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a) \(2^{4x+1}-8^{x+2}=0\)\(\Leftrightarrow2^{4x+1}-2^{3\left(x+2\right)}=0\)
\(\Leftrightarrow2^{4x+1}-2^{3x+6}=0\)\(\Leftrightarrow2^{4x+1}=2^{3x+6}\)
\(\Leftrightarrow4x+1=3x+6\)\(\Leftrightarrow4x-3x=6-1\)\(\Leftrightarrow x=5\)
Vậy \(x=5\)
b) \(3^2.9^{2x}=27^{x+3}\)\(\Leftrightarrow3^2.3^{2.2x}=3^{3\left(x+3\right)}\)\(\Leftrightarrow3^2.3^{4x}=3^{3x+9}\)
\(\Leftrightarrow3^{2+4x}=3^{3x+9}\)\(\Leftrightarrow2+4x=3x+9\)\(\Leftrightarrow4x-3x=9-2\)\(\Leftrightarrow x=7\)
Vậy \(x=7\)
c) \(8^{2x}.64^2=16^{x+4}\)\(\Leftrightarrow2^{3.2x}.2^{6.2}=2^{4\left(x+4\right)}\)\(\Leftrightarrow2^{6x}.2^{12}=2^{4\left(x+4\right)}\)
\(\Leftrightarrow2^{6x+12}=2^{4x+16}\)\(\Leftrightarrow6x+12=4x+16\)\(\Leftrightarrow6x-4x=16-12\)
\(\Leftrightarrow2x=4\)\(\Leftrightarrow x=2\)
Vậy \(x=2\)
a: \(\Leftrightarrow x^3=\dfrac{539}{64}\)
hay \(x=\dfrac{7\sqrt{11}}{4}\)
c: \(\Leftrightarrow2^{2x-1}=2^9\cdot2^2=2^{11}\)
=>2x-1=11
hay x=6
d: \(\Leftrightarrow x^{17}-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{0;1;-1\right\}\)
a) \(\left(19x+2\times5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right):14=5^2-4^2=25-16=9\)
\(\Rightarrow19x+50=126\)
\(\Rightarrow19x=76\Rightarrow x=4\)
Vậy x = 4
b) \(2\times3^2=10\times3^{12}+8\times27^4\)
\(\Rightarrow2\times3^2=10\times\left(3^3\right)^4+8\times27^4\)
\(\Rightarrow2\times3^2=27^4\times\left(10+8\right)\)
\(\Rightarrow18=27^4\times18\)
\(\Rightarrow27^4\times18-18=0\Rightarrow18\times\left(27^4-1\right)=0\)
=> Không thấy biến x đâu cả
c) Ta thấy 33 = 27
\(\Rightarrow3^{2x-5}=3^3\Rightarrow2x-5=3\Rightarrow2x=8\Rightarrow x=4\)
Vậy x = 4
d) \(3^{x+1}-x=80\Rightarrow3^{x+1}=81\)
Ta thấy 34 = 81
\(\Rightarrow3^{x+1}=3^4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a) (5x - 1) : 3 + 1 = 4
=> (5x - 1) : 3 = 3
=> (5x - 1) = 9
=> 5x - 1 = 9
=> 5x = 10
=> x = 2
b) 54 : (16 - x) - 1=5
=> 54:(16-x) = 6
=> 16-x = 9
=> x = 7
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
a)\(\left(x-2,5\right)^2=\frac{4}{9}\\ \left(x-\frac{5}{2}\right)^2=\left(\pm\frac{2}{3}\right)^2\\\Leftrightarrow\left\{{}\begin{matrix}x-\frac{5}{2}=\frac{2}{3}\\x-\frac{5}{2}=\frac{-2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{19}{6}\\x=\frac{11}{6}\end{matrix}\right. \)
vậy....
b)\(\left(2x+\frac{1}{3}\right)^3=\frac{-8}{27}\\ \left(2x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\\ 2x+\frac{1}{3}=\frac{-2}{3}\\ x=\frac{-1}{2}\)
vậy...
c/ 2x - 1 = \(5^{98}:5^{96}\)
2x - 1 = \(5^2\) = 25
2x = 25 + 1 = 26
x = 26 : 2
x = 13
d/ 7x + 3 = \(3^5.2^3.9\)
7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)
7x + 3 = \(17496\)
7x = 17496 - 3 = 17493
x = 17493 : 7
x = 2499
e/\(2^{2x+6}=1\)
\(2^{2x+6}=2^0\)
2x + 6 = 0
2x = 0 - 6 = - 6
x = - 6 : 2
x = - 3
j/ \(2^x=8\)
\(2^x=2^3\)
x = 3
g/ \(2^x:2^3=16\)
\(2^{x-3}=2^4\)
x - 3 = 4
x = 4 + 3
x = 7
h/ \(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)\) = 56
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
x = 3
Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!
a) \(3^{x-2}=27\cdot9\)
\(3^{x-2}=3^3\cdot3^2=3^5\)
\(\Rightarrow\)\(x-2=5\Rightarrow x=7\)
b) \(2^{x+1}+2^{x+3}=80\)
\(\Rightarrow2^{x+1}\left(1+2^2\right)=80\)
\(\Rightarrow2^{x+1}\cdot5=80\)
\(\Rightarrow2^{x+1}=16=2^4\)
\(\Rightarrow x+1=4\Rightarrow x=3\)
c) \(2^{2x-3}=16\cdot8\)
\(2^{2x-3}=2^4\cdot2^3=2^7\)
\(\Rightarrow2x-3=7\)
\(\Rightarrow2x=4\Rightarrow x=2\)
d) \(2^{x-2}\cdot2^x=64\)
\(\Rightarrow2^{x-2+x}=64=2^6\)
\(\Rightarrow x-2+x=6\)
\(\Rightarrow2x-2=6\)
\(\Rightarrow2x=8\Rightarrow x=4\)
Giải:
a) \(3^{x-2}=27.9\)
\(\Leftrightarrow3^{x-2}=3^3.3^2\)
\(\Leftrightarrow3^{x-2}=3^5\)
Vì \(3=3\)
Nên \(x-2=5\)
\(\Leftrightarrow x=5+2\)
\(\Leftrightarrow x=7\)
Vậy x = 7.
b) \(2^{x+1}+2^{x+3}=80\)
\(\Leftrightarrow2^{x+1}\left(1+2^2\right)=80\)
\(\Leftrightarrow2^{x+1}.5=80\)
\(\Leftrightarrow2^{x+1}=\dfrac{80}{5}=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
Vì \(2=2\)
Nên \(x+1=4\)
\(\Leftrightarrow x=4-1\)
\(\Leftrightarrow x=3\)
Vậy x = 3.
c) \(2^{2x-3}=16.8\)
\(\Leftrightarrow2^{2x-3}=2^4.2^3\)
\(\Leftrightarrow2^{2x-3}=2^7\)
Vì \(2=2\)
Nên \(2x-3=7\)
\(\Leftrightarrow2x=7+3=10\)
\(\Leftrightarrow x=\dfrac{10}{2}=5\)
Vậy x = 5.
d) \(2^{x-2}.2^x=64\)
\(2^{2x-2}=2^6\)
Vì \(2=2\)
Nên \(2x-2=6\)
\(\Leftrightarrow2x=6+2=8\)
\(\Leftrightarrow x=\dfrac{8}{2}=4\)
Vậy x = 4.
Chúc bạn học tốt!