\(\dfrac{2^7.9^2}{3^3.2^5}\)
2^7.9^2/3^3.2^5
\(\frac{2^7.9^2}{3^3.2^5}\)\(=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=2^2.3=4.3=12\)
\(\frac{2^7.9^2}{3^3.2^5}\)
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=2^2.3=12\)
Có gì sai sót thì bỏ qua nhé chị !
Ta có : \(\frac{2^7.9^2}{3^3.2^5}=\frac{2^5.2^2.81}{27.2^5}=\frac{2^5.2^2.27.3}{27.2^5}=2^2.3=12\)
Vậy \(\frac{2^7.9^2}{3^3.2^5}=12\)
\(\frac{2^7\cdot9^2}{3^3\cdot2^5}=\frac{2^7\cdot\left(3^2\right)^2}{3^3\cdot2^5}=\frac{2^7\cdot3^4}{3^3\cdot2^5}=2^2\cdot3=4\cdot3=12\)
Tính bằng cách hợp lí:
\(\frac{2^7.9^2}{3^3.2^5}\)
Ta có :\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}\)
\(=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=12\)
Trả lời:
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^2.\left(3^2\right)^2}{3^3}=\frac{2^2.3^4}{3^3}=\frac{2^2.3}{1}=4.3=12\)
Học tốt
thực hiện phép tính:
\(\frac{2^7.9^2}{3^3.2^5}\)
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=12\)
\(\frac{2^7.9^2}{3^3.2^5}\)
\(=\frac{2^{5+2}.\left(3^2\right)^2}{3^3.2^5}\)
\(=\frac{2^5.2^2.3^4}{3^3.2^5}\)
\(=2^2.3\)
\(=4.3\)
\(=12\)
^_^
\(\frac{2^7\cdot9^2}{3^3\cdot2^5}\)
⇒\(\frac{2^7\cdot\left(3^2\right)^2}{3^3\cdot2^5}\)
⇒\(\frac{2^7\cdot3^4}{3^3\cdot2^5}\)
⇒\(2^2\cdot2\)
⇒\(4\cdot2\)
➩\(8\)
Chúc bạn học tốt nhé Thuong Hoai!
2^3.2^2+7.9
Thực hiện các phép tính:
j) \(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}\)
k) \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}\)
\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)
a) \(\dfrac{2^7.9^3}{6^5.8^2}\)
b) \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}\)
a) \(\dfrac{2^7.9^3}{6^5.8^2}=\dfrac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\dfrac{2^7.3^6}{2^5.3^5.2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
a) \(\dfrac{2^7.9^3}{6^5.8^2}=\dfrac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\dfrac{2^7.3^6}{2^5.3^5.2^6}=\dfrac{2^7.3^6}{2^{11}.3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
b) \(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\dfrac{3^5}{0,2}=1215\)
Tìm x
\(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(2^{x-3}-3.2^x=-92\)
1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)
\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)
2) \(2^{x-3}-3.2^x=-92\)
\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)
\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)
\(\Rightarrow2^x=32\Rightarrow x=5\)
A=(-4/5+4/3)+(-5/4+14/5)-7/3
B=8/3.2/5.3/8.10.19/92
C=-5/7.2/11+-5/7.9/14+12/7
Mk sửa đề nha :
\(C=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{12}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
\(=\frac{-5}{7}.1+\frac{12}{7}\)
\(=\frac{-5}{7}+\frac{12}{7}\)
\(=1\)
Study well ! >_<