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20 tháng 6 2017

Ta có : (6 - x)2014 = (6 - x)2015

=> (6 - x)2014 - (6 - x)2015 = 0

<=> (6 - x)2014(1 - 6 - x) = 0

<=> \(\orbr{\begin{cases}\left(6-x\right)^{2014}=0\\1-6-x=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}6-x=0\\-5-x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=-5\end{cases}}\)

20 tháng 6 2017

sory bạn trừng hợp hai mk nhầm : 

 1 - (6 - x) = 0

=> 1 - 6 + x = 0

=> -5 + x = 0

=> x = 5 

28 tháng 7 2017

1. x3 - \(\dfrac{4}{25}\)x = 0
<=> x(x2 - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh

28 tháng 7 2017

1 ) \(x^3-\dfrac{4}{25}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)

Vậy .............

2 ) \(3^{4x+4}=9^{x+2}\)

\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)

\(\Leftrightarrow4x+4=2x+4\)

\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)

3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )

4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)

\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)

\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)

\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)

Vậy ......

5) \(2+4+6+...+2x=210\)

\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)

\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)

\(\Leftrightarrow1+2+3+...+x=105\)

\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)

\(\Leftrightarrow x\left(x+1\right)=210\)

Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)

\(\Leftrightarrow x=14\)

Vậy ......

6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Leftrightarrow x=17.\)

Vậy ...........

\(\)

3 tháng 9 2015

a) 1/5.8+1/8.11+1/11.14+......+1/x.(x+3)=101/1540

1/3.3.[1/5.8+1/8,11+1/11.14+......+1/x.(x+3)=101/1540

1/3.[3/5.8+3/8.11+3/11.14+........+3/x.(x+3)]=101/1540

1/3.[1/5-1/8+1/8-1/11+1/11-1/14+....+1/x-1/x+3=101/1540

1/3.[1/5-1/x+3]=101/1540

1/5-1/x+3=101/1540.3

1/5-1/x+3=303/1540

1/x+3=1/3-303/1540=1/308

=>x+3=308    =>x=305

              Vậy x=305

25 tháng 3 2016

1/3.3(1/5.8+1/8.11+1/11.14+.....1/x(x+1)_101/1540

1/3.(1/5-1/8+1/8-1/11+1/11-1/14+....1/x+1/x+3)=101/1540

1/3.(1/5-1/x+3)=101/1540

1/5-1/x+3=101/1540/1/3=303/1540

1/x+3=1/5-303/1540=1/308

x+3+308

x=305

16 tháng 8 2018

305 bạn nhé

16 tháng 8 2018

phần a là 05 nha phần b mk ko bít