tìm x biết
a)\(\dfrac{x^7}{81}=27\)
b)\(\dfrac{x^8}{9}=729\)
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a) \(\dfrac{x^7}{81}=27\) => \(x^7=81.27=3^4.3^3=3^7\)=> \(x=3\)
b) \(\dfrac{x^8}{9}=729\)=> \(x^8=9.729=\)(\(\pm\)\(3^2\)).(\(\pm\)\(3^6\))=(\(\pm\)\(3^{^8}\)) => x = \(\pm\)3
a) \(\frac{x^7}{81}=27\)
=> x7 = 27.81
=> x7 = 33.34
=> x7 = 37
=> x = 3
Vậy x = 3
b) \(\frac{x^8}{9}=729\)
=> x8 = 729.9
=> x8 = 36.32
=> x8 = 38 = (-3)8
=> \(x\in\left\{3;-3\right\}\)
Vậy \(x\in\left\{3;-3\right\}\)
\(\frac{x^7}{81}=27\Rightarrow x^7=27.81\Rightarrow x^7=3^3.3^4\Rightarrow x^7=3^7\Rightarrow x=3\)
\(\frac{x^8}{9}=729\Rightarrow x^8=729.9\Rightarrow x^8=3^6.3^2\Rightarrow x^8=3^8\Rightarrow x=3\)
x^7/81 = 27 x^8/9 = 729
x^7 : 81= 27 x^8 : 9 = 729
x^7 =27.81 x^8 = 729.9
x^7 =2187 x^8 =6561
x^7 = 3^7 x^8 = 3^8
=> x =3 => x = 3
\(\frac{x^7}{81}=27\Rightarrow x^7=27.81=2187\Rightarrow x=\sqrt[7]{2187}=3\)
\(\frac{x^8}{9}=729\Rightarrow x^8=729.9=6561\Rightarrow x=\sqrt[8]{6561}=3\)
a) Ta có: \(\frac{x^7}{81}=27\)
\(\Rightarrow x^7=27.81=2187\)
Mà \(2187=3^7\) \(\Rightarrow x^7=3^7\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x^8}{9}=729\)
\(\Rightarrow x^8=729.9=6561\)
Mà \(6561=3^8\) \(\Rightarrow x^8=3^8\Leftrightarrow x=3\)
Vậy x = 3
CHÚC BẠN HỌC TỐT
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
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
â)\(\dfrac{x^7}{81}=27\Leftrightarrow x^7=27\times81=2187\Leftrightarrow x=3\)
b)\(\dfrac{x^8}{9}=729\Leftrightarrow x^8=729\times9=6561\Leftrightarrow x=3\)
a, \(\dfrac{x^7}{81}=27\)=> x7=27. 81=>x7=2187=>x=3
b, \(\dfrac{x^8}{9}=729\)=>x8=729. 9= 6561=>x=3 hoặc -3