(1-1/2)x(1-1/3)x(1-1/4).....(1-1/1003).(1-1/1004)
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\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{1002}{1003}.\frac{1003}{1004}\)
\(=\frac{1}{1004}\)
Tk giúp !!
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+) \(5\frac{2}{3}x+1\frac{2}{3}=4\frac{1}{2}\Leftrightarrow\frac{17}{3}x+\frac{5}{3}=\frac{9}{2}\Leftrightarrow\frac{17}{3}x=\frac{17}{6}\Leftrightarrow x=\frac{1}{2}\)
+) \(\frac{x}{27}=\frac{-2}{9}\Leftrightarrow x=\frac{-2}{9}.27=-6\)
+) \(\left|x+1,5\right|=2\Leftrightarrow\orbr{\begin{cases}x+1,5=2\\x+1,5=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0,5\\x=-3,5\end{cases}}}\)
+) \(A=\left|x-1004\right|-\left|x+1003\right|\)
Ta có BĐT \(\left|x\right|-\left|y\right|\le\left|x-y\right|,\)dấu "=" xảy ra khi và chỉ khi x,y cùng dấu hay \(xy\ge0\)
Áp dụng: \(A=\left|x-1004\right|-\left|x+1003\right|\le\left|x-1004-x-1003\right|=\left|-2007\right|=2007\)
Vậy \(maxA=2007\Leftrightarrow\left(x-1004\right)\left(x+1003\right)\ge0\Leftrightarrow\orbr{\begin{cases}x\ge1004\\x\le-1003\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
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có tất cả số số hạng là
(1005-1)/1+1=1005 (số)
tổng của tất cả số hạng đó là
(1+1005)*1005/2=505515
đáp số 505515
![](https://rs.olm.vn/images/avt/0.png?1311)
Kết quả là 1/1004