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a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
Ta có:
\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot5^n\)
\(2^n\left(\frac{1}{2}+4\right)=9\cdot5^n\)
\(\frac{9}{2}\cdot2^n=9\cdot5^n\)
Tức: \(9\cdot\frac{1}{2}\cdot2^n=9\cdot5^n\)
Suy ra: \(2^{n-1}=5^n\)
Nhận thấy: \(n-1< n\)
Hơn nữa \(2< 5\)
Do đó: \(2^{n-1}< 5^n\)
Vậy không có n thỏa mãn
\(\left(5^{2x}\cdot5^{x+2}\right):25=125^2\)
\(5^{2x+x+2}=125^2\cdot25\)
\(5^{3x+2}=\left(5^3\right)^2\cdot5^2\)
\(5^{3x+2}=5^6\cdot5^2\)
\(5^{3x+2}=5^8\)
\(\Rightarrow3x+2=8\)
\(3x=8-2\)
\(3x=6\)
\(x=6:3\)
\(x=2\)
\(\frac{1}{2}x+\frac{3}{5}x=-\frac{2}{5}\\ x\left(\frac{1}{2}+\frac{3}{5}\right)=-\frac{2}{5}\\ x.\frac{11}{10}=-\frac{2}{5}\\ x=-\frac{4}{11}\)
\(\frac{1}{2}x+\frac{3}{5}x=\frac{-2}{5}\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x=\frac{-2}{5}\)
\(\frac{11}{10}x=\frac{-2}{5}\)
\(x=\frac{-2}{5}:\frac{11}{10}\)
\(x=-\frac{4}{11}\)
a) 2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
b) 2x . 24 = 26
=> x + 4 = 6
x = 6 - 4
x = 2
c) 5x + x = 39 - 311 : 39
6x = 39 - 32
6x = 39 - 9
6x = 30
x = 30 : 6
x = 5
d) 9x - 1 = 81
9x - 1 = 92
=> x - 1 = 2
x = 2 + 1
x = 3
e) 6x = 521 : 519 + 3 . 22 - 70
6x = 52 + 3 . 4 - 1
6x = 25 + 12 - 1
6x = 37 - 1
6x = 36
x = 36 : 6
x = 6
\(a)\frac{2}{3}x-\frac{1}{2}=\frac{5}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{5}{12}+\frac{1}{2}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}:\frac{2}{3}=\frac{11}{8}\)
\(b)\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Rightarrow\frac{14}{5}x-50=51.\frac{2}{3}=34\)
\(\Rightarrow\frac{14}{5}x=34+50=84\)
\(\Rightarrow x=84:\frac{14}{5}=30\)
a) 2/3.x - 1/2 = 5/12
2/3.x = 5/12 + 1/2
2/3.x = 11/12
x = 11/12 : 2/3
x = 11/8
b) \(\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)
\(\frac{14}{5}.x-50=51.\frac{2}{3}\)
\(\frac{14}{5}.x-50=34\)
\(\frac{14}{5}.x=34+50\)
\(\frac{14}{5}.x=84\)
\(x=84:\frac{14}{5}\)
\(x=30\)
\(\frac{1}{2}\cdot2^x+2^x\cdot2^2=2^8+2^5\)
\(2^x\left(\frac{1}{2}+4\right)=2^8+2^5\)
\(2^x\cdot\frac{9}{2}=288\)
\(2^x=64\)
\(2^x=2^6\)
\(x=6\)
\(9^x:3^x=3^7\)
\(3^{2x}:3^x=3^7\)
\(3^x=3^7\)
\(x=7\)
\(7^{x+2}+2\cdot7^{x-1}=345\)
\(7^x\cdot7^2+2\cdot7^x:7=345\)
\(7^x\left(7^2+\frac{2}{7}\right)=345\)
\(7^x\cdot\frac{345}{7}=345\)
\(7^x=7\)
\(x=1\)
a) 1/2.2^x + 2^x+2 = 256 + 32
1/2.2^x + 2^2.2^x=288
2^x(1/2+4)= 288
2^x.4,5=288
2^x= 288:4,5
2^x=64=2^6
x=6
320 : ( x - 1 ) = ( 53 - 52 ) : 4 + 15
<=> 320 : ( x - 1 ) = ( 125 - 25 ) : 4 + 15
<=> 320 : ( x - 1 ) = 100 : 4 + 15
<=> 320 : ( x - 1 ) = 40
<=> x - 1 = 8
<=> x = 9
240 : ( x - 5 ) = 22 . 52 - 20
<=> 240 : ( x - 5 ) = 4 . 25 - 20
<=> 240 : ( x - 5 ) = 80
<=> x - 5 = 3
<=> x = 8
70 : ( x - 3 ) = ( 34 - 1 ) : 4 - 16
<=> 70 : ( x - 3 ) = 80 : 4 - 16
<=> 70 : ( x - 3 ) = 4
<=> x - 3 = 35/2
<=> x = 41/2
a) 320: (x-1) = (53 - 52):4 + 15 b) 240: (x-5) = 22.52 - 20 c) 70:(x-3) = (34 - 1):4 - 16
320: (x-1) = (125- 25): 4 + 15 240: (x-5) = 4.25 - 20 70:(x-3) = (81-1): 4 - 16
320: (x-1) = 100:4 + 15 240: (x-5) = 100 - 20 70: (x-3) = 80: 4 - 16
320: (x-1) = 25 + 15 240: (x-5) = 80 70: (x-3) = 20 -16
320: (x-1) = 40 (x-5) = 240:80 70: (x-3) = 4
(x-1) = 320:40 (x-5) = 3 (x-3) = 70:4
(x-1) = 80 x = 5+3 (x-3) = 17,5
x = 80+1 x = 8 x = 17,5 : 3
x = 81 x = 5,83
\(\left(\right. x - 2 \left.\right)^{5} - \left(\right. x - 2 \left.\right)^{2} = 0\)
\(\left(\right. x - 2 \left.\right)^{2} \left[\right. \left(\right. x - 2 \left.\right)^{3} - 1 \left]\right. = 0\)
Ta có: \(\left(x-2\right)^5=\left(x-2\right)^2\)
=>\(\left(x-2\right)^5-\left(x-2\right)^2=0\)
=>\(\left(x-2\right)^2\cdot\left\lbrack\left(x-2\right)^3-1\right\rbrack=0\)
TH1: \(\left(x-2\right)^2=0\)
=>x-2=0
=>x=2
TH2: \(\left(x-2\right)^3-1=0\)
=>\(\left(x-2\right)^3=1\)
=>x-2=1
=>x=1+2
=>x=3