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\(5x+2x=6^2-5^0\)
\(7x=36-1\)
\(7x=35\)
\(x=5\)
vậy \(x=5\)
\(5x+x=150:2+3\)
\(6x=75+3\)
\(x=78:6\)
\(x=13\)
vậy \(x=13\)
những câu sau tương tự
1) 5x + 2x = 6^2 - 5^0
=> 7x = 36 - 1 = 35
=> x = 35 : 5 = 5.
Vậy x = 5
2) 5x + x = 150 : 2 + 3
=> 6x = 75+3 = 78
=> x = 78 : 6 =13.
Vậy x = 13
3) 6x + x = 5^11 : 5^9 - 3^1
=> 7x = 5^2 -3 = 25 - 3 = 22
=> x = 22 : 7 = 22 / 7.
Vậy x = 22/7
4) 5x + 3x = 3^6 : 3^3 .4 + 12
=> 8x = 3^3 .4 + 12= 27.4 + 12 = 108 + 12 =120
=> x = 120 : 8 = 15.
Vậy x = 15
5) 4x + 2x = 68 - 2^19 : 2^16
=> 6x = 68 - 2 ^3 = 68 - 8 =60
=> x = 60 : 6 =10.
Vậy x = 10
6) 5x + x = 39 - 3^11 : 3^9
=> 6x = 39 - 3^2 = 39 - 9 = 30
=> x = 30 : 6 =5.
Vậy x =5
7) 7x - x = 5^21 : 5^19 + 3. 2^2 - 7^0
=> 6x = 5^2 + 3.4 - 1 = 25 + 12 - 1 = 36
=> x = 36 : 6 = 6.
Vậy x = 6
8) 7x - 2x = 6^17 : 6^15 + 44 : 11
=> 5x = 6^2 + 4 = 36 + 4 = 40
=> x = 40 : 5 = 8 .
Vậy x = 8
(7x-11)3=25.22+200
(7x-11)3=25+2+200
(7x-11)3=27+200
(7x-11)3=128+200
(7x-11)3=328
xong hết bt
a.23x +1=128
= 23x x 2=128
=128:2=64=2 mũ 6
vậy x=2
b.(7x-11)3=25+52+200
(7x-11)3=257=6,3579 mũ 3
7x=17,3579
x=2,4797
c.(2x+1)+(2x+2)+(2x+3)+...+(2x+101)=5757
vế trái có 101 số hạng
VT =(2x +1+2x+101).101:2=(4x+102).101:2=5757
(4x+102).101 =5757.2=11514
(4x+102)=11514:101=114
4x=114-102=12
x=12:4=3
vậy x=3
(7x-11)3=25.52+200
=> (7x-11)3=800+200
=> (7x-11)3=1000
=> (7x-11)3=103
=> 7x - 11 = 10
=> 7x = 21
=> x = 3
(2x-15)5=(2x-15)3
=> (2x-15)5 - (2x-15)3 = 0
=> (2x-15)3 . [ (2x-15)2 - 1 ] = 0
=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}2x=15\\2x=16\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)
Mà x thuộc N
=> x = 8
(3x-5)10=(3x-5)9
=> (3x-5)10 - (3x-5)9 = 0
=> (3x-5)9 .[ (3x-5) - 1 ] = 0
=> \(\orbr{\begin{cases}\left(3x-5\right)^9=0\\\left(3x-5\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-5=0\\3x-5=1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=5\\3x=6\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
Mà x thuộc N
=> x = 2
a)(7x-11)^3=1000
(7x-11)^3=10^3
7x-11 =10
7x =10+11=21
x =21:7=3
\(5x+2x=6^2-5^0\Leftrightarrow5x+2x=6^2-1\)
\(\Leftrightarrow5x-2x=36-1\Leftrightarrow5x-2x=35\Leftrightarrow x\left(5+2\right)=35\)
\(\Leftrightarrow7x=35\Leftrightarrow x=35:7\Leftrightarrow x=5\)
\(5x+x=150:2+3\Leftrightarrow5x+x=75+3\)
\(\Leftrightarrow5x+x=78\Leftrightarrow x\left(5+1\right)=78\Leftrightarrow x6=78\)
\(\Leftrightarrow x=78:6\Leftrightarrow x=13\)
a) 5x + 2x = 62 - 50
7x = 35 => x = 5
b) 5x + x = 150 : 2 + 3
6x = 78 => x = 13
c) 6x + x = 511 : 59 + 31
7x = 28 => x = 4
d) 5x + 3x = 36 : 33 x 4 + 12
8x = 120 => x = 15
e) 4x + 2x = 68 - 219 : 216
6x = 60 => x = 10
f) 5x + x = 39 - 311 : 39
6x = 30 => x = 5
g) 7x - x = 521 : 519 + 3 x 22 - 70
6x = 36 => x = 6
h) 7x - 2x = 617 : 615 + 44 : 11
5x = 40 => x = 8
\(a,\left(7x-11\right)^3=2^5.5^2+200.\)
\(\left(7x+11\right)^3=32.25+200.\)
\(\left(7x+11\right)^3=800+200.\)
\(\left(7x-11\right)^3=1000.\)
\(\left(7x-11\right)^3=10^3.\)
\(\Rightarrow7x-11=10.\)
\(\Rightarrow x=\left(10+11\right):3=7\in Z.\)
Vậy.....
\(b,3^x+25=26.2^2+2.3^0.\)
\(3^x+25=26.4+2.\)
\(3^x+25=104+2.\)
\(3^x+25=106.\)
\(3^x=106-25.\)
\(3^x=81.\)
\(3^x=3^4\Rightarrow x=4\in Z.\)
Vậy.....
\(c,2^x+3.2=64.\)(có vấn đề).
\(d,5^{x+1}+5^x=750.\)
\(5^x.5^1+5^x+1=750.\)
\(5^x\left(5^1+1\right)=750.\)
\(5^x\left(5+1\right)=750.\)
\(5^x.6=750.\)
\(5^x=750:6.\)
\(5^x=125.\)
\(5^x=5^3\Rightarrow x=3\in Z.\)
Vậy.....
\(e,x^{15}=x.\)
\(\Rightarrow x\left(x^{14}-1\right)=0\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right..\)
\(f,\left(x-5\right)^4=\left(x-5\right)^6.\)
\(\Leftrightarrow\left(x-5\right)^4-\left(x-5^6\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(1-x+5\right)\left(1+x-5\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4\left(6-x\right)\left(x-4\right)=0.\)
\(\Leftrightarrow\left(x-5\right)^4=0\Rightarrow x-5=0\Rightarrow x=5\in Z.\)
\(6-x=0\Rightarrow x=6\in Z.\)
\(x-4=0\Rightarrow x=4\in Z.\)
Vậy.....
\(a)5x+x=39-3^{11}\div3^9\)9
\(5x+x1=39-3^2\)
\(5x+x1=39-9\)
\(5x+x1=30\)
\(x\left(1+5\right)=30\)
\(x6=30\)'
\(\Leftrightarrow x=30\div6\)
\(\Leftrightarrow x=5\)
\(b)7x-x=5^{21}\div5^{19}+3.2^2-7^0\)
\(7x-x=5^2+3.4-1\)
\(7x-x1=25+12-1\)
\(7x-x1=37-1\)
\(7x-x1=36\)
\(\Leftrightarrow x\left(7-1\right)=36\)
\(\Leftrightarrow x6=36\)
\(\Leftrightarrow x=36\div6\)
\(\Rightarrow x=6\)
\(c)5x+3x=3^6.4+12\)
\(5x+3x=729.4+12\)
\(5x+3x=2916+12\)
\(5x+3x=2928\)
\(\Leftrightarrow x\left(3+5\right)=2928\)
\(\Leftrightarrow x8=2928\)
\(\Leftrightarrow x=2928\div8\)
\(\Rightarrow x=366\)
\(d)7x-2x=6^{17}\div6^{15}+44\div11\)
\(7x-2x=6^2+44\div11\)
\(7x-2x=36+4\)
\(7x-2x=40\)
\(\Leftrightarrow x\left(7-2\right)=40\)
\(\Leftrightarrow x5=40\)
\(\Leftrightarrow x=40\div5\)
\(\Rightarrow x=8\)