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b, \(\left(x^2+2015\right).\left(x-2016\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2015=0\\x-2016=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x^2==-2015\\x=2016\end{cases}}\)( \(x^2=-2015\)loại do \(x^2\ge0\))
Vậy x= 2016
a, \(xy+3x-7y=21\)
\(\Leftrightarrow x.\left(y+3\right)-7y-21=0\)
\(\Leftrightarrow x.\left(y+3\right)-7.\left(y+3\right)=0\)
\(\Leftrightarrow\left(y+3\right).\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-3\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-7\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y+3=0\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x-7=0\\y+3\in Z\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}y=-3\\x-7\in Z\end{cases}}\\\hept{\begin{cases}x=7\\y+3\in Z\end{cases}}\end{cases}}\)
a, xy + 3x - 7y = 21
=> x(y + 3) - 7y - 21 = 21 - 21
=> x(y + 3) - (7y + 21) = 0
=> x(y + 3) - 7(y + 3) = 0
=> (x - 7)(y + 3) = 0
=> \(\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)
Vậy x = {7;-3}
b, (x2 + 2015)(x - 2016) = 0
\(\Rightarrow\orbr{\begin{cases}x^2+2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=2015\left(loại\right)\\x=2016\end{cases}}}\)
Vậy x = 2016
![](https://rs.olm.vn/images/avt/0.png?1311)
a. (9x + 2).3 = 60
<=> 9x + 2 = 20
<=> 9x = 18
<=> x = 2
b. 71 + (26 - 3x):5 = 75
<=> (26 - 3x) : 5 = 4
<=> 26 - 3x = 4/5
<=> 3x = 26 - 4/5
<=> x = 42/5
c. 2x = 32
<=> 2x = 25
<=> x = 5
d. (x - 6)2 = 9
<=> x - 6 = 3
<=> x = 9
a) \(\left(9x+2\right)\times3=60\)
\(\Rightarrow9x+2=60:3\)
\(\Rightarrow9x+2=20\)
\(\Rightarrow9x=20-2\)
\(\Rightarrow9x=18\)
\(\Rightarrow x=18:9\)
\(\Rightarrow x=2\)
Vậy x = 2
b) \(71+\left(26-3x\right):5=75\)
\(\Rightarrow\left(26-3x\right):5=75-71\)
\(\Rightarrow\left(26-3x\right):5=4\)
\(\Rightarrow26-3x=4\times5\)
\(\Rightarrow26-3x=20\)
\(\Rightarrow3x=26-20\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=6:3\)
\(\Rightarrow x=2\)
Vậy x = 2
c) \(2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5
d) \(\left(x-6\right)^2=9\)
\(\Rightarrow\orbr{\begin{cases}x-6=3\\x-6=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=3\end{cases}}\)
Vậy x = 9 hoặc x = 3
_Chúc bạn học tốt_
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 1170 : x - 13 = 339 . 3 . 10 2
1170:(x - 13) = 339 - 3.100
1170:(x - 13) = 39
x - 13 = 1170:39 = 30
x = 30+13 = 43
b, 3 x - 9 + 15 = 12 + 243 : 3 2
3(x - 9) + 15 = 12+243:9
3(x - 9)+15 = 12+27
3(x - 9)+15 = 39
3(x - 9) = 24
x - 9 = 8
x = 17
![](https://rs.olm.vn/images/avt/0.png?1311)
a . 3 x = 9 ⇔ 3 x = 3 2 ⇔ x = 2
b . 5 x = 5 3 ⇔ x = 3
c . 3 x + 1 = 3 2 ⇔ x + 1 = 2 ⇔ x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: 3 x = 3 2 nên x = 2.
b) Ta có: 5 x = 5 3 nên x = 3.
c) Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1.
![](https://rs.olm.vn/images/avt/0.png?1311)
TL :
\(12\left(x-81\right):3=4^3+2^3\)
\(12\left(x-81\right):3=64+8\)
\(12\left(x-81\right):3=72\)
\(12\left(x-81\right)=72.3\)
\(12\left(x-81\right)=216\)
\(x-81=216:12\)
\(x-81=18\)
\(\Rightarrow x=109\)
Câu b đang suy nghĩ. ( mik ngu vl )
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 9:
a) 15x + 40 = 15 + 20.8
15x + 40 = 15 + 160
15x + 40 = 175
15x = 175 - 40 = 135
x = 135 / 15 = 9
b) ( x-1 )( 5-x ) = 0
=> x-1 = 0 hoặc 5-x = 0
+) x-1 = 0 +) 5-x = 0 ( còn lại tự làm nhé ^^ )
c) x - 140 : 35 = 270
x - 140 = 270 . 35 = 9450
x = 9450 + 140 = 9590
d) ( 14 - 3x ) + ( 6 + x ) = 0
14 - 3x + 6 + x = 0
( 14 + 6 ) - ( 3x -x ) = 0
20 - 2x = 0
2x = 20
x = 10
3 + 6 + 9 + 12 + ... + 3x = 315
=> (3x + 3)[(3x - 3) : 3 + 1] : 2 = 315
=> 3((x + 1)(x - 1 + 1) = 315 . 2
=> 3x(x + 1) = 630
=> x(x + 1) = 630 : 3
=> x2 + x - 210 = 0
=> x2 + 15x - 14x - 210 = 0
=> x(x + 15) - 14(x + 15) = 0
=> (x - 14)(x + 15) = 0
=> \(\orbr{\begin{cases}x-14=0\\x+15=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=14\\x=-15\left(ktm\right)\end{cases}}\)
Vậy x = 14