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Vào lúc: 2019-11-18 20:18:05 Xem câu hỏi

Ta có

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)   và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\)  nên

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)

\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)

\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)

Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)

Vào lúc: 2019-11-18 19:53:57 Xem câu hỏi

Ta có

\(A=\frac{1}{14}+\frac{1}{29}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{1877}\)

\(=\frac{1}{1^2+2^2+3^2}+\frac{1}{2^2+3^2+4^2}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{24^2+25^2+26^2}\)

\(B=n^2+\left(n+1\right)^2+\left(n+2\right)^2=3n^2+6n+5\left(1\right)\)

+ Với \(n\ge1\)từ (1) ta có \(B\le3n^2+9n+6=3\left(n^2+3n+2\right)=3\left(n+1\right)\left(n+2\right)\)Từ đó

\(A>\frac{1}{3}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{3}C\)

Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)

\(\Rightarrow A>\frac{1}{3}\cdot\frac{6}{13}=\frac{2}{13}>0,15\)

+ Với \(n\ge1\)từ (1) ta có \(B>2n^2+6n+4=2\left(n^2+3n+2\right)=2\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A< \frac{1}{2}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{2}C\)

Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)

\(\Rightarrow A< \frac{1}{2}\cdot\frac{6}{13}=\frac{3}{13}< 0,25\)

Vậy \(0,15< A< 0,25\)

Vào lúc: 2019-11-17 19:47:02 Xem câu hỏi

MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên

\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)

\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{2008}{abc}\) ( với \(abc\ne0\))

Vào lúc: 2019-11-14 19:46:28 Xem câu hỏi

Ta có

\(B=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(b-a\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}-\frac{\left(x-c\right)\left(x-a\right)}{\left(a-c\right)\left(a-b\right)}+\frac{\left(x-a\right)\left(x-b\right)}{\left(a-c\right)\left(c-b\right)}\)

\(=\frac{\left(x-b\right)\left(x-c\right)-\left(x-c\right)\left(x-a\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-c\right)\left(x-a\right)-\left(x-b\right)\left(x-a\right)}{\left(b-c\right)\left(c-a\right)}\)

\(=\frac{\left(x-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(x-a\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)}\).

\(=\frac{x-c}{a-c}-\frac{x-a}{a-c}=\frac{x-c-x+a}{a-c}\)

\(=1\)

Vào lúc: 2019-11-14 19:12:37 Xem câu hỏi

Ta có

\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{b-a}{\left(a-b\right)\left(a-c\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a}\left(1\right)\)

Tương tự ta có

\(\frac{c-a}{\left(b-c\right)\left(b-a\right)}=\frac{1}{b-c}+\frac{1}{a-b}\left(2\right)\)

\(\frac{a-b}{\left(c-b\right)\left(c-a\right)}=\frac{1}{b-c}+\frac{1}{c-a}\left(3\right)\)

Từ (1) (2) và (3) ta có

\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-b\right)\left(c-a\right)}\)

\(=\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)

\(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\left(đpcm\right)\)

Vào lúc: 2019-11-13 20:43:52 Xem câu hỏi

Ta có

\(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\frac{2a+2b+2c}{a+b+c}\)

\(=2\)

Từ \(\frac{2b+c-a}{a}=2\Rightarrow2a=2b+c-a\Rightarrow3a-2b=c\)và \(3a-c=2b\)

Tương tự có \(3b-2c=a;3b-a=2c\) và \(3c-2a=b;3c-b=2a\)

Thay vào biểu thức M ta có

\(M=\frac{a\cdot b\cdot c}{2\cdot b\cdot2\cdot a\cdot2\cdot c}=\frac{1}{8}\)

Vào lúc: 2019-11-13 20:31:20 Xem câu hỏi

Ap dụng hằng đẳng thức.

\(A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{b^2}{\left(a-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(c-a\right)}+\frac{c^2}{\left(c-a\right)\left(b-c\right)}\)

\(=\frac{\left(a+b\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(b+c\right)\left(b-c\right)}{\left(b-c\right)\left(c-a\right)}\)

\(=\frac{a+b}{a-c}+\frac{b+c}{c-a}=\frac{a+b}{a-c}-\frac{b+c}{a-c}=1\left(đpcm\right)\)

Vào lúc: 2019-11-12 20:02:54 Xem câu hỏi

Ta có

\(\frac{a-b}{1+ab}=\frac{b-c}{1+bc}=\frac{a-c}{1+ac}\)       nên

\(\frac{a-b}{1+ab}+\frac{b-c}{1+bc}+\frac{c-a}{1+ca}=\frac{a-b}{1+ab}+\frac{b-a}{1+bc}+\frac{a-c}{1+bc}+\frac{c-a}{1+ca}\)

\(=\left(a-b\right)\left[\frac{1}{1+ab}-\frac{1}{1+bc}\right]+\left(c-a\right)\left[\frac{1}{1+ac}-\frac{1}{1+bc}\right]\)

\(=\frac{\left(a-b\right)\left(1+bc-1+ab\right)}{\left(1+ab\right)\left(1+bc\right)}+\frac{\left(c-a\right)\left(1+bc-1-ac\right)}{\left(1+ac\right)\left(1+bc\right)}\)

\(=\frac{b\left(c-a\right)\left(a-b\right)}{\left(1+ab\right)\left(1+bc\right)}+\frac{c\left(c-a\right)\left(b-a\right)}{\left(1+ac\right)\left(1+bc\right)}\)

\(=\frac{\left(a-b\right)\left(c-a\right)}{\left(1+bc\right)}\left[\frac{b}{1+ab}-\frac{c}{1+ac}\right]\)

\(=\frac{\left(a-b\right)\left(c-a\right)\left(b-c\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\left(đpcm\right)\)

Vào lúc: 2019-11-12 18:48:33 Xem câu hỏi

\(P=\frac{\left(2003^2\cdot2013+31\cdot2004-1\right)\left(2003\cdot2008+4\right)}{2004\cdot2005\cdot2006\cdot2007\cdot2008}\)

Đặt a=2004 ta có

\(P=\frac{\left[\left(x-1\right)^2\cdot\left(a+9\right)+31\cdot a-1\right]\left[\left(a-1\right)\left(a+4\right)+4\right]}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{\left[\left(a^2-2a+1\right)\left(a+9\right)+31a-1\right]\left[\left(a^2+3a-4\right)+4\right]}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{\left(a^3+9a^2-2a^2-18a+a+9+31a-1\right)\left(a^2+3a\right)}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{\left(a^3+7a^2+14a+8\right)\left(a^2+3a\right)}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}=1\)

Vậy \(P=1\)

Vào lúc: 2019-11-12 18:35:18 Xem câu hỏi

Ta có

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)       hay \(M>1\)

\(M=\left(1-\frac{a}{b+a}\right)+\left(1-\frac{c}{b+c}\right)+\left(1-\frac{a}{a+c}\right)< 3-\left(\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\right)\)

\(=3-1=2\)   hay \(M>2\)

Vậy \(1< M< 2\). Do đó M k thể là số nguyên dương

Vào lúc: 2019-11-11 20:04:12 Xem câu hỏi

Ta có

\(2^{17}+2^{14}\)

\(=2^{14}\left(2^3+1\right)\)

\(=2^{14}\cdot9\)

Vì \(9⋮9\Rightarrow2^{14}\cdot9⋮9\)

hay \(2^{17}+2^{14}⋮9\left(đpcm\right)\)

Vào lúc: 2019-11-11 19:44:38 Xem câu hỏi

\(\frac{x-3}{5}=\frac{5-2x}{-11}\)

\(-11\cdot\left(x-3\right)=5\cdot\left(5-2x\right)\)

\(-11x+33=25-10x\)

\(-11x+33-25+10x=0\)

\(8-x=0\)

\(x=8\)

Vào lúc: 2019-11-11 19:11:54 Xem câu hỏi

\(6x^2-7x-20\)

\(=6x^2+8x-15x-20\)

\(=2x\cdot\left(3x+4\right)-5\left(3x+4\right)\)

\(=\left(3x+4\right)\left(2x-5\right)\)

Vào lúc: 2019-11-11 15:59:14 Xem câu hỏi

\(\left(2x^3+7x^2-4x\right):\left(x+4\right)\)

\(=x\cdot\left(2x^2+7x^2-4x\right):\left(x+4\right)\)

\(=x\cdot\left(2x^2-x+8x-4\right):\left(x+4\right)\)

\(=x\cdot\left(2x-1\right)\left(x+4\right):\left(x+4\right)\)

\(=x\cdot\left(2x-1\right)\)

Vào lúc: 2019-11-10 19:38:11 Xem câu hỏi

\(\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)

\(\Leftrightarrow\frac{4}{4x-8+\frac{7}{x}}+\frac{3}{4x-10+\frac{7}{x}}=1\)

Đặt \(4x+\frac{7}{x}=a\)

\(\Rightarrow\frac{4}{a-8}+\frac{3}{a-10}=1\)

\(\Leftrightarrow a^2-23a+144=0\)

\(\Rightarrow\left(a-16\right)\left(a-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=16\\a=9\end{cases}}\)

đến đây tự giải nha

Vào lúc: 2019-11-10 19:19:12 Xem câu hỏi

ĐKXD \(x^2-6x+6\ge0\)

\(x^2-6x+9=4\sqrt{x^2-6x+6}\)

\(\Leftrightarrow\left(x^2-6x+6\right)-4\sqrt{x^2-6x+6}+3=0\)

Đặt \(a=\sqrt{x^2-6x+6}\left(a>0\right)\)

\(\Rightarrow a^2-4a+3=0\Leftrightarrow\left(a-3\right)\left(a-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{cases}}\)

\(+\sqrt{x^2-6x+6}=3\)

\(\Rightarrow x^2-6x+6=9\)

\(\Rightarrow\orbr{\begin{cases}x=3+2\sqrt{3}\\x=3-2\sqrt{3}\end{cases}}\)

\(+\sqrt{x^2-6x+6}=1\)

\(\Rightarrow x^2-6x+6=1\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vào lúc: 2019-11-09 20:34:01 Xem câu hỏi

\(5x=2y\Rightarrow\frac{x}{2}=\frac{y}{5}=\frac{2y}{10}=\frac{7x}{14}\)

áp dụng tính chất DTSBN ta có

\(\frac{2y}{10}=\frac{7x}{14}=\frac{2y-7x}{10-14}=-\frac{20}{-4}=5\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=5\\\frac{y}{5}=5\end{cases}\Rightarrow\hept{\begin{cases}x=10\\y=25\end{cases}}}\)

Vào lúc: 2019-11-09 16:05:30 Xem câu hỏi

\(A=\frac{10^4\cdot81-16\cdot15^2}{4^4\cdot675}\)

\(=\frac{2^4\cdot5^4\cdot3^3-2^4\cdot3^2\cdot5^2}{2^8\cdot3^3\cdot5^2}\)

\(=\frac{2^4\cdot5^2\cdot3\left(5^2\cdot3-3\right)}{2^8\cdot3^3\cdot5^2}\)

\(=\frac{5^2\cdot3-3}{2^4\cdot3^2}\)

\(=\frac{3\cdot\left(5^2-1\right)}{2^4\cdot3^2}=\frac{24}{48}=\frac{1}{2}\)

Vào lúc: 2019-11-09 15:36:42 Xem câu hỏi

A B C D 1 2 1 2

Ta có

\(\widehat{A}+\widehat{B}+\widehat{C}=180^O\)

\(\Rightarrow\widehat{C}=180^O-\widehat{A}-\widehat{B}\)

\(\Rightarrow\widehat{C}=180-75-65=40\)

Vì AD là tia phân giác của góc A

\(\Rightarrow\hept{\begin{cases}A_1=37,5\\A_2=37,5\end{cases}}\)

Ta có

\(A_1+C+D_1=180\)

\(\Rightarrow D_1=180-C-A_1\)

\(\Rightarrow D_1=180-40-37,5=102,5^O\)

Tương tự \(\Delta ABD\)ta có

\(D_1=77,5^o\)

Vậy......

Vào lúc: 2019-11-08 19:29:48 Xem câu hỏi

Ta có

\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}=\frac{a^2+ab-bc-ab}{\left(a+b\right)\left(a+c\right)}=\frac{a\cdot\left(a+b\right)-b\cdot\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}=\frac{a}{a+c}-\frac{b}{a+b}\left(1\right)\)

tương tự

\(\frac{b^2-bc}{\left(a+b\right)\left(b+c\right)}=\frac{b}{a+b}-\frac{c}{b+c}\left(2\right)\)

\(\frac{c^2-ab}{\left(c+a\right)\left(b+c\right)}=\frac{c}{c+b}-\frac{a}{a+b}\left(3\right)\)

Cộng (1);(2) và (3) ta có

\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}=\frac{a}{a+c}-\frac{b}{a+b}+\frac{b}{a+b}-\frac{c}{b+c}+\frac{c}{c+b}-\frac{a}{a+b}=0 \)

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