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\(S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
Bài này khá ez thôi:
a) bạn sửa lại đề rồi làm theo cách làm của b,c,d nhé
b) Ta có: \(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\left(\forall x\right)\)
\(\Rightarrow5x\ge0\Rightarrow x\ge0\) khi đó:
\(PT\Leftrightarrow x+1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow x=5\)
c,d tương tự nhé
c,\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}+\right|+...+\left|x+\frac{1}{97.99}\right|\ge0\forall x\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)Khi đó:
\(x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)
\(\Rightarrow49x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{49}{99}\)
\(\text{Đ}\text{ặt}:A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(A=\frac{100}{101}:2=\frac{50}{101}\)
\(\Rightarrow\frac{1}{3}x.x=\frac{50}{101}\)
\(x.\left(\frac{1}{3}.1\right)=\frac{50}{101}\)
\(x.\frac{1}{3}=\frac{50}{101}\)
$x=\frac{50}{101}:\frac{1}{3}=\frac{150}{101}$
\(.\frac{1}{3}x.x=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(\frac{100}{101}\right)\)
\(\frac{1}{3}xx=\frac{50}{101}\)
\(x.x=\frac{150}{101}\)
còn lại tự tính
Ta có\(\frac{1}{1\cdot3}\) +\(\frac{1}{3\cdot5}\)+\(\frac{1}{5\cdot7}\)+.....+\(\frac{1}{x\cdot\left(x+2\right)}\)=\(\frac{16}{34}\)
=> 2(\(\frac{1}{1\cdot3}\)+\(\frac{1}{3\cdot5}\)+\(\frac{1}{5\cdot7}\)+......+\(\frac{1}{x+\left(x+2\right)}\)) = \(\frac{16}{34}\)*2
=> \(\frac{2}{1\cdot3}\)+\(\frac{2}{3\cdot5}\)+\(\frac{2}{5\cdot7}\)+.....+\(\frac{2}{x\cdot\left(x+2\right)}\)= \(\frac{32}{34}\)
1-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{x}\)-\(\frac{1}{x+2}\)=\(\frac{32}{34}\)
1-\(\frac{1}{x+2}\)=\(\frac{32}{34}\)
\(\frac{1}{x+2}\)= 1-\(\frac{32}{34}\)
\(\frac{1}{x+2}\)= \(\frac{1}{17}\)
=> x+2=17
x=17-2
x=15
a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)
Vậy \(x=1999\)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)
\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)
Vậy \(x=45\)
\(\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+...+\left|x+\dfrac{1}{99.101}\right|=100x\)
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{1.3}\right|\ge0\\\left|x+\dfrac{1}{3.5}\right|\ge0\\\left|x+\dfrac{1}{99.101}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+\left|x+\dfrac{1}{5.7}\right|+...+ \left|x+\dfrac{1}{99.101}\right|\ge0\)\(\Rightarrow100x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+x+\dfrac{1}{5.7}+...+x+\dfrac{1}{99.101}=100x\)\(\Rightarrow50x+\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}=100x\)
\(\Rightarrow50x+\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=100x\)
\(\Rightarrow50x+\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=100x\)
\(\Rightarrow50x+\dfrac{50}{101}=500x\)
\(\Rightarrow50x=\dfrac{50}{101}\)
\(\Rightarrow x=\dfrac{1}{101}\)
\(a,\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x^2=8^2\)
\(\Leftrightarrow x=\pm8\)
\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Leftrightarrow x=15\)
Vậy x = 15
Bài cuối tương tự
\(a.\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right).\left(2x+1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49.\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
\(b.\)
\(1-3+3^2-3^3+...+\left(-3^x\right)=\frac{1-9^{1006}}{4}\)
Đặt \(A=1-3+3^2-3^3+...+\left(-3^x\right)\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...+\left(-3^{x+1}\right)\)
\(\Rightarrow3A+A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow4A=1+\left(-3^{x+1}\right)\)
\(\Rightarrow A=\frac{1+\left(-3^{x+1}\right)}{4}\)
\(\Rightarrow\frac{1+\left(-3^{x+1}\right)}{4}=\frac{1-9^{1006}}{4}\)
\(\Rightarrow-3^{x+1}=-9^{1006}\)
\(\Rightarrow-3^{x+1}=-3^{2012}\)
\(\Rightarrow x+1=2012\)
\(\Rightarrow x=2012-1\)
\(\Rightarrow x=2011\)
Vậy : \(x=2011\)
a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)
A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)
A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)
A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)
A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)
A = \(34+\frac{34}{103}=\frac{3536}{103}\)
X+(1/1.3+1/3.5+1/5.7+...+1/99.101)=100
X+(2/1.3+2/3.5+2/5.7+...+2/99.101)=100
X+(1 -1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)=100
X+(1-1/101)=100
X+100/101=100
X=100-100/101
X=10000/101