Giúp mik vs, cần gấp:
Tìm x ( "." là dấu nhân nhé)
a. x . (x + 1) - x - 1 = 0
b.(6 - x)2014 = ( 6 - x)2015
c.34x+4 = 9x + 2
d. 3 . ( 1/5.8 + 1/8.11 + 1/11.14 + ...+ 1/97.100 + x) = 319/100
e. 1/21 + 1/28 + 1/36 + ... + 2/x(x+1) = 2/9
Tìm x biết: a) 34x+4=9x+2
b) 3.(1/5.8 + 1/8.11 + 1/11.14 +........+ 1/97.100+x) =319/100
c) (6-x)2014=(6-x)2015
d) 2+4+6+...+2.x=210
e) 1/21+1/28+1/36+...+2/x.(x+1) = 2/9
Ta có : (6 - x)2014 = (6 - x)2015
=> (6 - x)2014 - (6 - x)2015 = 0
<=> (6 - x)2014(1 - 6 - x) = 0
<=> \(\orbr{\begin{cases}\left(6-x\right)^{2014}=0\\1-6-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}6-x=0\\-5-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-5\end{cases}}\)
sory bạn trừng hợp hai mk nhầm :
1 - (6 - x) = 0
=> 1 - 6 + x = 0
=> -5 + x = 0
=> x = 5
Ta có : 2 + 4 + 6 + ...... + 2x = 210
Nên : số số hạng là ; (2x - 2) : 2 + 1 = x - 1 + 1 = x (số)
Ta có : tổng là : (2x + 2) * x : 2 = 210
=> (2x2 + 2x) : 2 = 210
=> x2 + x = 210
=> x(x + 1) = 20.21
=> x = 20
bài 1 Rút gọn B=(1 -1/3).(1 -1/6).(1 -1/10).(1 -1/15)...(1 -1/190)
Bài 2 Tìm x biết
a) 1/21+ 1/28+ 1/36+..+1/x(x+1)=2/9
b)1/5.8 + 1/8.11+ 1/11.14+..+1/x(x+3)=101/1540
tìm x; 3.x/2.5 + 3.x/5.8 + 3.x/8.11 + 3.x/11.14= 1/21
làm giúp mình với nha, mình đang cần gấp lắm
3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21
=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21
=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21
x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21
x . ( 1/2 - 1/14 ) = 1/21
x . 3/7 = 1/21
x = 1/21 : 3/7
=> x = 1/9
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
[ @Phan Đức Gia Linh _ Xin cảm ơn những bạn đã quan tâm tới câu hỏi của mình! ]
Tìm x, biết:
1) x\(^3\) - \(\dfrac{4}{25}\)x = 0
2) 3\(^{4x+4}\) = 9\(^{x+2}\)
3) 3 . \(\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)\) = \(\dfrac{319}{100}\)
4) (6 - x) \(^{2014}\) = (6 - x) \(^{2015}\)
5) 2 + 4 + 6 + ... + 2x = 210
6) \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) + ... + \(\dfrac{2}{x.\left(x+1\right)}\) = \(\dfrac{2}{9}\)
Các bạn cố gắng giải đầy đủ giúp mình!
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
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 ...........
\(\)
2. 34x+4 = 9x+2
<=> 32(2x+2) = 9x+2
<=> (32)2x+2 = 9x+2
<=> 92x+2 = 9x+2
<=> 2x + 2 = x + 2
<=> 2x - x = 2 - 2
<=> x = 0
@Phan Đức Gia Linh
tìm x :
a, 1/5.8+1/8.11+11/11.14+...+1/x.(x+3)=101/1540
b, 1+1/3+1/6+1/10+...+1/x.(x+1):2=1 và 1991/1993
tìm x biết
1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/15401+1/3+1/6+1/10+...+1/x(x+1):2=1 1991/1993Tìm x biết:
3.x/2.5+3.x/5.8+3.x/8.11+3.x/11.14=1/21
trả lời hộ mik mik cho 1 link
3.x/2.5+3.x/5.8+3.x/8.11+3.x/11.14=1/21
tìm x biết
a 1/5.8+1/8.11+1/11.14+...1/x(x+3)=101/1540
b 1+1/3+1/6+1/10+...1/x(x+1)=\(1\frac{1991}{1993}\)
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(=3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(\Rightarrow x=305\)