(\(\dfrac{-3}{4}\))3x - 1 = \(\dfrac{256}{81}\)
Có thể chia ra 2 trường hợp
tìm x:
\(a,5^x.\left(5^2\right)^3=625\)
\(b,\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{4}\right)^{-2}-\left(\dfrac{-3}{5}\right)^4\)
\(c,\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(d,172x^2-7^9:98^3=2^{-3}\)
tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)
Tìm x , biết :
a.\((\dfrac{-3}{4})^{3x-1}=\dfrac{256}{81}\)
\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(\Leftrightarrow\left[\left(-\dfrac{3}{4}\right)^4\right]^{3x-1}=\dfrac{256}{81}\)
\(\Leftrightarrow\left[\left(\dfrac{3}{4}\right)^{-4}\right]^{3x-1}=\dfrac{256}{81}\)
\(\Leftrightarrow3x-1=-4\)
\(\Leftrightarrow x=-1\)
kq đã được kiểm nghiệm bằng máy tính
giải các phương trình sau
a) \(2^{x^2-1}=256\)
b) \(3^{x^2+3x}=81\)
c) \(2^{x^2-5x}=64\)
d) \(\left(\dfrac{1}{3}\right)^x=243\)
e) \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)
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
Tìm x:
a/\(\left(\dfrac{12}{25}\right)^x\)=\(\left(\dfrac{3}{5}\right)^2-\left(-\dfrac{3}{5}\right)^4\)
b/\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
c/172\(x^2-7^9:98^3=2^{-3}\)
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
Tìm x biết:
a) \(\dfrac{-32}{\left(-2\right)^x}=4\) f) \(\left(3x-1\right)^3=\dfrac{-8}{27}\)
b) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\) g) \(\left(2x+3\right)^2=\dfrac{9}{121}\)
c) \(\dfrac{1}{9}.27^x=3^x\) h) \(5^x+5^{x+2}=650\)
d) \(9^x:3^3=\dfrac{1}{243}\) i) \(\left(x-7\right)^{x+1}-\left(x-7\right)=0\)
e) \(\dfrac{x7}{81}=27\) m) \(\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
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 =.=
a,2.(\(\dfrac{1}{4}\)+x)\(^3\)=(\(-\dfrac{27}{4}\))
b,(x+\(\dfrac{1}{2}\))\(^3\):3=\(\dfrac{-1}{81}\)
c,(\(\dfrac{2}{3}\)-x)\(^2\)=1:\(\dfrac{4}{9}\)
d,(2x-\(\dfrac{1}{5}\))\(^2\)+\(\dfrac{16}{25}\)=1
e,(\(\dfrac{2}{5}\)-3x)\(^2\)-\(\dfrac{1}{5}\)=\(\dfrac{4}{25}\)
mọi người giúp mình với...
a/(1+3x)^4=256
\(\dfrac{\text{A= 3^15.29+3^15.88}}{3^{13}.81}\)
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, \(A=\dfrac{3^{15}.29+3^{15}.88}{3^{13}.81}\)
\(A=\dfrac{3^{15}\left(29+88\right)}{3^{13}.81}\)
\(A=\dfrac{3^{15}.117}{3^{13}.3^4}\)
\(A=\dfrac{3^{15}.3^2.13}{3^{17}}\)
\(A=\dfrac{3^{17}.13}{3^{17}}=13\)
\(\left(1+3x\right)^4=256\)
\(\Leftrightarrow\left(1+3x\right)^4=\left(\pm4\right)^4\)
\(\Leftrightarrow1+3x=4\Rightarrow3x=3\Rightarrow x=1\)
\(\Leftrightarrow1+3x=-4\Rightarrow3x=-5\Rightarrow x=\dfrac{-5}{3}\)
a)( \(\dfrac{-3}{4}\))\(^{^{ }3x-1}\)=\(\dfrac{256}{81}\)
b) (x-7)\(^{x+1}\) - (x-7)\(^{x+11}\) = 0
c)(x - \(\dfrac{1}{2}\))\(^2\) + (y + \(\dfrac{1}{2}\))\(^2\) =0
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