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a) (x : 3 - 4) . 5 = 15
x : 3 - 4 = 15 : 5
x : 3 - 4 = 3
x : 3 = 3 + 4
x : 3 = 7
x = 7 . 3
x = 21.
b) 128 - 3 . (x + 4) = 23
3 . (x + 4) = 128 - 23
3 . (x + 4) = 105
x + 4 = 105 : 3
x + 4 = 35
x = 35 - 4
x = 31.
c) (3x - 24) . 73 = 2 . 74
3x - 24 = 2 . 74 : 73
3x - 24 = 2 . 7
3x - 16 = 14
3x = 14 + 16
3x = 30
x = 30 : 3
x = 10.
d) 5x + x = 39 - 311 : 39
5x + x = 39 - 32
5x + x = 39 - 9
6x = 30
x = 30 : 6
x = 5.
a) \(\left(\dfrac{x}{3}-4\right).5=15\)
\(\Leftrightarrow\dfrac{x}{3}-4=\dfrac{15}{5}\)
\(\Leftrightarrow\dfrac{x}{3}-4=3\)
\(\Leftrightarrow\dfrac{x}{3}=3+4\)
\(\Leftrightarrow\dfrac{x}{3}=7\)
\(\Leftrightarrow x=7.3\)
\(\Leftrightarrow x=21\)
Vậy \(x=21\)
b) \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=128-23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=\dfrac{105}{3}\)
\(\Leftrightarrow x+4=35\)
\(\Leftrightarrow x=35-4\)
\(\Leftrightarrow x=31\)
Vậy \(x=31\)
c) \(\left(3x-2^4\right).7^3=2.7^4\)
\(\Leftrightarrow3x-2^4=\dfrac{2.7^4}{7^3}\)
\(\Leftrightarrow3x-2^4=2.7\)
\(\Leftrightarrow3x-2^4=14\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=14+16\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=\dfrac{30}{3}\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\)
d) \(5x+x=39-3^{11}:3^9\)
\(\Leftrightarrow6x=39-3^{11-9}\)
\(\Leftrightarrow6x=39-3^2\)
\(\Leftrightarrow6x=39-9\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=\dfrac{30}{6}\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
a, Ta thấy A chia hết cho 7 (nguyên tố)
Có : 7^2;7^3;....;7^10 đều chia hết cho 49 mà 7 ko chia hết cho 49
=> A ko chia hết cho 49
=> A chia hết cho 7 (nguyên tố ) mà A ko chia hết cho 49=7^2
=> A ko phải là số cp
Tương tự câu a : b, b chia hết cho 11 (nguyên tố) nhưng ko chia hết cho 11^2 => b ko chính phương
c, Vì 10^10 có tận cùng là 0
=> c có tận cùng là 8
=> c ko chính phương
k mk nha
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a, Ta có: \(142-\left(12x+30\right)=10^{5-3}\)
\(=>142-12x-30=10^2\)
\(112-12x=100\)
\(=>12x=112-100=12=>x=1\)
Vậy số cần tìm là 1;
b, Ta có: \(=>\left(5x+3^4\right)=6^9:6^8.3^4\)
\(=>5x+3^4=6.3^4=>5x=6.3^4-3^4\)
\(=>5x=5.3^4=>x=3^4=81\)
Vậy x=81;
CHÚC BẠN HỌC TỐT.......
Ta có: \(71.2-6(2x+5)=10^5:10^3\)
\(\Rightarrow142-12x-30=10^2\)
\(\Rightarrow142-30-10^2=12x\)
\(\Rightarrow142-30-100=12x\)
\(\Rightarrow12=12x\)
\(\Rightarrow x=\dfrac{12}{12}\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
4(x - 3) = 72 - 110
=> 4x - 12 = 49 - 1
=> 4x = 48 - 12
=> 4x = 24
=> x = 6
5x + x = 39 - 311 : 39
=> 6x = 39 - 32
=> 6x = 39 - 9
=> 6x = 30
=> x = 5
\(4\left(x-3\right)=7^2-1^{10}\)
\(4x-4\cdot3=49-1\)
\(4x\cdot12=48\)
\(4x=48:12\)
\(4x=4\)
\(x=4:4\)
\(x=1\)
\(5x+x=39-3^{11}:3^9\)
\(6x=39-3^{11-9}\)
\(6x=39-3^2\)
\(6x=39-9\)
\(6x=30\)
\(x=30:6\)
\(x=5\)