A=5/9/10:3/2-(2/1/3.4/1/2-2.2/1/3):7/4
Những câu hỏi liên quan
A=5/9/10:3/2-(2/1/3.4/1/2-2.2/1/3):7/4
Tính giá trị các biểu thức sau.
a) A = 7^ 0 + 7^ 1 + 7^ 2
b) B =(7^5 + 7^9 ).(5^4 + 5^6 ).(2^3.4-2.2^4 )
c) C =3 ^2 .[(5^2 – 3) : 11] – 2^ 4 + 2.5^3
d) D = 9 ^2 − {5^ 2 − [5^ 2 − 2(4.5 − 3^ 2 )]}
Tính:
a) 2^5+5.13-3.2^3
b) 5^13:5^10-5^2.2^2
c) 4^5:4^3-3^9:3^7+5^0
LÀM 1 CÂU THÔI CŨNG ĐC Ạ!
a) \(2^5+5.13-3.2^3\)
\(=32+5.13-3.8\)
\(=32+65-24\)
\(=97-24\)
\(=73\)
b) \(5^{13}:5^{10}-5^2.2^2\)
\(=5^3-25.4\)
\(=125-100\)
\(=25\)
c) \(4^5:4^3-3^9:3^7+5^0\)
\(=4^2-3^2+1\)
\(=16-9+1\)
\(=7+1\)
\(=8\)
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So Sanh
a, A=1/2.2/3+3/10 B=7/8-1/4.3/2
b,C=2/3.4/5+8/15-1/5 D=1/2.5/6+2/3.3/4
1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
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Có phải ở sách NCPT ko bn
Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{203}{3^{100}}< 3\)
\(B< \frac{3}{4}\left(đpcm\right)\)
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Bài 1: Tính giá trị biể thức
a,(3.4^2.2^7)^2 : (9^2.10-9^2)
b,24^3:3^4-32^12:16^12
c,(2^4.5^2.11^2.7) : (2^3.5^3.7^2.11)
a) S1 = 1+2^2+2^3+......+2^20
b) S2 =1+3^2+3^3+.......+3^15
c)S3 =6+5^2+5^3+........+5^10
d)S4= 1.2+2.3+3.4+......+99.100
e)S5 = 2.2+2.5+3.8+.....+50.149
a) 156-(x+61)=82 b) (x-35)-120=0 c) 124-(x-118)=217 d) 2x-49=5.3^2 e) 200-(2x+6)=4^3 f) 2(x-51)=2.2^3+20 g) 4(x-3)=7^2-1^10 h) 3^2(x+4)-5^2=5.2^2 i) 9^x-1=9
1 tính
a. \((\left|-5\right|7^4+3^2.2\dfrac{2}{5}):(7^5.125.7^3.50)\)
b. \((\dfrac{3^2}{9}.\dfrac{3^3}{81})^{12}:(\dfrac{3^6}{81^2})^{10}\)
c. \([(\dfrac{1}{2})^2.(\dfrac{1}{3})^4.\dfrac{2}{7}]:\left(\dfrac{1}{3}\right)^{-2}.\dfrac{-2^2}{7}\)
b. \(\left(\dfrac{3^2}{9}.\dfrac{3^3}{81}\right)^{12}:\left(\dfrac{3^6}{81^2}\right)^{10}\)
\(=\left(1.\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left(\dfrac{1}{3}\right)^{12}:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left[\left(\dfrac{1}{3}\right)^2\right]^6:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left(\dfrac{1}{9}\right)^6:\left(\dfrac{1}{9}\right)^{10}\)
\(=\left(\dfrac{1}{9}\right)^{-4}=6561\)
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