CMR: \(\dfrac{1}{a}.a+10=\dfrac{1}{a}-\dfrac{1}{a}+1\)
giúp mk với a+10 ở trong ngoặc nha, mk k bấm đk do liệt số 9 và 10 rồi
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Các bạn giúp với :Bài 1:a, CMR: A dfrac{3}{1^2.2^2}+dfrac{5}{2^2.3^2}+dfrac{7}{3^2.4^2}+...+dfrac{21}{10^2.11^2} 1b, Cho B dfrac{3}{4}+dfrac{8}{9}+dfrac{15}{16}+dfrac{24}{25}+...+dfrac{2499}{2500}. CMR: B không phải là số nguyên.c, So sánh: C dfrac{2}{2^1}+dfrac{3}{2^2}+dfrac{4}{2^3}+...+dfrac{2021}{2^{2020}} với 3.
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Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
10) tính giá trị biểu thức
a) \(\dfrac{1}{2}\) x \(\dfrac{3}{4}\) + \(\dfrac{1}{2}\)
B) \(\dfrac{3}{4}\) : \(\dfrac{2}{3}\) - \(\dfrac{1}{6}\)
( ghi chi tiết giúp mk với )
a: =1/2(3/4+1)=1/2x7/4=7/8
b: =9/8-1/6=27/24-4/24=23/24
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a.\(\dfrac{1}{2}\times\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{1}{2}\times\left(\dfrac{3}{4}+1\right)=\dfrac{1}{2}\times\dfrac{7}{4}=\dfrac{7}{8}\)
b.\(\dfrac{3}{4}:\dfrac{2}{3}-\dfrac{1}{6}=\dfrac{3}{4}\times\dfrac{3}{2}-\dfrac{1}{6}=\dfrac{9}{8}-\dfrac{1}{6}=\dfrac{23}{24}\)
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Xem thêm câu trả lời
Cho A = \(\dfrac{n^9+1}{n^{10}+1}\) và B = \(\dfrac{n^8+1}{n^9+1}\) trong đó n\(\in\)N; n>1. Hãy so sánh nghịch đảo của A và B rồi so sánh A với B
A = \(\dfrac{n^9+1}{n^{10}+1}\)
\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n - \(\dfrac{n-1}{n^9+1}\)
B = \(\dfrac{n^8+1}{n^9+1}\)
\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) = n - \(\dfrac{n-1}{n^8+1}\)
Vì n > 1 ⇒ n - 1> 0
\(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)
⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)
⇒ A < B
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Bài 2: So sánh các phân số sau mà không quy đồng mẫu số và tử số:
a)\(\dfrac{2025}{2022}\)và\(\dfrac{2024}{2021}\)
b)\(\dfrac{8^9+1}{8^{10}+1}\)và \(\dfrac{8^8+1}{8^9+1}\)
giúp mk với nhé
a)\(\dfrac{3}{5x8}+\dfrac{3}{8x11}+...+\dfrac{3}{2006x2009}\)
b)\(\dfrac{1}{6x10}+\dfrac{1}{10x14}+...+\dfrac{1}{402x406}\)
c)\(\dfrac{10}{7x12}+\dfrac{10}{12x17}+...+\dfrac{10}{502x507}\)
Chú ý: x ở đề bài là dấu nhân nha các bạn.Mong các bạn giúp mk nhanh 1 chút vì mk đag cần gấp
a)
\(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2006.2009}\)
\(=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+....+\frac{2009-2006}{2006.2009}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)
\(=\frac{1}{5}-\frac{1}{2009}=\frac{2004}{10045}\)
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b)
\(B=\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{402.406}\)
\(\Rightarrow 4B=\frac{4}{6.10}+\frac{4}{10.14}+...+\frac{4}{402.406}\)
\(4B=\frac{10-6}{6.10}+\frac{14-10}{10.14}+...+\frac{406-402}{402.406}\)
\(4B=\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{402}-\frac{1}{406}\)
\(4B=\frac{1}{6}-\frac{1}{406}=\frac{100}{609}\Rightarrow B=\frac{25}{609}\)
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c)
\(C=\frac{10}{7,12}+\frac{10}{12.17}+...+\frac{10}{502.507}\)
\(\Rightarrow \frac{C}{2}=\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{502.507}\)
\(\frac{C}{2}=\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{507-502}{502.507}\)
\(\frac{C}{2}=\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{502}-\frac{1}{507}\)
\(\frac{C}{2}=\frac{1}{7}-\frac{1}{507}=\frac{500}{3549}\)
\(\Rightarrow C=\frac{1000}{3549}\)
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So sánh:
A) \(\dfrac{n+1}{n+2}\) và \(\dfrac{n}{n+3}\)
B) A= \(\dfrac{10^{11}-1}{10^{12}-1}\) và B= \(\dfrac{10^{10}+1}{10^{11}+1}\)
Mọi người giúp mình với mình đang cần gấp!
Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
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So sánh A và B:
A= \(\dfrac{10^{2020}+1}{10^{2021}+1}\) B=\(\dfrac{10^{2021}+1}{10^{2022}+1}\)
Giúp mình với!
Ta có:
\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )
Suy ra: \(A>B\)
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7 tháng 5 2021 lúc 13:27
So sánh:
a/ \(A=\dfrac{17^{18}+1}{17^{19}+1};B=\dfrac{17^{17}+1}{17^{18}+1}\)
b/ \(A=\dfrac{10^8-2}{10^8+2};B=\dfrac{10^8}{10^8+4}\)
c/ \(A=\dfrac{20^{10}+1}{20^{10}-1};B=\dfrac{20^{10}-1}{20^{10}-3}\)
GIÚP MÌNH VỚI
Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
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Tìm x
a)( x-dfrac{1}{3}) (x+dfrac{2}{5}) 0
b)(x+dfrac{3}{5})(x+1) 0
c)dfrac{3}{7}x-dfrac{2}{5}xdfrac{-17}{35}
d)(dfrac{3}{4}x-dfrac{9}{10})(dfrac{1}{3}+dfrac{-3}{5}x) 0
Làm hộ mk nha câu nào cx đk làm đk mấy câu cx đk
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Tìm x
a)( \(x-\dfrac{1}{3}\)) (\(x+\dfrac{2}{5}\)) >0
b)(\(x+\dfrac{3}{5}\))(\(x+1\)) <0
c)\(\dfrac{3}{7}x-\dfrac{2}{5}x=\dfrac{-17}{35}\)
d)(\(\dfrac{3}{4}x-\dfrac{9}{10}\))(\(\dfrac{1}{3}+\dfrac{-3}{5}x\)) =0
Làm hộ mk nha câu nào cx đk làm đk mấy câu cx đk
a, \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)
+,Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow x>\dfrac{1}{3}\)
+, Xét \(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow x< -\dfrac{2}{5}\)
Vậy...........
b, \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
Vì \(x+\dfrac{3}{5}< x+1\) với mọi \(x\in R\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\)
Vậy...........
c, \(\dfrac{3}{7}x-\dfrac{2}{5}x=\dfrac{-17}{35}\)
\(\Rightarrow\dfrac{1}{35}x=\dfrac{-17}{35}\)
\(\Rightarrow x=-17\)
d, \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}+\dfrac{-3}{5}x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\-\dfrac{3}{5}x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!
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a/ \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{5}\right)>0\)
TH1:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{5}>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x>\dfrac{1}{3}\)
TH2:\(\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{5}< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< -\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow x< -\dfrac{2}{5}\)
Vậy \(x>\dfrac{1}{3}\) hoặc \(x< -\dfrac{2}{5}\) thì tm
b/ \(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
TH1:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< -\dfrac{3}{5}\\x>-1\end{matrix}\right.\) \(\Rightarrow-1< x< -\dfrac{3}{5}\)
TH2:\(\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{3}{5}\\x< -1\end{matrix}\right.\)(vô lý)
Vậy....................
c/ \(\dfrac{3}{7}x-\dfrac{2}{5}x=-\dfrac{17}{35}\)
\(\Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{5}\right)x=-\dfrac{17}{35}\)
\(\Rightarrow\dfrac{1}{35}x=-\dfrac{17}{35}\)
\(\Rightarrow x=-\dfrac{17}{35}:\dfrac{1}{35}=-17\)
Vậy.............
d/ \(\left(\dfrac{3}{4}x-\dfrac{9}{10}\right)\left(\dfrac{1}{3}+\dfrac{-3}{5}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{9}{10}=0\\\dfrac{1}{3}-\dfrac{3}{5}x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{10}\\\dfrac{3}{5}x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=\dfrac{5}{9}\end{matrix}\right.\)
Vậy.....................
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