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\(72-7\left(13-x\right)=14\)
\(7\left(13-x\right)=58\)
\(13-x=\dfrac{58}{7}\)
\(x=13-\dfrac{58}{7}\)
\(x=\dfrac{91}{7}-\dfrac{58}{7}\)
\(x=\dfrac{33}{7}\)
`72-7(13-x)=14`
`=>7(13-x)=72-14`
`=>7(13-x)=58`
`=>13-x=58:7`
`=>13-x=58/7`
`=>x=13-58/7`
`=>x=33/7`
![](https://rs.olm.vn/images/avt/0.png?1311)
Bai 2 :
\(\hept{\begin{cases}x+y=7\\x-7=13\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=7\\x=20\end{cases}}}\)
Thay x vào phương trình đầu ta có :
\(20+y=7\Leftrightarrow y=-13\)
Vậy \(\left\{x;y\right\}=\left\{20;-13\right\}\)
Thử \(20-13=7\); \(20-7=13\)( thỏa mãn )
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1: Ta có: \(2155-\left(174+2155\right)+\left(-68+174\right)\)
\(=2155-174-2155-68+174\)
=-68
2) Ta có: \(-25\cdot72+25\cdot21-49\cdot25\)
\(=-25\cdot\left(72-21+49\right)\)
\(=-25\cdot100=-2500\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3^{2020}=3^{2018}\cdot3^2=3^{2018}\cdot9\)
mà 9<10
nên \(3^{2020}< 10\cdot3^{2018}\)
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Đáp án là A
Ta có: M = 14 - 23 + (5 - 14) - (5 - 23) + 17 = 14 - 23 + 5 - 14 - 5 + 23 + 17
= (14 - 14) + (5 - 5) - (23 - 23) + 17 = 0 + 0 - 0 + 17 = 17
N = 24 - (72 - 13 + 24) - (72 - 13) = 24 - 72 + 13 - 24 - 72 + 13
= (24 - 24) - (72 + 72) + (13 + 13) = 0 - 144 + 26 = -118
Do đó M > N
72-7.(13-x)=14
=>7.(13-x)=72-14
=>7.(13-x)=58
=>13-x=58:7
=>13-x=\(\dfrac{58}{7}\)
\(=>x=13-\dfrac{58}{7}\)
\(=>x=\dfrac{33}{7}\)
Vậy ................
\(72-7(13-x)=14\)
\(7(13-x)=72-14\)
\(13-x=58:7\)
\(13-x=\) \(\dfrac{58}{7}\)
\(x=13-\dfrac{58}{7}\)
\(x=\dfrac{33}{7}\)