Bài 1: Tìm x:
A, \(\frac{1}{1.2}+...+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}< x=99>\)
B, \(\frac{1}{1.3}+\frac{1}{1.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{24}{49}< x=47>\)
P/S: những cái ngoặc kép bên cạnh là kết quả của x nhé!
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14 tháng 4 2019
A, 1/(1.2)+...+1/[x(x+1)]=99
=>1-1/2+...+1/x+1/(x-1)=99
=>1-1/(x-1)=99
=>1/(x-1)=-98
=>1/(x-1)=-98/1
=>1.(-98)=(x-1).1(tích chéo)
=>x-1=-98
=>x=-97
DP
22 tháng 7 2017
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)
\(\Leftrightarrow3x+\frac{3}{2}=1\)
\(\Leftrightarrow3x=-\frac{1}{2}\)
\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)
Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Leftrightarrow x=99\)
22 tháng 7 2017
a) => ( x + 1/2 ) . 3 = 1
=> 3x + 3/2 = 1
=> 3x = 1 - 3/2
=> 3x = -1/2
=> x = -1/2 : 3 = -1/6
17 tháng 7 2019
b \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
=>\(\frac{1}{5}-\frac{1}{x+1}\)\(=\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
=>\(\frac{1}{x+1}=\frac{1}{100}\)
=> x+1 =100
=>x=99
XO
17 tháng 7 2019
b) \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{19}{100}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(\Rightarrow x=99\)
c) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x\left(x+2\right)}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{49}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{50}{99}\)
\(\Rightarrow50.\left(x+2\right)=99\)
\(\Rightarrow x+2=\frac{99}{50}\)
\(\Rightarrow x=-\frac{1}{99}\)
d) Ta có : 6 = 1.6 = 2.3 = (-2) . (-3)
Lâp bảng xét 6 trường hợp:
| \(2x+1\) | \(1\) | \(6\) | \(2\) | \(3\) | \(-2\) | \(-3\) |
| \(y-2\) | \(6\) | \(1\) | \(3\) | \(2\) | \(-3\) | \(-2\) |
| \(x\) | \(0\) | \(\frac{5}{2}\) | \(\frac{1}{2}\) | \(1\) | \(-\frac{3}{2}\) | \(-2\) |
| \(y\) | \(8\) | \(3\) | \(5\) | \(4\) | \(-1\) | \(0\) |
Vậy các cặp (x,y) \(\inℤ\)thỏa mãn là : (0;4) ; (1; 4) ; (-2 ; 0)
e) \(x^2-3xy+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)+3y-x=1\)
\(\Rightarrow x\left(x-3y\right)-\left(x-3y\right)=1\)
\(\Rightarrow\left(x-3y\right)\left(x-1\right)=1\)
Lại có : 1 = 1.1 = (-1) . (-1)
Lập bảng xét các trường hợp :
| \(x-1\) | \(1\) | \(-1\) |
| \(x-3y\) | \(1\) | \(-1\) |
| \(x\) | \(2\) | \(0\) |
| \(y\) | \(\frac{1}{3}\) | \(\frac{1}{3}\) |
Vậy các cặp(x,y) thỏa mãn là : \(\left(2;\frac{1}{3}\right);\left(0;\frac{1}{3}\right)\)
23 tháng 5 2016
Nhận xét :
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
23 tháng 5 2016
b)
Tương tự câu a) , phương trình tương đương với :
\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)
\(\Rightarrow x=\frac{97}{195}\)
HT
1 tháng 6 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)
\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)
\(A=\frac{x}{2x+1}\)
Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)
IM
4 tháng 11 2016
Ta có :
\(\begin{cases}\left|x+\frac{1}{1.2}\right|\ge0\\\left|x+\frac{1}{2.3}\right|\ge0\\...\\\left|x+\frac{1}{99.100}\right|\ge0\end{cases}\)\(\left(\forall x\right)\)
\(\Rightarrow100x>0\)
=> x > 0
=> \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+....+\left|x+\frac{1}{99.100}\right|\)
\(=x+\frac{1}{1.2}+x+\frac{1}{2.3}+.....+x+\frac{1}{99.100}=100x\)
\(\Rightarrow100x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=0\)
Dễ thấy VT \(\ne\)VP
=> \(x\in\varnothing\)
4 tháng 11 2016
Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;...;\left|x+\frac{1}{99.100}\right|\ge0\)
=> \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)
=> \(100x\ge0\Rightarrow x\ge0\)
=> \(\left|x+\frac{1}{1.2}\right|=\left(x+\frac{1}{1.2}\right);\left|x+\frac{1}{2.3}\right|=\left(x+\frac{1}{2.3}\right);...;\left|x+\frac{1}{99.100}\right|=\left(x+\frac{1}{99.100}\right)\)=> \(\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+...+\left(x+\frac{1}{99.100}\right)=100x\)
=> 99x + \(\frac{99}{100}\) = 100x
=> x = \(\frac{99}{100}\)
26 tháng 3 2020
a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{x}{3}\)= \(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)
Với \(\frac{x}{3}\) = \(\frac{33}{100}\)
\(\Rightarrow\)100x= 33.3
\(\Rightarrow\)100x=99
\(\Rightarrow\)x=\(\frac{99}{100}\)
Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)
\(\Rightarrow\)100x=-33.3
\(\Rightarrow\)100x=-99
\(\Rightarrow\)x=\(\frac{-99}{100}\)
Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)
26 tháng 3 2020
b, \(\frac{4}{1.5}\)+ \(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)
\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)
Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)
\(\Rightarrow\)(5x-4).101=100.101
\(\Rightarrow\)505x-404=10100
\(\Rightarrow\)505x=10504
\(\Rightarrow\)x=\(\frac{104}{5}\)
Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)
\(\Rightarrow\)(5x-4). 101=-100.101
\(\Rightarrow\)505x-404=-10100
\(\Rightarrow\)505x=-9696
\(\Rightarrow\)x=\(\frac{-96}{5}\)
Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)
a. \(\frac{1}{1.2}+...+\frac{1}{x.\left(x+1\right)}=99\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+...+\frac{1}{x}-\frac{1}{x+1}=99\)
\(\Rightarrow1-\frac{1}{x+1}=99\)
\(\Rightarrow\frac{1}{x+1}=1-99=-98\)
\(\Rightarrow x=\frac{1}{-98}-1\)
\(\Rightarrow x=-\frac{99}{98}\)
P/s : Bạn ơi đề sai, x sai hay mk sai ạ???