Chứng minh : \(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{64}{125}\)
Chứng minh : \(\frac{\left(5^4-5^3\right)}{125^4}=\frac{64}{125}\)
Tìm x
a)\(^{3^x}+^{3^{x+2}}=810\)
b)\(\left(x+\frac{2017}{2018}\right)^6=0\)
Ta có : 3x + 3x + 2 = 810
=> 3x(1 + 32) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
ta có \(3^3+3^x+2=810\)
=>\(3^x\left(1+3^2\right)=810\)
=>\(3^x.10=810\)
=>\(3^x=81\)
=>\(3^x=3^4\)
=>x=4
Vậy x=4
\(\frac{\left(5^4-5^3\right)}{125^4}-\frac{64}{125}\)
\(\frac{\left(5^4-5^3\right)}{125^4}-\frac{64}{125}\)
\(=\frac{\left(625-125\right)}{500}-\frac{64}{125}\)
\(=\frac{500}{500}-\frac{64}{125}\)
\(=0-0,51\)
\(=-0,51\)
Chứng minh rằng:
a)\(12^8.9^{12}=18^{16}\)
b)\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
c)\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
Làm nhanh giúp mình nhá. Thanks. ^_^
a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
\(VP=18^{16}=2^{16}\cdot3^{32}\)
=> VT=VP
b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
(đề sai)
c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)
128.912=186
=216.38.324=216.332
=216.332=186
Chứng minh rằng :
\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(\frac{\left(5^4-5^3\right)}{125^4}=\frac{64}{125}\)
Làm nhanh thì 2 hoặc 3 thích nhé. Gấp lắm
Cái dấu sọc dọc xuống kế bên các phân số ko liên quan
\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left(3^3.\left(3-1\right)\right)^2}=\frac{3^6}{3^6.2^2}=\frac{1}{4}.\)
con thứ 2 làm tương tự. hình như là đề con thứ 2 sai em ơi
Đề 2 e chỉnh lại \(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{64}{125}\)
cmr :\(\frac{\left(5^4-5^3\right)^3}{125^5}\) =\(\frac{64}{25^3}\)
Tìm x, biết:
a) \(\left(\frac{-3}{4}\right)^{3x-1}=\frac{-27}{64}\)
b) \(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{265}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+3\right)^2}=\frac{64}{27}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
Tim x biet
k) \(\left[\left(3,75:\frac{1}{4}+2\frac{2}{5}.125\%\right)-\left(\frac{7}{2}.0,8-1,2:\frac{3}{2}\right)\right]:\left(1\frac{1}{2}+0,75\right)x=64\)
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=-\frac{64}{125}\)
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}=\left(\frac{-4}{5}\right)^3\)
=> \(\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)
=> \(\frac{2}{3}x=\frac{3}{5}-\frac{-4}{5}\)
=> \(\frac{2}{3}.x=\frac{3}{5}+\frac{4}{5}=\frac{7}{5}\)
=> \(x=\frac{7}{5}:\frac{2}{3}\)
=> \(x=\frac{7}{5}.\frac{3}{2}=\frac{21}{10}\)
\(\left(\frac{3}{5}-\frac{2}{3}x\right)^3=\frac{-64}{125}=\left(\frac{-4}{5}\right)^3\)
\(\Rightarrow\frac{3}{5}-\frac{2}{3}x=\frac{-4}{5}\)
\(\Rightarrow\frac{2}{3}x=\frac{3}{5}-\frac{-4}{5}=\frac{7}{5}\)
\(\Rightarrow x=\frac{7}{5}.\frac{3}{2}=\frac{21}{10}\)
so sánh:
\(\frac{\left(5^4-5^3\right)^3}{125^5}\) và\(\frac{64}{25^3}\)
Ta có:
\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{\left(5^3\right)^3.\left(5-1\right)^3}{\left(5^3\right)^5}=\frac{5^9.4^3}{5^{15}}=\frac{4^3}{5^6}=\frac{64}{5^6}\) (1)
\(\frac{64}{25^3}=\frac{64}{\left(5^2\right)^3}=\frac{64}{5^6}\) (2)
Từ (1) và (2) =>\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^3}\)