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
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 : \(\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{64}{125}\)
a) \(\frac{\left(-1\right)^3}{15}+\left(-\frac{2}{3}\right):2\frac{2}{3}-\left|-\frac{5}{6}\right|\)
b) \(1\frac{5}{13}-0,\left(3\right)-\left(1\frac{4}{9}+\frac{18}{13}-\frac{1}{3}\right)\)
c) \(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{5}-125\frac{1}{3}\)
d) \(\frac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
Tim x biet
k) \(\left[\left(3,75:\frac{1}{4}+2\frac{2}{5}.125\%\right)-\left(\frac{7}{2}.0,8-1,2:\frac{3}{2}\right)\right]:\left(1\frac{1}{2}+0,75\right)x=64\)
Thuc hien phep tinh:
a/\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}\)+ \(\frac{0,6-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-0,16-\frac{4}{125}-\frac{4}{625}}\)
b/ \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
giúp tui với : -8 . 25 . ( -2 ) . ( -25 ) .4 .125
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) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
Thực hiện phép tính (Tính hợp lý nếu có thể)
g) \(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
h) \(10.\sqrt{0,01}.\sqrt{\frac{16}{9}+3\sqrt{49}-\frac{1}{6}\sqrt{4}}\)
i) \(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}\)
k) \((2\frac{1}{3}+3\frac{1}{2}):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7\frac{1}{2}\)
n) \(4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
m) \([1,5+2\frac{1}{2}-\left(2\sqrt{2}\right)^2]:[4\frac{1}{2}-\sqrt{1,96}+0,9]\)
o) \(\frac{5}{21}.\left(4\frac{1}{5}.7\frac{3}{4}+5\frac{1}{4}.4,2\right)\)
p) \(\left(\frac{2}{5}+\frac{2}{7}-\frac{2}{11}\right):\left(\frac{3}{7}-\frac{3}{11}+\frac{3}{5}\right)\)
Làm nhanh làm đúng mình Tick nha :))
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
\(^{\left(\frac{5^4-5^3}{125^4}\right)^3}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\frac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\frac{1}{3}\right)^3.\left(1\frac{1}{2}\right)^4.\left(-1\right)^5}\)