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Ta có : 3x - 7/3 - 2x - 1/2 = 7 .
=> x ( 3 - 2 ) - ( 7/3 + 1/2 ) = 7 .
=> x - ( 14/6 + 3/6 ) = 7 .
=> x - 17/6 = 7 .
=> x = 7 + 17/6 .
=> x = 59/6 .
vậy x = 59/6 .
\(3x-\frac{7}{3}-2x-\frac{1}{2}=7\)
\(\Leftrightarrow\left(3x-2x\right)-\left(\frac{7}{3}+\frac{1}{2}\right)=7\)
\(\Leftrightarrow x-\frac{17}{6}=7\)
\(\Leftrightarrow x=7+\frac{17}{6}\)
\(\Leftrightarrow x=\frac{59}{6}\)
1,
x10 = x
=> x10 - x = 0
=> x(x9 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
KL: x thuộc {1; 0}
2,
\(S=2+2^2+2^3+...+2^{2016}\)
=> \(2S=2^2+2^3+2^4+...+2^{2017}\)
=> \(2S-S=\left(2^2+2^3+2^4+...+2^{2017}\right)-\left(2+2^2+2^3+...+2^{2016}\right)\)
=> \(S=2^{2017}-2\)
Bài 1:
x10 = x => x= { -1;1}
Bài 2:
\(S=2+2^2+2^3+...+2^{2016}\)
\(2S=2^2+2^3+2^4+2^{2017}\)
\(2S-S=2^{2017}-2\)
Vậy \(S=2^{2017}-2\)
\(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{x\left(x+2\right)}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{2}-\frac{1}{x+2}=\frac{4}{9}\)
<=> \(\frac{1}{x+2}=\frac{1}{18}\)
=> \(x+2=18\)
<=> \(x=16\)
Vậy...
\(\frac{2}{2.3}\) + \(\frac{2}{3.4}\) + \(\frac{2}{4.5}\) + .......+ \(\frac{2}{x.\left(x+1\right)}\) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + .......+ \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
2 . ( \(\frac{1}{2}\) - \(\frac{1}{x+1}\) ) = \(\frac{2017}{2019}\)
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{2019}\) : 2
\(\frac{1}{2}\) - \(\frac{1}{x+1}\) = \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2}\) - \(\frac{2017}{4038}\)
\(\frac{1}{x+1}\) = \(\frac{1}{2019}\)
<=> x + 1 = 2019 => x = 2018
vậy x = 2018
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2017}{4038}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2019}\)
\(\Rightarrow x+1=2019\)
\(\Leftrightarrow x=2018\)
Vậy \(x=2018\)
\(\text{ ( 2x - 4 ) . ( 3 - x ) = 0 }\)
\(\Rightarrow\orbr{\begin{cases}2x-4=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
\(\Rightarrow x\in\left\{2;3\right\}\)
chúc bạn học tốt !!
x2-2x+3
=x2-x-x+1+2
=x.(x-1)-(x-1)+2
=(x-1)(x-1)+2
Để x2-2x+3 chia hết cho x-1 thì:
(x-1)(x-1)+2 chia hết cho x-1
=>2 chia hết cho x-1
=>x-1 thuộc Ư(2)={1;-1;2;-2}
Ta có bàng sau:
| x-1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
Vậy x={2;0;3;-1}
3/4 +1/4x+x-7/6x=5/12
<=> 1/12x=-1/3
<=> x=-4
hok tốt
mik vs nha
thank nha thực ra tại lúc đó mk thấy khó quá nên đăng lên nhưng bây h mk làm đc r kết quả giống bn đó