Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dạng 3 :
a) 3x - 10 = 2x + 13
=> 3x - 2x = 13 - 10
=> x = 3
b) x + 12 = -5 - x
=> x + x = -5 - 12
=> 2x = -17
=> x = -8,5
c) x + 5 = 10 - x
=> x + x = 10 - 5
=> 2x = 5
=> x = 2,5
d) 6x + 23 = 2x - 12
=> 2x - 6x = 23 + 12
=> -4x = 35
=> x = -8,75
e) 12 - x = x + 1
=> x + x = 12 - 1
=> 2x = 11
=> x = 5,5
f) 14 + 4x = 3x + 20
=> 4x - 3x = 20 - 14
=> x = 6
a, \(2.x^x=10.3^{12}+8.27^4\)
\(2.x^x=10.3^{12}+8.3^{12}\)
\(2.x^x=3^{12}.\left(10+8\right)\)
\(2.x^x=3^{12}.18\)
\(2.x^x=3^{12}.2.3^3\)
\(2.x^x=3^{15}.2\)
\(x^x=3^{15}\)( Hình như sai đề )
b,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
Bài 1:
a) \(33^{2x}:11^{2x}=81\)\(\Leftrightarrow\left(33:11\right)^{2x}=81\)
\(\Leftrightarrow3^{2x}=3^4\)\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(\frac{x}{-5}=\frac{4}{21}\)\(\Leftrightarrow21x=-20\)\(\Leftrightarrow x=\frac{-20}{21}\)
Vậy \(x=\frac{-20}{21}\)
Bài 2:
\(A=\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)
\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+\left(3^2+3^6+3^{10}+3^{14}\right)}\)
\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+3^2.\left(1+3^4+3^8+3^{12}\right)}\)
\(=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right).\left(1+3^2\right)}=\frac{1}{1+3^2}=\frac{1}{1+9}=\frac{1}{10}\)
\(33^{2x}:11^{2x}=81\)!
\(\left(33:11\right)^{2x}=81\)
\(3^{2x}=81\)
\(3^{2x}=3^4\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
vậy \(x=2\)
\(\frac{x}{-5}=\frac{4}{21}\)
x.21=-5.4
x.21=-20
x=-20:21
\(x=-\frac{20}{21}\)
vậy \(x=-\frac{20}{21}\)
***********************************************************
a) Ta có: \(\dfrac{x+12}{10-x}=-\dfrac{x-10+22}{x-10}=-1+\dfrac{22}{x-10}\)
Vì \(\left(x+12\right)⋮\left(10-x\right)\) nên \(22⋮\left(x-10\right)\)
Do đó ta có bảng:
x-10 | -22 | -11 | -2 | -1 | 1 | 2 | 22 |
x | -12 | -1 | 8 | 9 | 11 | 12 | 32 |
Vậy \(x\in\left\{-12;-1;8;9;11;12;32\right\}\)
c) \(\left(x-3\right)⋮\left(x+1\right)\)
=> \(\left(x-3\right)-\left(x+1\right)⋮\left(x+1\right)\)
=> \(\left(x-3-x-1\right)⋮\left(x+1\right)\)
=>\(-4⋮\left(x+1\right)\)
=> x+1\(\in\) ư(-4)= \(\left\{\pm1,\pm2,\pm4\right\}\)
ta có bảng sau
x+1 | -4 | -2 | -1 | 1 | 2 | 4 |
x | -5 | -3 | -2 | 0 | 1 | 3 |
vậy x\(\in\left\{-5,-3;-2;0;1;3\right\}\)
-2x/5 - x/4 = -12 - 3 = -15
-8x/20 - 5x/20 = -300/20
suy ra : -8x - 5x = -300
x ( -8 -5 ) = -300
Xx ( -13 ) = 300/13
x + 2x + 3 = 12
3x = 12 - 3
3x = 9
x = 3
x + 2.x + 3 = 12
x.(2 + 1) + 3 = 12
x.3 + 3 = 12
x.3 = 12 − 3
x.3 = 9
x = 9 : 3
x = 3
Vậy x = 3