Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) 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, 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) 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.
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
Mình chỉ giải câu a thôi,mấy câu còn lại dễ.
a)Ta có:\(\dfrac{x}{27}=\dfrac{-3}{x}\)
=>\(x^2=-3\cdot27=-81\)(Nhân chéo)
Mà x2>0 với mọi x nên :
Không có giá trị nào thỏa mãn điều kiện của x
Tìm x biết :
a) \(\dfrac{x}{27}=-\dfrac{3}{x}\) \(\Rightarrow2x=-3.27\Rightarrow2x=-81\Rightarrow x=-40,5\)
b) \(-\dfrac{9}{x}=-\dfrac{x}{\dfrac{4}{49}}\Rightarrow2x=-9.\left(-\dfrac{4}{9}\right)\Rightarrow2x=4\Rightarrow x=2\)
c) \(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\) ( mk nghĩ bn chép sai đề bài câu này )
\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{8}{15}-\dfrac{7}{19}\)
\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{257}{285}\)
\(\Rightarrow\left[{}\begin{matrix}7x-\dfrac{5}{3}=-\dfrac{257}{285}\\7x-\dfrac{5}{3}=\dfrac{257}{285}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{218}{1995}\\x=\dfrac{244.}{665}\end{matrix}\right.\)
d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\)
\(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\)
\(\left|\dfrac{1}{23}x\right|=-\dfrac{43}{95}-\dfrac{18}{90}\)
\(\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{23}x=\dfrac{62}{95}\\\dfrac{1}{23}x=-\dfrac{62}{95}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\dfrac{1}{95}\\x=-15\dfrac{1}{95}\end{matrix}\right.\)
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
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