K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
24 tháng 1 2017
7) ( 2x2+ 1)3= 729
=> ( 2x2+ 1)3 = 36
=> ( 2x2+ 1)3= 93
=> 2x2+ 1= 9
=> 2x2= 8
=> x2= 4
=> x= 2
8) ( 3x2- 43)3= 125
=> (3x2- 43)3= 53
=> 3x2- 43= 5
=> 3x2= 48
=> x2= 16
=> x= 4
Vậy x=4
6 tháng 3 2019
a, \(3x\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}\right)=\frac{1}{21}\)
\(3x.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(x.\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\frac{3x}{7}=\frac{1}{21}\Rightarrow x=\frac{1}{9}\)
DN
6 tháng 3 2019
\(a,\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)
\(\Leftrightarrow3x\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{3x}{11.14}\right)=\frac{1}{21}\)
\(\Leftrightarrow3x.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x.\frac{3}{7}=\frac{1}{21}\)
\(\Leftrightarrow\frac{3x}{7}=\frac{1}{21}\)
\(\Leftrightarrow63x=7\Leftrightarrow x=\frac{1}{9}\)
\(\Leftrightarrow21.9x=7\)
\(\Leftrightarrow189x=7\)
\(\Leftrightarrow x=\frac{1}{27}\)
b, Gọi số ki-lo-gam dưa là a kg(a>0)
Theo bài ra ta có:
\(\frac{3}{4}a=3\frac{1}{2}\Leftrightarrow\frac{3}{4}a=\frac{7}{2}\Leftrightarrow a=\frac{11}{4}\left(kg\right)\)
Vậy số ki-lo-gam dưa là \(\frac{11}{4}\) kg
I
4
30 tháng 8 2017
a, dễ, tự làm
b, \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+.........+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.........+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.....+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
\(\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy ...
30 tháng 8 2017
a) (x-2)3 = (x-2)2
<=> (x-2)3-(x-2)2 = 0
<=> (x-2)2(x-2-1) = 0
<=> \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\dfrac{3x}{2.5}+\dfrac{3x}{5.8}+...+\dfrac{3x}{11.14}=\dfrac{1}{21}\)
<=> \(x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)
<=> \(x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)
<=> \(x=\dfrac{1}{21}:\dfrac{3}{7}\)
<=> \(x=\dfrac{1}{9}\)
U
7
(3x - 1) * 2 - 21 = 43
(3x - 1) * 2 = 43 + 21
(3x - 1) * 2 = 64
3x - 1 = 64 : 2
3x - 1 = 32
3x = 32 + 1
3x = 33
x = 33 : 3
x = 11
Vậy x = 11
\(\left(3x-1\right)^2-21=43\)
=>\(\left(3x-1\right)^2=21+43=64\)
=>\(\left[\begin{array}{l}3x-1=8\\ 3x-1=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=9\\ 3x=-7\end{array}\right.=>\left[\begin{array}{l}x=3\\ x=-\frac73\end{array}\right.\)