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ta có : 1/1.6+1/6.11+1/11.16+....+1/96.101
= 1/5.5/1.6+ 1/5.5/6.11+1/5.5/11.16+...+1/5.5/96.101
=1/5 . ( 5/1.6+5/6.11+5/11.16+...+5/96.101)
=1/5 . ( 1/1-1/6 +1/6-1/11+1/11-1/16+....+1/96-1/101)
=1/5 . (1/1-1/101)
=1/5 . 100/101
= 20/101
5A=\( 1-{1\over 6}+{1\over 6}-{1\over 11}+...{1\over 96}-{1\over 101}\)
=\(1- {1 \over 101}={100 \over 101}\)
suy ra A =\({20 \over 101}\)
\(C=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(C=\frac{1}{5}\left(1-\frac{1}{101}\right)\)
\(C=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
\(5C=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\)
\(5C=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)
\(5C=1-\frac{1}{101}\)
\(C=\frac{100}{\frac{101}{5}}\)
a)
=\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5\left(7^3-7^4\right)}{7^3.3^2}\)
=\(\frac{3^4\left(3-1\right)}{^{ }3^4\left(9+3\right)}-\frac{5.7^3-5.7^4}{7^3.3^2}\)
=\(\frac{1}{6}-\frac{7^3.5\left(1-7\right)}{7^3.3^2}=\frac{1}{6}-\frac{30}{9}=-\frac{19}{6}\)
Vậy A=\(-\frac{19}{6}\)
câu b lúc nã mk làm sai rui
dây mới đúng
=\(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
=\(\frac{1}{5}\left(1-\frac{1}{101}\right)=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
Do \(\left|a\right|\ge0\) nên:
a) \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\ge0\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\) (100 số hạng x)
\(\Leftrightarrow100x+5050=101x\Leftrightarrow201x=5050\Leftrightarrow x=\frac{5050}{201}\)
b) Đề sai nhé!
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
Ta có: \(\left|x+\frac{1}{101}\right|\ge0\); \(\left|x+\frac{2}{101}\right|\) \(\ge0\); ...; \(\left|x+\frac{100}{101}\right|\ge0\)
\(\Rightarrow101x\ge0\)
và \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|\ge0\)
\(\Rightarrow\left|x+\frac{1}{101}\right|=x+\frac{1}{101}\); \(\left|x+\frac{2}{101}\right|=x+\frac{2}{101}\); ...; \(\left|x+\frac{100}{101}\right|=x+\frac{100}{101}\)
Thay vào đề bài ta đc:
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)
\(\Rightarrow\) \(100x\) + \(\left(\frac{1+2+...+101}{101}\right)=101x\)
\(\Rightarrow100x+101=101x\)
\(\Rightarrow x=101\)
Vậy \(x=101.\)
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x (1)
điều kiện:101x\(\ge\) 0 \(\Rightarrow\) x\(\ge\) 0
từ (1) \(\Rightarrow\) \(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}\)=101x
\(\Rightarrow\) 100x+(\(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\))=101x
\(\Rightarrow\) 100x+\(\frac{5050}{101}\)=101x
\(\Rightarrow\) \(\frac{5050}{101}\)=101x-100x
\(\Rightarrow\) x=50
k bt mk lm sai hay lm đúng nữa
nếu mk lm sai thì thôi nha!
bạn ơi hình như đề sai ở chỗ cuối cùng kia kìa chỗ đó có phải : x . x ( 1 + 5 )
Đúng ko bạn ?????
Sai đề