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Trả lời :
a, \(3^6\text{ x }3^2+2\text{ x }81^2=3^8+2\text{ x }\left(3^4\right)^2=3^8+2\text{ x }3^8=\left(2+1\right)\text{ x }3^8=3\text{ x }3^8=3^9\)
b, \(\left(6^3\text{ x }8^4\right)\text{ : }12^3\)
\(=6^3\text{ x }8^4\text{ : }12^3\)
\(=\left(2\cdot3\right)^3\text{ x }\left(2^3\right)^4\text{ : }\left(2^2\cdot3\right)^3\)
\(=2^3\cdot3^3\text{ x }2^{12}\text{ : }\left(2^6\cdot3^3\right)\)
\(=2^3\cdot3^3\text{ x }2^{12}\text{ : }2^6\text{ : }3^3\)
\(=2^9\)
Câu c không hiểu đề !
Lời giải:
\(2^{27}.3^{18}=(2^3)^9.(3^2)^9=8^9.9^9=(8.9)^9=72^9\)
\(25^4.2^8=25^4.(2^2)^4=25^4.4^4=(25.4)^4=100^4\)
\((-9)^4.27^2=9^4.27^2=(3^2)^4.(3^3)^2=3^8.3^6=3^{8+6}=3^{14}\)
\(-5^4:25=-5^4:(5^2)=-5^{4-2}=-5^2\)
a) Ta thấy: có 5 thừa số (-5) nên tích mang dấu "-" nên:
(-5).(-5).(-5).(-5).(-5) = -55
b) (-2).(-2).(-2).(-3).(-3).(-3)
= (-2).(-3).(-2).(-3).(-2).(-3)
=6.6.6 = 63
hoặc: ta thấy tích có 6 thừa số nguyên âm nên tích mang dấu "+"
(-2).(-2).(-2).(-3).(-3).(-3)
= 23.33
Câu 1 :
a) \(4.\left(\frac{1}{32}\right)^{-2}:\left(2^3.\frac{1}{16}\right)\)
\(=2^2.32^2:\left(\frac{1}{8}.16\right)=\left(2.32\right)^2:2=64^2:2\)
\(=2048=2^{11}\)
b) \(5^2.3^5.\left(\frac{3}{5}\right)^2\)
\(=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)
VIẾT CÁC BIỂU THỨC DƯỚI DẠNG LUỸ THỪA CỦA 1 SỐ HỮU TỈ
\(a,4\cdot\left(\frac{1}{32}\right)^{-2}:\left(2^3\cdot\frac{1}{16}\right)\\ =4\cdot1024:\left(8\cdot\frac{1}{16}\right)\\ =4\cdot1024:\frac{1}{2}\\ =2\cdot1024\\ =2\cdot2^{10}\\ =2^{11}\)
\(b,5^2\cdot3^5\cdot\left(\frac{3}{5}\right)^2\\ =5^2\cdot\left(\frac{3}{5}\right)^2\cdot3^5\\ =3^2\cdot3^5\\ =3^7\)
2 SO SÁNH
\(a,10^{20}\text{ và }9^{10}\)
Có: \(9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(\Rightarrow10^{20}>3^{20}\\ \text{hay}\text{ }10^{20}>9^{10}\)
\(b,\left(-5\right)^3\text{ và }\left(-3\right)^{50}\)
Có: \(\left(-3\right)^{50}=3^{50}\)
\(\Rightarrow\left(-5\right)^3< 3^{50}\\ \text{hay }\left(-5\right)^3< \left(-3\right)^{50}\)
\(c,64^3\text{ và }16^{12}\)
Có: \(64^3=\left(4^3\right)^3=4^9;16^{12}=\left(4^2\right)^{12}=4^{24}\)
\(\Rightarrow4^9< 4^{24}\\ hay\text{ }64^3< 16^{12}\)
\(d,\left(\frac{1}{16}\right)^{10}\text{ và }\left(\frac{1}{2}\right)^{50}\)
Có: \(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5\cdot10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\\ \text{hay }\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
1,
\(A=2^0+2^1+2^2+..+2^{2006}\)
\(=1+2+2^2+...+2^{2016}\)
\(2A=2+2^2+2^3+..+2^{2007}\)
\(2A-A=\left(2+2^2+2^3+..+2^{2007}\right)-\left(1+2+2^2+..+2^{2006}\right)\)
\(A=2^{2017}-1\)
\(B=1+3+3^2+..+3^{100}\)
\(3B=3+3^2+3^3+..+3^{101}\)
\(3B-B=\left(3+3^2+..+3^{101}\right)-\left(1+3+..+3^{100}\right)\)
\(2B=3^{101}-1\)
\(\Rightarrow B=\frac{3^{100}-1}{2}\)
\(D=1+5+5^2+...+5^{2000}\)
\(5D=5+5^2+5^3+...+5^{2001}\)
\(5D-D=\left(5+5^2+..+5^{2001}\right)-\left(1+5+...+5^{2000}\right)\)
\(4D=5^{2001}-1\)
\(D=\frac{5^{2001}-1}{4}\)
a) \(\left(-8\right).\left(-3\right)^3.\left(+125\right)\\ =\left(-2\right)^3.\left(-3\right)^3.\left(+5\right)^3\\ =\left[\left(-2\right).\left(-3\right).\left(+5\right)\right]^3\\ =30^3\)
b) \(27.\left(-2\right)^3.\left(-7\right).\left(+49\right)\\ =3^3.\left(-2\right)^3.\left(-7\right).\left(-7\right)^2\\ =\left[3.\left(-2\right)\right]^3.\left[\left(-7\right).\left(-7\right)^2\right]\\ =\left(-6\right)^3.\left(-7\right)^3\\ =\left[\left(-6\right).\left(-7\right)\right]^3\\ =42^3\)
Trả lời
(2+3)4.42= 54.42= 625.16=10000
52.22.4= 25.4.4=400
Học tốt. k mik nha