\(\left(\frac13+\frac16+\frac{1}{12}+\cdots+\frac{1}{192}\right)\)x 3
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Câu 1:
c: \(\frac19+\frac28+\frac37+\cdots+\frac91\)
\(=\left(\frac19+1\right)+\left(\frac28+1\right)+\cdots+\left(\frac82+1\right)+1\)
\(=\frac{10}{2}+\frac{10}{3}+\cdots+\frac{10}{10}=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)
Ta có: \(\left(\frac12+\frac13+\frac14+\cdots+\frac{1}{10}\right)\cdot x=\frac19+\frac28+\frac37+\cdots+\frac91\)
=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)=10\left(\frac12+\frac13+\cdots+\frac{1}{10}\right)\)
=>x=10
Câu 2:
d: \(\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023\cdot2024}\)
\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{2021\cdot2022\cdot2023}-\frac{1}{2022\cdot2023\cdot2024}\right)\)
\(=\frac13\left(\frac{1}{1\cdot2\cdot3}-\frac{1}{2022\cdot2023\cdot2024}\right)\)

a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)

* Chứng minh \(\frac16
Ta có: \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)
\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)
\(\Rightarrow F<\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\cdots+\frac{1}{99\cdot100}\)
\(\) \(\Rightarrow F<\frac14-\frac15+\frac15-\frac16+\frac16-\frac17+\cdots+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow F<\frac14-\frac{1}{100}\)
\(\Rightarrow F<\frac{12}{25}\)
Mà \(\frac16=\frac{12}{72}<\frac{12}{25}\)
\(\Rightarrow\frac16 (1)
* Chứng minh \(F<\frac14\)
Lại có: \(\) \(F=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\cdots+\frac{1}{100^2}\)
\(F=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+\cdots+\frac{1}{100\cdot100}\)
\(\Rightarrow F>\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{100\cdot101}\)
\(\Rightarrow F>\frac15-\frac16+\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow F=\frac15-\frac{1}{101}\)
\(\Rightarrow F>\frac{96}{505}\)
Mà \(\frac14=\frac{96}{384}<\frac{96}{505}\)
\(\Rightarrow F<\frac14\) (2)
Từ (1) và (2) suy ra: \(\frac16
Vậy \(\frac16

a: \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{99\cdot101}\)
\(=1-\frac13+\frac13-\frac15+\cdots+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b: \(B=\frac12-\left(\frac{1}{5\cdot11}+\frac{1}{11\cdot17}+\frac{1}{17\cdot23}+\frac{1}{23\cdot29}+\frac{1}{29\cdot35}\right)\)
\(=\frac12-\frac16\left(\frac{6}{5\cdot11}+\frac{6}{11\cdot17}+\frac{6}{17\cdot23}+\frac{6}{23\cdot29}+\frac{6}{29\cdot35}\right)\)
\(=\frac12-\frac16\left(\frac15-\frac{1}{11}+\frac{1}{11}-\frac{1}{17}+\cdots+\frac{1}{29}-\frac{1}{35}\right)\)
\(=\frac12-\frac16\left(\frac15-\frac{1}{35}\right)=\frac12-\frac16\cdot\frac{6}{35}=\frac12-\frac{1}{35}=\frac{33}{70}\)



Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0

ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
Kết quả là \(\frac{127}{64}\) nha bạn!
lời giải rõ ràng hơn đc hem