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Tìm x ∈ N
a) 2x chia hết cho 12 ⇒ 2x ∈ B(12)
2x chia hết cho 30 ⇒ 2x ∈ B(30)
Mà x có hai chữ số ⇒ 10 ≤ x ≤ 99
\(\Rightarrow2x\in BC\left(12;30\right)\)
Mà: \(B\left(12\right)=\left\{0;12;24;36;48;60;72;84;96;108;...\right\}\)
\(B\left(30\right)=\left\{0;30;60;90;120;...\right\}\)
\(\Rightarrow BC\left(12;30\right)=\left\{0;60;...\right\}\)
\(\Rightarrow2x=60\)
\(\Rightarrow x=\dfrac{60}{2}\\ \Rightarrow x=30\)
b) \(9^{x+2}-9^{x+1}+9^x=657\)
\(\Rightarrow9^x\cdot\left(9^2-9+1\right)=957\)
\(\Rightarrow9^x\cdot\left(81-8\right)=657\)
\(\Rightarrow9^x\cdot73=657\)
\(\Rightarrow9^x=9\)
\(\Rightarrow9^x=9^1\)
\(\Rightarrow x=1\)
bạn có thể giải giùm mk bài tính nhanh đc ko??? Mk đang cần gấp á. Cảm ơn bạn nhiều nha!
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a, 7\(x\).(2\(x\) + 10) =0
\(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(x\in\) {-5; 0}
b, -9\(x\) : (2\(x\) - 10) = 0
9\(x\) = 0
\(x\) = 0
c, (4 - \(x\)).(\(x\) + 3) = 0
\(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy \(x\in\) {-3; 4}
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(9x-2:3^2=3^4\)
\(9x-2:9=81\)
\(2:9=9x-81\)
\(\dfrac{2}{9}=9x-81\)
\(9x=81+\dfrac{2}{9}\)
\(9x=\dfrac{731}{9}\)
\(x=\dfrac{731}{9}:9\)
\(x=\dfrac{731}{81}\)
\(a.5x-5^2=10\) \(b.9x-2:3^2=3^4\)
\(5x=10+5^2\) \(9x-2=3^4.3^2\)
\(5x=35\) \(9x-2=729\)
\(x=35:5=7\) \(9x=729+2=731\)
\(x=731:9\)
\(x=\dfrac{731}{81}\)
\(c=10x+\left(2^2\right).5=10^2\)
\(10x+20=100\)
\(10x=100-20\)
\(10x=80\)
\(x=80:10=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5.2² + (x + 3) = 5²
5.4 + x + 3 = 25
20 + x + 3 = 25
x + 23 = 25
x = 25 - 23
x = 2
b) 2³ + (x - 3²) = 5³ - 4³
8 + (x - 9) = 125 - 64
8 + x - 9 = 61
x - 1 = 61
x = 61 + 1
x = 62
c) 4.(x - 5) - 2³ = 2⁴.3
4x - 20 - 8 = 16.3
4x - 28 = 48
4x = 48 + 28
4x = 76
x = 76 : 4
x = 19
d) 5.(x + 7) - 10 = 2³.5
5x + 35 - 10 = 8.5
5x + 25 = 40
5x = 40 - 25
5x = 15
x = 15 : 5
x = 3
e) 7² - 7.(13 - x) = 14
49 - 91 + 7x = 14
7x - 42 = 14
7x = 14 + 42
7x = 56
x = 56 : 7
x = 8
a) \(5\cdot2^2+\left(x+3\right)=5^2\)
\(\Rightarrow x+3=5^2-5\cdot2^2\)
\(\Rightarrow x+3=25-5\cdot4\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=5-3\)
\(\Rightarrow x=2\)
b) \(2^3+\left(x-3^2\right)=5^3-4^3\)
\(\Rightarrow8+\left(x-9\right)=125-64\)
\(\Rightarrow8+x-9=61\)
\(\Rightarrow x-1=61\)
\(\Rightarrow x=61+1\)
\(\Rightarrow x=62\)
c) \(4\left(x-5\right)-2^3=2^4\cdot3\)
\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)
\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)
\(\Rightarrow4\cdot\left(x-5\right)=56\)
\(\Rightarrow x-5=56:4\)
\(\Rightarrow x-5=14\)
\(\Rightarrow x=19\)
d) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow5\left(x+7\right)=50\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
e) \(7^2-7\left(13-x\right)=14\)
\(\Rightarrow7\left(13-x\right)=7^2-14\)
\(\Rightarrow7\left(13-x\right)=49-14\)
\(\Rightarrow7\left(13-x\right)=35\)
\(\Rightarrow13-x=5\)
\(\Rightarrow x=13-5\)
\(\Rightarrow x=8\)
f) \(5x-5^2=10\)
\(\Rightarrow5x=10+5^2\)
\(\Rightarrow5x=10+25\)
\(\Rightarrow5x=35\)
\(\Rightarrow x=\dfrac{35}{5}\)
\(\Rightarrow x=7\)
g) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x=3^4+2\cdot3^2\)
\(\Rightarrow9x=81+2\cdot9\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
h) \(10x+2^2\cdot5=10^2\)
\(\Rightarrow10x=10^2-2^2\cdot5\)
\(\Rightarrow10x=100-4\cdot5\)
\(\Rightarrow10x=80\)
\(\Rightarrow x=\dfrac{80}{10}\)
\(\Rightarrow x=8\)
i) \(125-5\left(4+x\right)=15\)
\(\Rightarrow5\left(4+x\right)=125-5\)
\(\Rightarrow5\left(4+x\right)=120\)
\(\Rightarrow4+x=\dfrac{120}{5}\)
\(\Rightarrow4+x=24\)
\(\Rightarrow x=24-4\)
\(\Rightarrow x=20\)
j) \(2^6+\left(5+x\right)=3^4\)
\(\Rightarrow5+x=3^4-2^6\)
\(\Rightarrow5+x=81-64\)
\(\Rightarrow5+x=17\)
\(\Rightarrow x=17-5\)
\(\Rightarrow x=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(3x^2-9x=0\)
\(\Leftrightarrow3x\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x=0\\x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\end{array}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
b ) \(\left(x-3\right)\left(x-5\right)=0\)
\(\Rightarrow x-3\) và \(x-5\) trái dấu
Mà \(x-3>x-5\) nên \(x-3>0\) và \(x-5< 0\)
\(\begin{cases}x-3>0\Rightarrow x>3\\x-5< 0\Rightarrow x< 5\end{cases}\)
\(\Rightarrow3< x< 5;x\) nguyên \(\Rightarrow x=4\).
Vậy \(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 7(x - 1) + 2x(x - 1) = 0
<=> (2x + 7)(x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-7\\x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) /7x+14/=63
/7x/ =63-14
/x/ =49:7
x =7
2) /18+9x/=81
/9x/=81-18
/x/=63:9
x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5\left(x+7\right)-10=2^3\cdot5\)
\(\Rightarrow5\left(x+7\right)-10=40\)
\(\Rightarrow5\left(x+7\right)=40+10\)
\(\Rightarrow x+7=\dfrac{50}{5}\)
\(\Rightarrow x+7=10\)
\(\Rightarrow x=10-7\)
\(\Rightarrow x=3\)
b) \(9x-2\cdot3^2=3^4\)
\(\Rightarrow9x-18=81\)
\(\Rightarrow9x=81+18\)
\(\Rightarrow9x=99\)
\(\Rightarrow x=\dfrac{99}{9}\)
\(\Rightarrow x=11\)
c) \(5^{25}\cdot5^{x-1}=5^{25}\)
\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)
\(\Rightarrow5^{x-1}=1\)
\(\Rightarrow5^{x-1}=5^0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(\Rightarrow2.\left[5+\left(9x-63\right)\right]=10\)
\(\Rightarrow5+\left(9x-63\right)=10:2=5\)
\(\Rightarrow9x-63=5-5=0\)
\(\Rightarrow9x=63\)
\(\Rightarrow x=63:9=7\)