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bài 2
\(2^x\cdot7=224\)
\(\Rightarrow2^x=224\div7\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(\left(3^x+5\right)^2=289\)
\(\Rightarrow\left(3^x+5\right)^2=17^2\)
\(\Rightarrow3^x+5=17\)
\(\Rightarrow3^x=17-5\)
\(\Rightarrow3^x=12\)
\(2^3.19-2^3.14+1^{2018}\)
\(=2^3\left(19-14\right)+1\)
\(=2^3.5+1\)
\(=41\)
\(10^2-\left[60:\left(5^6:5^4-3.5\right)\right]\)
\(=10^2-\left[60:\left(5^2-3.5\right)\right]\)
\(=10^2-\left[60:10\right]\)
\(=10^2-6\)
\(=94\)
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7
a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)
\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)
b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)
\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)
\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)
c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)
\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)
\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)
d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)
e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)
\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)
a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)
\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{303}{610}\)
\(\Rightarrow B=\frac{101}{610}\)
b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)
\(\Rightarrow C=\frac{408}{205}\)
c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)
a) 2 + 6 + 10 + 14 +...+202
= 2.1 + 2.3 + 2.5 + 2.7 +...+2.101
=2.(1+3+5+7+...+101)
=2.[(1+101).51:2]
=2.2601
=5202
b) Đặt A=1+2+22+23+...+265
=> 2A=2+22+23+24+...+266
=>2A-A=266-1
A=266-1
+) Số số hạng của dãy là : \(\left(202-2\right):4+1=51\) (số)
Tổng của dãy là : \(\frac{\left(202+2\right)\times51}{2}=5202\)
+) Đặt \(A=1+2+2^2+2^3+...+2^{65}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{66}\)
\(\Rightarrow2A-A=A=\left(2+2^2+2^3+2^4+...+2^{66}\right)-\left(1+2+2^2+2^3+...+2^{65}\right)\)
\(\Rightarrow A=2^{66}-1\)
+) Đặt \(B=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5B=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5B-B=4B=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(\Rightarrow4B=5^{101}-5\)
\(\Rightarrow B=\frac{5^{101}-5}{4}\)
_Chúc bạn học tốt_