10+2x = 64 : 22
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a) 22+x=52
x=52-22
x=30
b) 2x-3=45
2x=45+3
2x=48
x=48:2
x=24
c)4x = 64
4x=43
\(\Rightarrow x=3\)
\(a.x-2=\left(19-x\right).2\)
\(x-2=38-2x\)
\(x+2x=38+2\)
\(3x=40\)
\(x=\frac{40}{3}\)
\(b.x+22+2x=64\)
\(3x=64-22\)
\(3x=42\)
\(x=14\)
a)16. x2 = 64
x2 = 64 : 16
x2 = 4
x2 = 22
⇒ x = 2
b) (5.x - 2) - 64 = -36
(5.x - 2) = -36 + 64
5.x - 2 = 28
5.x = 28 + 2
5.x = 30
x = 30 : 5
x = 6
c) (2x - 10).(5 - x) = 0
TH1: 2x - 10 = 0
2x = 0 + 10
2x = 10
x = 10 : 2
x = 5
TH2: 5 - x = 0
x = 5 - 0
x = 5
⇒ Vậy x = 5.
10(x-20)=10
x - 20 = 10-10
x - 20 = 0
x = 0 + 20
x = 20
597 - 3x = 9.2
597 -3x = 18
3x = 597 - 18
3x = 579
x = 579 : 3
x = 193
10 + 2x = 1024 : 64
10 + 2x = 16
2x = 16 - 10
2x = 6
x = 6 : 2
x = 3
\(\left\{{}\begin{matrix}4x+4y=64\\-2x+5y=10\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}4x+4y=64\\4x-10y=-20\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}14y=64-\left(-20\right)\\x=\dfrac{64-4y}{4}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=6\\x=10\end{matrix}\right.\)
Vậy nghiệm của hệ phương trình (x;y) = (10;6)
\(\left\{{}\begin{matrix}4x+4y=64\\-2x+5y=10\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}4x+4y=64\\-4x+10y=20\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}14y=64\\-4x+10y=20\end{matrix}\right.\)
⇔
\(a.12\times125\times54\)
\(=1500\times54\)
\(=81000\)
\(b.64\times125\times54\)
\(=8000\times54\)
\(=432000\)
\(c.36\times\left[143+57\right]+64+\left(143+57\right)\)
\(=36\times200+64+200\)
\(=7200+64+200\)
\(=7264+200\)
\(=7464\)
10 + 2x = 64 : 22
10 + 2 x = 64 : 4
10 + 2x = 16
2x = 16 - 10
2x = 6
x = 6 : 2
x = 3
Vậy x = 3
=))
\(10+2x=64:2^2\)
\(\Leftrightarrow10+2x=64:4\)
\(\Leftrightarrow10+2x=16\)
\(\Leftrightarrow2x=16-10\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=6:2\)
\(\Leftrightarrow x=3\)
~ Rất vui vì giúp đc bn ~