Chứng minh rằng
a, \(7^6+7^5-7^4\) chia hết cho 55
b, \(81^7-27^9-9^{13}\) chia hết cho 405
c, \(16^5+2^{15}\) chia hết cho 33
Chứng minh rằng
a, \(7^6+7^5-7^4\) chia hết cho 55
b, \(81^7-27^9-9^{13}\) chia hết cho 405
c, \(16^5+2^{15}\) chia hết cho 33
b) 817 - 279 -913 chia hết cho 405
Ta có: 817 - 279 -913 = 328- 327-326
= 326(32-3-1)
= 326. 5 = 322. 405 chia hết cho 405 (đpcm)
a)
\(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55\) chia hết cho 55
=>\(7^6+7^5-7^4\) chia hết cho 55
c)
\(16^5+2^{15}=2^{4.5}+2^{15}=2^{20}+2^{15}=2^{15}.\left(2^5+1\right)=2^{15}.33\)chia hết cho 33
Do đó: \(16^5+2^{15}\) chia hết cho 33
So sánh A với \(-\dfrac{1}{2}\) biết:
A=\(\left(\dfrac{1}{2^2}-1\right)\)\(\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(1-\dfrac{1}{100^2}\right)\)
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
\(=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)\) ( do có 99 cặp số )
\(=-\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=-\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{99}{100}.\dfrac{101}{100}\)
\(=-\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=-\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{-101}{200}< \dfrac{-100}{200}=\dfrac{-1}{2}\)
Vậy \(A< \dfrac{-1}{2}\)
Đề sai rồi kìa bn
Đúg ra phải là 1/100^2 -1 chứ
Chứng minh rằng:
M = \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{100}{3^{100}}\)<\(\dfrac{3}{4}\)
\(M=\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\)
\(\Rightarrow3M=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
\(\Rightarrow3M-M=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+...+\dfrac{100}{3^{100}}\right)\)
\(\Rightarrow2M=1+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)
\(\Rightarrow2M=1+\dfrac{1}{2}-\dfrac{1}{3^{99}.2}-\dfrac{100}{3^{100}}\)
\(\Rightarrow M=\dfrac{3}{4}-\dfrac{1}{3^{99}.4}-\dfrac{50}{3^{100}}< \dfrac{3}{4}\)
Vậy...
\(\dfrac{45^{10}.5^{20}}{75^{15}}\)Giải rõ ra hộ mình ~~~~
45^10*5^20/75^15
=(3^2*5)^10*5^20/(3*5^2)^15
=3^20*5^10*5^20/3^15*5^30
=3^20*5^30/3^15*5^30
=3^5=243
(1/5).5^7
x2-1/10=23/90
\(x^2-\dfrac{1}{10}=\dfrac{23}{90}\)
\(x^2-\dfrac{9}{90}=\dfrac{23}{90}\)
\(x^2=\dfrac{16}{45}\)
\(\Leftrightarrow x=\sqrt{\dfrac{16}{45}}\)
\(\Leftrightarrow x=-\sqrt{\dfrac{16}{45}}\)
tính:
\(\dfrac{\left(-6\right)^6}{216}\) \(\dfrac{64}{\left(-4\right)^5}\) \(\dfrac{900}{\left(-30\right)^3}\) \(\dfrac{225}{\left(-15\right)^3}\)
giúp giùm nha cần gấp lắm
\(\text{a) }\dfrac{\left(-6\right)^6}{216}=\dfrac{6^6}{216}=\dfrac{6^6}{6^3}=6^3=216\)
\(\text{b) }\dfrac{64}{\left(-4\right)^5}=-\dfrac{64}{4^5}=-\dfrac{4^3}{4^5}=-\dfrac{1}{4^2}=-\dfrac{1}{16}\)
\(\text{c) }\dfrac{900}{\left(-30\right)^3}=-\dfrac{900}{30^3}=-\dfrac{30^2}{30^3}=-\dfrac{1}{30}\)
\(\text{d) }\dfrac{225}{15^3}=\dfrac{15^2}{15^3}=\dfrac{1}{15}\)
Đề nghị bạn trình bày câu hỏi rõ ràng hơn nữa
\(5^x+5^{x+1}+5^{x-2}=151\)
\(5^{x-1}+5^{x-2}+5^{x-3}=155\)
\(5^{2+x}+5^{3+x}=750\)
a, \(5^x+5^{x+1}+5^{x-2}=151\)
\(\Rightarrow5^x.\left(1+5+5^{-2}\right)=151\)
\(\Rightarrow5^x.6,04=151\Rightarrow5^x=25=5^2\)
Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=2\)
b, \(5^{x-1}+5^{x-2}+5^{x-3}=155\)
\(\Rightarrow5^x.\left(5^{-1}+5^{-2}+5^{-3}\right)=155\)
\(\Rightarrow5^x.0,248=155\Rightarrow5^x=625=5^4\)
Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=4\)
c, \(5^{2+x}+5^{3+x}=750\) \(\Rightarrow5^x.\left(5^2+5^3\right)=750\) \(\Rightarrow5^x.150=750\Rightarrow5^x=5=5^1\) Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=1\) Chúc bạn học tốt!!!\(•5^x+5^{x+1}+5^{x-2}=151\\ 5^x\left(1+5+\dfrac{1}{25}\right)=151\\ 5^x=25\\ \Rightarrow x=2\)
\(•5^{x-1}+5^{x-2}+5^{x-3}=155\\ 5^x.\left(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}\right)=155\\ 5^x=625\\ \Rightarrow x=4\)
\(•5^{2+x}+5^{3+x}=750\\ 5^x\left(25+125\right)=750\\ 5^x=5\\ \Rightarrow x=1\)
Đề bài: Tính nhanh:
\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\dfrac{2.6}{3.5}=\dfrac{12}{15}=\dfrac{4}{5}\)
Chúc bạn học tốt!!!
cho mình xin đề toán bài 50,51,52 sách bài tập trang 17 lớp 7 tập 1
và bài 57,58,59 sách bài tập trang 18 lớp 7 tập 1
Bạn gõ tên mấy bài tập đó ra rồi tìm trên hoc24 cx có đó bạn