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\(2^x+2^{x+1}=96\)
\(\Rightarrow2^x\left(1+2\right)=96\)
\(\Rightarrow2^x.3=96\)
\(2^x=96:3\)
\(2^x=32\)
\(\Rightarrow2^x=2^5\)
\(x=5\)
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
a)
\(\dfrac{7}{5}+\dfrac{5}{6}:5-\dfrac{3}{8}\cdot\left(-3\right)\\ =\dfrac{7}{5}+\dfrac{1}{6}+\dfrac{9}{8}\\ =\dfrac{168+20+135}{120}\\ =\dfrac{323}{120}\)
x23:216=x15.x5
x23: 63= x20
==> 63=x23:x20
==> 63=x3
==> x=6
(5x.5x+2):53=125
==> 52x+2:53=53
==> 52x+2=53.53
==> 52x+2=56
==> 2x+2=6
2x=6–2
2x=4
x=4:2
x=2
(3x.3x+4):33=243
==> 32x+4:33=35
32x+4=35.33
32x+4=38
==> 2x+4=8
2x=8–4
2x=4
x=4:2
x=2
3x+3x+1+3x+2=243
3x.1+3x.31+3x.32=243
3x.(1+31+32)=243
3x. (1+3+9)=243
3x.13=243
3x=243:13
3x=...
a)5x-2 - 32 = 24 - ( 28 * 24- 210 * 22 )
=> 5x-2-32=16
=> 5x-2=25
=> (5x)2=25
=> 5x=5
=> x=1
b) ( 3 * x - 24 ) * 73 = 8 * 74
=> (3.x-16)=8.74:73
=> 3x-16=56
=> 3x=72
=> x=24
c) x15 = x
=> x15:x=1
=> x14=1
=> x=1
d) ( x - 5 )4 =( x - 5 )6
=> (x-5)6:(x-5)4=1
=> (x-5)2=1
=> x-5=1
=> x=6
a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
a) \(\frac{18^4.3^2.8^3}{27^3.16^2}=\frac{\left(2.3^2\right)^4.3^2.\left(2^3\right)^3}{\left(3^3\right)^3.\left(2^4\right)^2}=\frac{2^4.2^9.3^8.3^2}{3^9.2^8}=\frac{2^{13}.3^{10}}{3^9.2^8}=3.2^5=96\)
b) \(\frac{35^5.9^3.8^5}{81^4.32^5}=\frac{35^5.\left(3^2\right)^3.\left(2^3\right)^5}{\left(3^4\right)^4.\left(2^5\right)^5}=\frac{35^5.3^6.2^{15}}{3^{16}.2^{25}}=\frac{35^5}{3^{10}.2^{10}}=\frac{35^5}{6^{10}}\)
c) \(\frac{48^5.18^2}{81^2.34^4}=\frac{\left(2^4.3\right)^5.\left(2.3^2\right)^2}{\left(3^4\right)^2.\left(2.17\right)^4}=\frac{2^{20}.3^5.2^2.3^4}{3^8.2^4.17^4}=\frac{2^{22}.3^9}{3^8.2^4.17^4}=\frac{2^{18}.3}{17^4}\)
d) \(\frac{54^7.27^3.16^2}{243^2.64^3}=\frac{\left(2.3^3\right)^7.\left(3^3\right)^3.\left(2^4\right)^2}{\left(3^5\right)^2.\left(2^6\right)^3}=\frac{2^7.3^{21}.3^9.2^8}{3^{10}.2^{18}}=\frac{2^{15}.3^{30}}{3^{10}.2^{18}}=\frac{3^{20}}{2^3}\)