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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: \(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5\cdot10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\\ \text{hay }\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
\(50-\left[\left[20-2^3\right]:2+34\right]=50-\left[\left[20-8\right]:2+34\right]\)
\(=50-\left[12:2+34\right]=50-\left[6+34\right]=50-40=10\)
\(P=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(\Rightarrow P=\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{9}{9}-\frac{1}{9}\right)\left(\frac{16}{16}-\frac{1}{16}\right)...\left(\frac{2500}{2500}-\frac{1}{2500}\right)\)
\(\Rightarrow P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)
\(\Rightarrow P=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
\(\Rightarrow P=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{\left(2.3.4...50\right)\left(2.3.4...50\right)}\)
\(\Rightarrow P=\frac{51}{50.2}=\frac{51}{100}>\frac{50}{100}=\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Ta có:
\(P=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{50^2}\right)\)
\(P=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{2500}\right)\)
\(P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{2499}{2500}\)
\(P=\frac{3.8.15.....2499}{4.9.16.....2500}\)
Tới chỗ này rồi tiếp tục rút gọn
Kết quả cuối cùng là: \(P>\frac{1}{2}\)
Xin lỗi nha, tớ ko có giỏi ở phần rút gọn.
\(50-\left[\left(50-2^3\cdot5\right)\div2+3\right]=50-\left[10\left(5-2^2\right)\div2+3\right]=50-\left(5\cdot1+3\right)=50-8=42\)
=37 nha bn