DA
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
22 tháng 5 2022
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
17 tháng 8 2019
e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
NH
7 tháng 7 2019
\(a,\left(x+1\right)^2=81\)
\(\left(x+1\right)^2=9^2\) Hoặc \(\left(x+1\right)^2=\left(-9\right)^2\)
\(\left(x+1\right)=9\) \(x+1=-9\)
\(x=8\) \(x=-10\)
b,\(\left(x+5\right)^{^{ }3}=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(x+5=-4\)
=> \(x=-9\)
c,\(\left(2x-3\right)^2=9\)
=>\(\left(2x-3\right)^2=3^2\)Hoặc \(\left(2x-3\right)^2=\left(-3\right)^2\)
\(2x-3=3\) \(2x-3=-3\)
\(2x=6\) \(2x=0\)
=> \(\hept{\begin{cases}x=3\\x=0\end{cases}}\)
d, \(\left(4x+1\right)^3=27\)
\(\left(4x+1\right)^{^{ }3}=3^3\)
\(4x+1=3\)
\(4x=2\)
\(x=\frac{1}{2}\)
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{8^6}{4}=\frac{\left(2^3\right)^6}{2^2}=\frac{2^{18}}{2^2}=2^{16}\)
NH
7 tháng 7 2019
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{15}+4^{10}}{4^6+4^{11}}=\frac{4^{10}.4^5+4^{10}}{4^6+4^6.4^5}=\frac{4^{10}.\left(4^5+1\right)}{4^6.\left(4^5+1\right)}=\frac{4^{10}}{4^6}=4^4=256\)
phần D trên mk làm sai xin lỗi nha
QD
27 tháng 7 2017
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 =.=
DG
1 tháng 8 2017
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
6 tháng 9 2019
a) \(8< 2^x\le2^9.2^{-5}\)
\(\Leftrightarrow2^3< x\le2^{9-5}\)
\(\Leftrightarrow2^3< 2^x\le2^4\)
\(\Leftrightarrow3< x\le4\Leftrightarrow x=4\)
b) \(27< 81^3:3^x< 243\)
\(\Leftrightarrow3^2< \left(3^4\right)^3:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12}:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12-x}< 3^5\)
\(\Leftrightarrow2< 12-x< 5\)
\(\Leftrightarrow\hept{\begin{cases}x=8\\x=9\end{cases}}\)
26 tháng 7 2016
a,\(8< 2^x\le2^9.2^{-5}\)
\(2^3< 2^x\le2^4\)
\(\Rightarrow x=4\)
b, \(27< 81^3.3^x< 243\)
\(3^3< 3^{12-x}< 3^5\)
\(\Rightarrow3< 12-x< 5\)
12-x=4
x=8
c,\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^2\)
\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)
\(\Rightarrow x>5\)
x=6;7;8........
28 tháng 3 2018
a/ \(\dfrac{x}{9}=\dfrac{16}{x}\)
\(\Leftrightarrow x^2=9.16\)
\(\Leftrightarrow x^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right.\)
Vậy ...
b/ \(x^3+27=0\)
\(\Leftrightarrow x^3=-27\)
\(\Leftrightarrow x^3=\left(-3\right)^3\)
\(\Leftrightarrow x=-3\)
Vậy ...
c/ \(\left|x\left(x^2-\dfrac{5}{4}\right)=x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{5}{4}\right)=x\\x\left(x^2-\dfrac{5}{4}\right)=-x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{5}{4}x=x\\x^3-\dfrac{5}{4}x=-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-\left(\dfrac{5}{4}x+x\right)=0\\x^3-\left(\dfrac{5}{4}x-x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^3-\dfrac{9}{4}x=0\\x^3-\dfrac{1}{4}x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x^2-\dfrac{9}{4}\right)=0\\x\left(x^2-\dfrac{1}{4}\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{4}=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
c/ Với mọi x ta có :
\(\left|x-5\right|=\left|5-x\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|x-5\right|=\left|x+3\right|+\left|5-x\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|\left(x+3\right)+\left(5-x\right)\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge\left|8\right|\)
\(\Leftrightarrow\left|x+3\right|+\left|5-x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(x+3\right)\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3\\5\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3\\5\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-3\le x\le5\\x\in\varnothing\end{matrix}\right.\)
Vậy ...
\(8\left(x+2\right)^5=27\left(x+2\right)^8\)
\(2^3\left(x+2\right)^5=3^3\left(x+2\right)^8\)
\(3^3\left(x+2\right)^3=2^3\)
\(\left(x+2\right)^3=(\frac{2}{3})^3\)
\(x+2=\frac{2}{3}\)
\(x=\frac{2}{3}-2=-\frac{4}{3}\)