X x 7 = 14 x 2
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7 x 2 = 14 7 x 4 = 28 7 x 6 = 42 7 x 3 = 21
2 x 7 = 14 4 x 7 = 28 6 x 7 = 42 3 x 7 = 21
14 : 7 = 2 28 : 7 = 4 42 : 7 = 6 21 : 7 = 3
14 : 2 = 7 28 : 4 = 7 42 : 6 = 7 21 : 3 = 7
4/5 x 15/11 x 2/9 = 8/33
4/7 x 2/5 + 4/7 x 3/5 = 4/7
( 5/7 + 7/8 ) x 2 = 89/28
2/14 + 3/7 - 3/14 x 2 = 1/7
tk mình nhé, mk nhanh nhất
Học sinh nhẩm và ghi kết quả như sau:
7 x 5 = 35
7 x 6 = 42
7 x 2 = 14
7 x 4 = 28
35 : 7 =5
42 : 7 = 6
14 : 7 = 2
28: 7 =4
35 : 5 = 7
42 : 6 = 7
14 : 2 = 7
28 : 4 = 7
a: \(2^7+\left(x-3^7\right)=5^7-4^7\)
=>\(128+x-2187=78125-16384\)
=>\(x-2059=61741\)
=>\(x=61741+2059=63800\)
c: \(7^2-\left(x+15\right)=5\cdot2^2\)
=>49-(x+15)=5*4=20
=>x+15=29
=>x=14
d: 7^3-7(13-x)=14
=>343-7(13-x)=14
=>7(13-x)=343-14=329
=>13-x=47
=>x=13-47=-34
A = 7/7.17 + 7/17.27 + 7/27.37 + ............ +7/1997.2007
A=7/10 ( 10/7.17 + 10/17.27 + 10/27.37 + ................+10/1997.2007)
A= 7/10 ( 1/7 -1/17 + 1/17 - 1/27 + 1/27 - 1/37 +...............+ 1/1997 - 1/2007)
A= 7/10 (1/7 - 1/2007)
A= 7/10 . 2000/14049
A=200/2007
bây h mk có vc rùi tích đúng nha tối mk lm típ cho
\(\frac{1}{2}\cdot\frac{18}{14}+\frac{1}{2}\cdot\frac{3}{14}-\frac{1}{2}\cdot\frac{7}{14}\)
\(=\frac{1}{2}\left(\frac{18}{14}+\frac{3}{14}-\frac{7}{14}\right)\)
\(=\frac{1}{2}\cdot1=\frac{1}{2}\)
1, -15 + \(x\) = -14 - (-57)
-15 + \(x\) = -14 + 57
\(x\) = -14 + 57 + 15
\(x\) = 43 + 15
\(x\) = 58
2, (-14) + \(x\) - 7 = -10
-14 + \(x\) - 7 = -10
- 21 + \(x\) = - 10
\(x\) = -10 + 21
\(x\) = 11
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3