dề bài đâu mà tìm?

ủa mù à

\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)

\(\Leftrightarrow\left[\left(-\dfrac{3}{4}\right)^4\right]^{3x-1}=\dfrac{256}{81}\)

\(\Leftrightarrow\left[\left(\dfrac{3}{4}\right)^{-4}\right]^{3x-1}=\dfrac{256}{81}\)

\(\Leftrightarrow3x-1=-4\)

\(\Leftrightarrow x=-1\)

kq đã được kiểm nghiệm bằng máy tính

a: \(2^{x^2-1}=256\)

=>\(2^{x^2-1}=2^8\)

=>\(x^2-1=8\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

b: \(3^{x^2+3x}=81\)

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

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

c: \(2^{x^2-5x}=64\)

=>\(2^{x^2-5x}=2^6\)

=>\(x^2-5x=6\)

=>\(x^2-5x-6=0\)

=>(x-6)(x+1)=0

=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^x=243\)

=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)

=>x=-5

e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)

=>\(3^{-x-5}=3^{2x+1}\)

=>-x-5=2x+1

=>-3x=6

=>x=-2

a: \(\Leftrightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{9}{25}-\dfrac{81}{625}=\dfrac{144}{625}\)

=>x=2

b: =>3x-1=-4

=>3x=-3

hay x=-1

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

Ai giúp mình với, ai có câu trả lời chính xác mình sẽ kết bạn với nick đó

a, \(\left(1+3x\right)^4=256\)

\(\left(1+3x\right)^4=4^4\)

\(\Rightarrow1+3x=4\)

\(\Rightarrow3x=3\)

\(x=1\)

b, \(A=\dfrac{3^{15}.29+3^{15}.88}{3^{13}.81}\)

\(A=\dfrac{3^{15}\left(29+88\right)}{3^{13}.81}\)

\(A=\dfrac{3^{15}.117}{3^{13}.3^4}\)

\(A=\dfrac{3^{15}.3^2.13}{3^{17}}\)

\(A=\dfrac{3^{17}.13}{3^{17}}=13\)

a: \(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)

\(\Leftrightarrow\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^{-4}\)

=>3x-1=-4

=>3x=-3

hay x=-1

b: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=-1\\x-7=1\end{matrix}\right.\Leftrightarrow x\in\left\{7;6;8\right\}\)

c: \(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)

=>x-1/2=0 và y+1/2=0

=>x=1/2 và y=-1/2