Rút gọn biểu thức : 6-(y+15)+17 bài này zúp mik nhoa:3
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Bài 1 : Bài giải
\(\frac{28^{15}\cdot3^{17}}{84^{16}}=\frac{\left(2^2\cdot7\right)^{15}\cdot3^{17}}{\left(2^2\cdot3\cdot7\right)^{16}}=\frac{2^{30}\cdot7^{15}\cdot3^{17}}{2^{32}\cdot3^{16}\cdot7^{16}}=\frac{3}{2^2\cdot7}=\frac{3}{4\cdot7}=\frac{3}{28}\)
Bài 2 : Bài giải
\(\frac{3^6\cdot21^{12}}{175^9\cdot7^3}=\frac{3^6\cdot\left(3\cdot7\right)^{12}}{\left(5^2\cdot7\right)^9\cdot7^3}=\frac{3^6\cdot3^{12}\cdot7^{12}}{5^{18}\cdot7^9\cdot7^3}=\frac{3^{18}\cdot7^{12}}{5^{18}\cdot7^{12}}=\frac{3^{18}}{5^{18}}\)
\(\frac{3^{10}\cdot6^7\cdot4}{10^9\cdot5^8}=\frac{3^{10}\cdot\left(2\cdot3\right)^7\cdot2^2}{\left(2\cdot5\right)^9\cdot5^8}=\frac{3^{10}\cdot2^7\cdot3^7\cdot2^2}{2^9\cdot5^9\cdot5^8}=\frac{3^{17}\cdot2^9}{2^9\cdot5^{17}}=\frac{3^{17}}{5^{17}}\)
Ta có : \(3^{17}\cdot5^{18}=3^{17}\cdot5^{17}\cdot5=\left(3\cdot5\right)^{17}\cdot5=15^{17}\cdot5\)
\(3^{18}\cdot5^{17}=3\cdot3^{17}\cdot5^{17}=3\cdot\left(3\cdot5\right)^{17}=3\cdot15^{17}\)
\(\text{ Vì }5\cdot15^{17}>3\cdot15^{17}\text{ }\Rightarrow\text{ }3^{17}\cdot5^{18}>3^{18}\cdot5^{17}\text{ }\Rightarrow\text{ }\frac{3^{18}}{5^{18}}< \frac{3^{17}}{5^{17}}\)
28^15x3^17/84^16
=28^15x3^17/(28x3)^16
=28^15x3^17/28^16x28^16
=3/28
\(A=\frac{2^{15}.3^{12}-3^{11}.2^{17}}{2^{15}.3^{11}+3^{11}.2^{17}}\)
\(A=\frac{2^{15}.3^{11}.\left(3-2^2\right)}{2^{15}.3^{11}.\left(1+2^2\right)}\)
\(A=\frac{3-2^2}{1+2^2}\)
\(A=\frac{-1}{5}\)
3√2 - 5√18 + 6√72 - 4√98 = 3√2-5.3√2+6.2.3√2-4.7/3.3√2
= 3√2(1-5+12-28/3)
= 3√2.(-4/3)
= -4√2
Bây giờ mình sẽ viết đầy đủ hơn nhé:
a) \(\frac{28^{15}.3^{17}}{84^{16}}=\frac{\left(28.3\right)^{15}.3^2}{\left(28.3\right)^{15}.84}=\frac{9}{84}=\frac{3}{28}\)
b)\(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+2^2.3^2}{5.3^8+20}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}\)\(=\frac{9}{5}\)
\(A=\frac{\left(9^2\right)^{11}.3^{17}}{\left(3^3\right)^{10}.9^{15}}=\frac{9^{22}.3^{17}}{3^{30}.9^{15}}=\frac{9^7}{3^{12}}=9\)
6-(y+15)+17
=6-y-15+17
=6-15+17-y
=8-y
vậy kết quả là 8-y
k cho mik zới
ố, ghê nhoa