm sau
\(\dfrac{15.8+15.4}{12.3}=\dfrac{15.\left(8+4\right)}{12.3}=\dfrac{15.12}{12.3}=\dfrac{3.5.12}{12.3}\)
\(=\dfrac{5}{1}=5\)
b,
\(\dfrac{2.\left(-13\right).9.10}{-3.4.\left(-5\right).26}\)=\(\dfrac{2.\left(-13\right).\left(-3\right).\left(-3\right).\left(-2\right).\left(-5\right)}{-3.2.2_{ }.\left(-5\right).\left(-2\right).\left(-13\right)}\)=\(\dfrac{-3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{19}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9.\left(1+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Tu so:
15.8+15.4
=5.3.2.4+5.3.4
=12.10+12.5
=12.15
=>\(\frac{12.15}{12.3}=\frac{12}{12}.\frac{15}{3}=\frac{15}{3}=5\)
\(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{15.12}{12.3}=\frac{5}{1}=5\)
\(\frac{15.8+15.4}{12.3}=\frac{15.\left(8+4\right)}{12.3}=\frac{5.12}{4.3}=\frac{5}{1}=5\)