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KT
19 tháng 2 2018
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
19 tháng 2 2018
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)
27 tháng 3 2020
\(\left|5x-3\right|-3x=12\)
\(\Leftrightarrow\left|5x-3\right|=12+3x\)
\(\Leftrightarrow\hept{\begin{cases}5x-3=12+3x\\-\left(5x-3\right)=12+3x\end{cases}\Rightarrow\hept{\begin{cases}5x-3x=12+3\\-5x+3=12+3x\end{cases}\Rightarrow}\hept{\begin{cases}2x=15\\-5x-3x=12-3\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\-8x=9\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=\frac{-9}{8}\end{cases}}}\)
S
19 tháng 7 2018
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{100}\cdot\frac{100.101}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)
\(=\frac{1}{2}\left(2+3+...+101\right)=\frac{1}{2}\cdot\frac{100.103}{2}=25.103=2575\)
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=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)
25 tháng 8 2017
Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
26 tháng 8 2017
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
\(\left(\frac12-1\right)\left(\frac13-1\right)\left(\frac14-1\right).\ldots\left(\frac{1}{100}-1\right)\)
=\(\left(-\frac12\right)\left(-\frac23\right)\left(-\frac34\right).\ldots\left(-\frac{99}{100}\right)\)
vì ở đây có 100 phân số nên phép nhân của biểu thức này là phân số dương
biểu thức trở thành
\(\frac12x\frac23x.\ldots x\frac{99}{100}\)
=1/100
ê nhầm, vì ở đây có 99 phân số nên đáp án là -1/100 nhé sory