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Bài 1 :
n \(\in\) {3;4;5}
Bài 2 :
a) A < B
b) 2300 = 4150
Bài 3 :
x \(\in\) {-1; 0 ;1}
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/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
a,3x.3=243
3x+1=35
x+1=5
x=5-1
x=4
Vậy x=4
b,x20=x
Vô lí vì 1 ko bằng 20
Vậy x E 0
c,2x.162=1024
2x.28=210
2x+8=210
x+8=10
x=10-8
x=2
Vậy x=2
d,64.4x=168
43.4x=416
43+x=416
3+x=16
x=16-3
x=13
Vậy x=13
a) \(\left(x-6\right)^3=\left(x-6\right)^2\Leftrightarrow\orbr{\begin{cases}x-6=1\Leftrightarrow x=7\\x-6=0\Leftrightarrow x=6\end{cases}}\)
b) \(\left(7.x-11\right)^3=2^5.5^2+200\)
\(\Leftrightarrow\left(7.x-11\right)^3=800+200\)
\(\Leftrightarrow\left(7.x-11\right)^3=1000\)
\(\Leftrightarrow\left(7.x-11\right)^3=10^3\)
\(\Leftrightarrow7x-11=10\Leftrightarrow7x=21\Leftrightarrow x=3\)
c) \(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(\Leftrightarrow3+2^{x-1}=24-\left[4^2-3\right]\)
\(\Leftrightarrow3+2^{x-1}=24-13\)
\(\Leftrightarrow3+2^{x-1}=11\)
\(\Leftrightarrow2^{x-1}=8\Leftrightarrow2^{x-1}=2^3\Leftrightarrow x-1=3\Leftrightarrow x=4\)
a)\(3^x.3=243\Leftrightarrow3^x=81\Leftrightarrow3^x=3^4\Leftrightarrow x=4\)
b) \(2^x.16^2=1024\Leftrightarrow2^x.256=1024\Leftrightarrow2^x=4\Leftrightarrow2^x=2^2\Leftrightarrow x=2\)
c) \(64:4^x=16^8\Leftrightarrow4^x=67108864\Leftrightarrow4^x=4^{13}\Leftrightarrow x=13\)
d) \(2^x=16\Leftrightarrow2^x=2^4\Leftrightarrow x=4\)
Trả lời:
H.(7x-11)3 =25.52 +200
=(7x-11)3 =32.25 +200 =(7x-11)3 =800 +200
=(7x-11)3 =1000 =(7x-11)3 = 103
= 7x-11 = 10 = 7x = 10 + 11
= 7x = 21 = x = 21:7
= x = 3
I.3x +25 = 26.22+2.30
=3x +25 = 26.4 +2.1 =3x +25 = 106
=3x = 106-25 =3x = 81
=3x = 34 => x =4
K.27.3x= 243
= 3x =243:27
= 3x = 9
= 3x = 32
=> x = 2
Mấy câu khác cứ thế làm nha
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)