\(-1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{101}\left(1+2=3+..+101\right)\) Tính
\(F=1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{101}\left(1+2+3+...+101\right)\)
f=1-(2.3/4+3.4/6+...+101.102/202)
A=2.3/4+3.4/6+...+101.102/202)
2a=2.3/2+3.4/3+...+101.102/101
2a=3+4+...+102
2a=100.105/2
a=100.105/4
a=2625
f=1-2625
f=-(2624)
Tính \(H=\left[101-\frac{1}{9}-\frac{2}{10}-\frac{3}{11}-...-\frac{101}{109}\right]:\left[\frac{1}{45}+\frac{1}{50}+\frac{1}{55}+...+\frac{1}{545}\right]\)
$\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right)\div \left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{101}\right)-2
5. Tìm x biết:
a, \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+...+\left|x+10\right|=11x+1\)
b, \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+....+\left|x+\frac{100}{101}\right|\)=101x
Tĩm X?
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!
\(!x+\frac{1}{101}!+!x+\frac{2}{101}!+...+!x+\frac{100}{101}!=101x\) (1)
VT tổng các số không âm => VT>=0 vậy \(VP\ge0\Rightarrow x\ge0\)
với x>=0 biểu thức trong GT tuyệt đối >0 => bỏ dấu trị tuyệt đối biểu thức không đối
do vậy ta có (1) \(\Leftrightarrow\left(x+\frac{1}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow100.x+\left(\frac{1}{101}+...+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow x=\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}=\frac{1}{101}\left(1+2+...+100\right)=\frac{1}{101}\left(\frac{100.101}{2}\right)=50\)
đáp số: x=50
B1:Tính hợp lí
a) \(1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}\left(1+2+3\right)-...-\frac{1}{101}\left(1+2+...+101\right)\)
B2
Chứng minh \(1.3.5....99=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}\)
Giải nhanh nhé .Mình đag cần gấp .Cảm ơn!
Tớ có cái này đố các cậu
a)\(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
b)\(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
c)\(\frac{-3^3}{25}.\frac{75}{-21}.\frac{50}{35}\)
d)\(\frac{25.48-25.18}{20.5^3}\)
e)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2003}\right)\)
f)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{200}\right)\)
Chúc các cậu may mắn!!
a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)
\(=\frac{53}{101}.\frac{-97}{97}\)
\(=\frac{53}{101}.\left(-1\right)\)
\(=\frac{-53}{101}\)
b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)
\(=0\)
c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)
\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)
\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)
\(=\frac{3.3.5.2}{\left(-7\right).7}\)
\(=\frac{90}{-49}\)
d) \(\frac{25.48-25.18}{20.5^3}\)
\(=\frac{25\left(48-18\right)}{10.2.125}\)
\(=\frac{25.10.3}{10.2.25.5}\)
\(=\frac{3}{10}\)
\(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)
\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)
\(=\frac{53}{101}.\left(-1\right)=\frac{-53}{101}\)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-2\)
Tính giá trị biểu thức sau
\(\left(100+\frac{99}{2}+\frac{98}{3}+.....+\frac{1}{100}\right)\div\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{101}\right)-2\)
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~