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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
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
Mình cũng không rõ :) , nhưng tại cái đề thầy mình cho Kết quả là 7
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
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
Đặt \(A=\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)....\left(1+\dfrac{7}{2900}\right)\)
\(B=\left(81-\dfrac{3}{4}\right)\left(81-\dfrac{3^2}{5}\right)\left(81-\dfrac{3^3}{6}\right)....\left(81-\dfrac{3^{2014}}{2017}\right)\)
Ta có:
\(A=\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right).....\left(1+\dfrac{7}{2900}\right)\)
\(A=\dfrac{16}{9}.\dfrac{27}{20}.\dfrac{40}{33}.....\dfrac{2907}{2900}\)
\(A=\dfrac{2.8}{1.9}.\dfrac{3.9}{2.10}.\dfrac{4.10}{3.11}.....\dfrac{51.57}{50.58}\)
\(A=\dfrac{2.3.4.5.6....56.57}{1.2.3.4.5.....57.58}=\dfrac{1}{58}\)
\(B=\left(81-\dfrac{3}{4}\right)\left(81-\dfrac{3^2}{5}\right).....\left(81-\dfrac{3^{2014}}{2017}\right)\)
Vì trong dãy số trên có một thừa số là \(\left(81-\dfrac{3^6}{9}\right)=\left(81-81\right)=0\)
\(\Rightarrow B=0\)
Vì \(a=A+B\Rightarrow a=\dfrac{1}{58}+0=\dfrac{1}{58}\)(1)
Thay (1) vào đa thức \(f\left(x\right)=5x-29a\) ta được:
\(f\left(x\right)=5x-29.\dfrac{1}{58}=5x-\dfrac{1}{2}\)
Ta lại có:
\(f\left(x\right)=0\Leftrightarrow5x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{10}\)
Vậy nghiệm của đa thức trên là \(\dfrac{1}{10}\)
Chúc bạn học tốt!!!
a, Theo đề ta có:
\(2.3^x-405=3^{x-1}\)
=> \(2.3^x-405=3^x:3\)
=> \(405=(2.3^x)-(3^x:3)\)
=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)
=> \(405=3^x(2-\dfrac{1}{3})\)
=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)
=> \(405=3^x.\dfrac{5}{3}\)
=> \(3^x=405:\dfrac{5}{3}\)
=>\(3^x=405.\dfrac{3}{5}\)
=> \(3^x=81.3\)
=> \(3^x=243\)
=> \(3^x=3^5\)
=> x=5
Vậy:..............................
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
1) MK không chép lại đề nữa nhé!
3x = (32 ) -6 . (33 ) -5 . (34)8
3 x = 3-12.3-15 . 332
3x= 3-27.332
3x= 35
=>x= 5
Vậy x=5
3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
Vậy x = -7 hoặc x = 11
4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)
\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)
5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)
\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
\(\dfrac{7x}{81}=27\)
\(\Leftrightarrow7x=27.81\)
\(\Leftrightarrow x=\dfrac{2187}{7}\)
\(x.7:81=27\)
\(x.7=27.81\)
\(x.7=2187\)
\(x=\dfrac{2187}{7}\)