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a, 38 : 34 = 38-4 = 34
b, 108 : 102 = 108-2 = 106
c, a6 : a = a6-1 = a5
a) 3 mũ 15 : 3 mũ 15 = 3 mũ 10
b) 4 mũ 6 : 4 mũ 6 = 1
c) 9 mũ 8 : 3 mũ 2 = 9 mũ 8 : 9 = 9 mũ 7
bài mình làm 100% đúng nhé
bài này cô mình chữa rồi
a: \(=2^6\cdot2^{20}=2^{26}\)
b: \(=3^9\cdot3^8\cdot3^2\cdot2\cdot13=3^{19}\cdot2\cdot13\)
1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)
2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)
Ta có : \(\frac{8^x}{2^x}=\left(\frac{8}{2}\right)^x=4^x=4\)
=> \(x=1\)
Vậy ...
a: \(7\cdot3^x=5\cdot3^7+2\cdot3^7\)
\(\Leftrightarrow7\cdot3^x=7\cdot3^7\)
=>3x=37
hay x=7
b: \(4^{x+3}-3\cdot4^{x+1}=13\cdot4^{11}\)
\(\Leftrightarrow4^{x+1}\left(4^2-3\right)=13\cdot4^{11}\)
=>x+1=11
hay x=10
d: \(\left(x-1\right)^{13}=\left(x-1\right)^{12}\)
\(\Leftrightarrow\left(x-1\right)^{12}\left(x-2\right)=0\)
hay \(x\in\left\{1;2\right\}\)
a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)
b) \(\left(4x-1\right)^2=16x^2-8x+1\)
c) \(\left(x+2\right)^2=x^2+4x+4\)
d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)
e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)
f) \(25x^2-4=\left(5x-2\right)\left(5x+2\right)\)
g) \(\left(x+1\right)^3=x^3+3x^2+3x+1\)
h) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)
k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)
y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)
\(B=4sin^4\dfrac{\pi}{16}+2cos\dfrac{\pi}{8}\)
\(=4sin^4\dfrac{\pi}{16}-4sin^2\dfrac{\pi}{16}+2\)
\(=4sin^2\dfrac{\pi}{16}\left(sin^2\dfrac{\pi}{16}-1\right)+2\)
\(=-4sin^2\dfrac{\pi}{16}.cos^2\dfrac{\pi}{16}+2\)
\(=-sin^2\dfrac{\pi}{8}+2\)
\(=\dfrac{1}{2}\left(1-2sin^2\dfrac{\pi}{8}\right)+\dfrac{3}{2}\)
\(=\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{3}{2}\)
\(=\dfrac{\sqrt{2}+6}{4}\)
a: \(2^{10}:2^8=2^2=4\)
b: \(4^6:4^3=4^3=64\)
c: \(8^5:8^4=8^{5-4}=8\)
c: \(7^4:7^4=7^0=1\)