a, 2^4-x=32=2⁵

=> 4-x=5

<=> x=-1

b, (x+1,5)²+(y-2,5)¹⁰=0

Vì (x+1,5)2\(\ge\)0,   (y-2,5)¹⁰\(\ge\)0

\(\Rightarrow\hept{\begin{cases}x+1,5=0\\y-2,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1,5\\y=2,5\end{cases}}}\)

a)\(2^{4-x}\)=32

=>\(2^{4-x}\)=32=\(2^5\)

=>4-x=5

=>x=4-5=-1

=>x=-1

Ta có : \(\frac{\left(4^x\right)^2}{2^x}=8\)

\(\Rightarrow4^{2x}=8.2^x\)

\(\Rightarrow4^{2x}=2^3.2^x\)

\(\Rightarrow\left(2^2\right)^{2x}=2^{x+3}\)

\(\Rightarrow2^{4x}=2^{x+3}\)

=> 4x = x + 3

=> 3x = 3

=> x = 1

Vậy x = 1. 

\(A,B,C,D\inℤ\)

a,

\(\dfrac{5^x}{125}=5^4\\ 5^x:5^3=5^4\\ 5^x=5^4\cdot5^3\\ 5^x=5^7\\ \Rightarrow x=7\)

b,

\(\dfrac{3^x}{3}+3^{x-2}=4\\ 3^{x-1}+3^{x-2}=3^1+3^0\\ \Rightarrow x=2\)

c,

\(\left(x+\dfrac{2006}{2007}\right)^6=0\\ \Rightarrow x+\dfrac{2006}{2007}=0\\ x=0-\dfrac{2006}{2007}\\ x=\dfrac{-2006}{2007}\)

d,

\(\left(x-\dfrac{1}{5}\right)^3=\dfrac{8}{125}\\ \left(x-\dfrac{1}{5}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow x-\dfrac{1}{5}=\dfrac{2}{5}\\ x=\dfrac{2}{5}+\dfrac{1}{5}\\ x=\dfrac{3}{5}\)

e,

\(3^x+3^{x-2}=810\\ 3^x\left(1+3^2\right)=810\\ 3^x\cdot10=810\\ 3^x=810:10\\ 3^x=81\\ 3^x=3^4\\ \Rightarrow x=4\)

g,

\(5^{x+2}+5^{x+1}+5^x=19375\\ 5^x\left(5^2+5+1\right)=19375\\ 5^x\cdot31=19375\\ 5^x=19375:31\\ 5^x=625\\ 5^x=5^4\\ \Rightarrow x=4\)

cảm ơn nha bn. mk kết bn vs nhau nhé

a: =>1/6x=-49/60

=>x=-49/60:1/6=-49/60*6=-49/10

b: =>3/2x-1/5=3/2 hoặc 3/2x-1/5=-3/2

=>x=17/15 hoặc x=-13/15

c: =>1,25-4/5x=-5

=>4/5x=1,25+5=6,25

=>x=125/16

d: =>2^x*17=544

=>2^x=32

=>x=5

i: =>1/3x-4=4/5 hoặc 1/3x-4=-4/5

=>1/3x=4,8 hoặc 1/3x=-0,8+4=3,2

=>x=14,4 hoặc x=9,6

j: =>(2x-1)(2x+1)=0

=>x=1/2 hoặc x=-1/2

giúp đi ạ

a)\(x^2=0\\ \Leftrightarrow x=0\)

vậy...

b)\(x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

vậy...

c)\(x^2=2\\ \Rightarrow x^2=\left(\pm\sqrt{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

vậy...

d)\(x^2=6\left(x>0\right)\\ \Rightarrow x^2=\left(\pm\sqrt{6}\right)^2\\ màx>0\\ \Rightarrow x=\sqrt{6}\)

vậy...

e)\(x^2=7\left(x< 0\right)\)

\(wtf\) ????? thông minh đấy \(x^2\ge0\) mà điều kiện lại là x < 0 ??? :D

rỗng r

f) \(\left(x+1\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

vậy....

g)\(\left(x-2\right)^2=2\\ \Rightarrow\left(x-2\right)^2=\left(\pm\sqrt{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}+2\\x=-\sqrt{2}+2\end{matrix}\right.\)

tự tính :D

vậy..

h)\(\left(x+\sqrt{3}\right)^2=5\\ \Leftrightarrow\left(x+\sqrt{3}\right)^2=\left(\pm\sqrt{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\sqrt{3}=\sqrt{5}\\x+\sqrt{3}=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\\x=\end{matrix}\right.\)

tự tính lười lắm

a) \(\frac{x-3}{x+5}=\frac{5}{7}\)

\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)

\(\Rightarrow7x-21=5x+25\)

\(\Rightarrow7x-5x=21+25\)

\(\Rightarrow2x=46\)

\(\Rightarrow x=23\)

Vậy \(x=23\)

b) \(\frac{7}{x-1}=\frac{x+1}{9}\)

\(\Rightarrow\left(x-1\right).\left(x+1\right)=7.9\)

\(\Rightarrow\left(x-1\right)x-\left(x+1\right)=7.9\)

\(\Rightarrow x^2-x-x-1=63\)

\(\Rightarrow x^2-1=63\)

\(\Rightarrow x^2=64\)

\(\Rightarrow x=8\) hoặc \(x=-8\)

Vậy \(x=8\) hoặc \(x=-8\)

c) \(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow x+4=\pm10\)

+) \(x+4=10\Rightarrow x=6\)

+) \(x+4=-10\Rightarrow x=-16\)

Vậy \(x\in\left\{6;-16\right\}\)