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a) \(\left(4x-13\right)^4+4^3=145\)
\(\Rightarrow\left(4x-13\right)^4+64=145\)
\(\Rightarrow\left(4x-13\right)^4=81\)
\(\Rightarrow4x-13=\pm3\)
+) \(4x-13=3\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
+) \(4x-13=-3\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=4\) hoặc \(x=\frac{5}{2}\)
b) \(3^{x+2}-3^x=72\)
\(\Rightarrow3^x.3^2-3^x=72\)
\(\Rightarrow3^x.\left(3^2-1\right)=72\)
\(\Rightarrow3^x.8=72\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
c) \(2^{x+2}-2^{x-1}=224\)
\(\Rightarrow2^{x-1+3}-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.2^3-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.\left(2^3-1\right)=224\)
\(\Rightarrow2^{x-1}.7=224\)
\(\Rightarrow2^{x-1}=32\)
\(\Rightarrow2^{x-1}=2^5\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
Vậy x = 6
( 4 x - 13 ) 4 + 4 3 = 145
( 4 x - 13 ) 4 + 64 = 145
( 4 x - 13 ) 4 = 81
( 4 x - 13 ) 4 = 3 4
=> 4 x - 13 = 3
4 x = 16
x = 4
3 x + 2 - 3 x = 72
3 x ( 3 2 - 1 ) = 72
3 x . 8 = 72
3 x = 9
3 x = 3 2
=> x = 2
2 x + 2 - 2 x - 1 = 224
2 x ( 2 2 - 2 -1 ) = 224
2 x . 3 , 5 = 224
2 x = 64
2 x = 2 6
=> x = 6
( 4 x - 13 ) 4 + 4 3 = 145
( 4 x - 13 ) 4 + 64 = 145
( 4 x - 13 ) 4 = 81
( 4 x - 13 ) 4 = 3 4
=> 4 x - 13 = 3
4 x = 16
x = 4
3 x + 2 - 3 x = 72
3 x ( 3 2 - 1 ) = 72
3 x . 8 = 72
3 x = 9
3 x = 3 2
=> x = 2
2 x + 2 - 2 x - 1 = 224
2 x ( 2 2 - 2 -1 ) = 224
2 x . 3 , 5 = 224
2 x = 64
2 x = 2 6
=> x = 6
a) \(\left(x+3\right):4=12\)
\(x+3=12\cdot4\)
\(x+3=48\)
\(\Rightarrow x=45\)
b) 52015 : x = 52013
=> x = 52015 : 52013
=> x = 52
c) 21 + 3 . ( x - 7 ) = 72
3 . ( x - 7 ) = 72 - 21
3 . ( x - 7 ) = 51
x - 7 = 51 : 3
x - 7 = 17
=> x = 24
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,\(3^{2x+2}=9^{x+3}\)
\(3^{2x+2}=3^{2x+3}\)
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(1999^{2x-6}=1\)
\(\Rightarrow1999^{2x-1}=1999^0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
c) \(x^{2002}=x\)
\(\Rightarrow x^{2002}-x=0\)
\(\Rightarrow x.\left(x^{2001}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^{2001}-1=0\)
+) \(x=0\)
+) \(x^{2001}-1=0\Rightarrow x^{2001}=1\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d) \(\left(x-1\right)^2=9\)
\(\Rightarrow x-1=\pm3\)
+) \(x-1=3\Rightarrow x=4\)
+) \(x-1=-3\Rightarrow x=-2\)
Vậy \(x\in\left\{4;-2\right\}\)
e) \(\left(2x-3\right)^2=81\)
\(\Rightarrow2x-3=\pm9\)
+) \(2x-3=9\Rightarrow2x=12\Rightarrow x=6\)
+) \(2x-3=-9\Rightarrow2x=-6\Rightarrow x=-3\)
Vậy \(x\in\left\{6;-3\right\}\)
Các phần khác làm tương tự
\(a,2x-138=2^3:\left(-3\right)^2\)
\(\Rightarrow2x-138=8:9\)
\(\Rightarrow2x=\frac{8}{9}+138\)
\(\Rightarrow2x=\frac{1250}{9}\)
\(\Rightarrow x=\frac{626}{9}\)
\(10+2x=\left(-4\right)^5:\left(-4\right)^3\)
\(10+2x=-1024:\left(-64\right)\)
\(10+2x=16\)
\(2x=16-10\)
\(2x=6\)
\(x=6:2=3\)
a) 10 - (x-4)=14
<=> 10 - x + 4 = 14
<=> -x = 0
<=> x = 0
Vậy x=0
b) \(\left|x+2\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=5\\x+2=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-7\end{cases}}}\)
Vậy x=3; x=-7
c) \(\frac{1}{2}x+\frac{2}{3}\left(x-1\right)=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{2}{3}x-\frac{2}{3}=\frac{1}{3}\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x=\(\frac{1}{2}\)
d)\(2x^2-72=0\)
\(\Leftrightarrow2x^2=72\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy x=6 ; x=-6
e) \(\left[\left(3x-5x\right)8\right]:4=18\)
\(\Leftrightarrow\left(3x-5x\right)8=72\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\frac{9}{2}=-4,5\)
Vậy x=-4,5
\(3^{2x}-3^x=72\)
Ta có: 31 = 3; 32 = 9; 33 = 27; 34 = 81 mà 81 - 9 = 72 nên x = 2.
=> \(3^{2.2}-3^2=72\)
32x - 3x = 72
<=> (u)2 - u = 72
<=> u = 9 hoặc u = -8
<=> 3x = 9 => x = 2
<=> 3x = -8 => ko có gt thỏa mãn
=> x = 2