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\(4^x+4^{x+3}=2160\)
\(4^x\left(1+4^3\right)=2160\)
\(4^x\cdot65=2160\)
\(4^x=2160\text{ : }65\)
\(4^x=33,2307692\)
\(\Rightarrow\text{ Đề sai}\)
a) (2x-1)\(^2\)+\(\left|2y-x\right|\)=0
Vì (2x-1)\(^2\)\(\ge\)0 với mọi x
\(\left|2y-x\right|\)\(\ge\)0 với mọi y
\(\Rightarrow\)\(\left\{\begin{matrix}2x-1=0\\2y-x=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{\begin{matrix}x=\frac{1}{2}\\2y-\frac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{4}\end{matrix}\right.\)
Vậy .....
b)\(\left|x-\frac{1}{3}\right|\)+\(\frac{4}{5}\)=\(\frac{14}{5}\)
\(\Rightarrow\)\(\left|x-\frac{1}{3}\right|\)=2
\(\Rightarrow\)\(\left[\begin{matrix}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{matrix}\right.\)\(\Rightarrow\)\(\left[\begin{matrix}x=\frac{7}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)
Vậy ....
a, => (-2)^x = -(2^2)^6.(2^3)^15
=> (-2)^x = -2^12.2^15 = -2^27 = (-2)^27
=> x = 27
b, Vì |x+5| và (3y-4)^2012 đều >= 0
=> |x+5|+(3y-4)^2012 >= 0
Dấu "=" xảy ra <=> x+5=0 và 3y-4=0 <=> x=-5 và y=4/3
c, => (2x-1)^2+|2y-x| = 12-5.2^2+8 = 0
Vì (2x-1)^2 và |2y-x| đều >= 0
=> (2x-1)^2+|2y-x| >= 0
Dấu "=" xảy ra <=> 2x-1=0 và 2y-x=0 <=> x=1/2 và y=1/4
Tk mk nha
b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)
\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)
\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)
\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)
\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)
\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)
\(2^x=\frac{2^0}{2^3}=2^{-3}\)
\(\Rightarrow x=-3\)
a) \(4^x+4^{x+3}=4160\)
\(\Rightarrow4^x+4^x.4^3=4160\)
\(\Rightarrow4^x.\left(1+4^3\right)=4160\)
\(\Rightarrow4^x.65=4160\)
\(\Rightarrow4^x=64\)
\(\Rightarrow4^x=4^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)
\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)
\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)
\(\Rightarrow2^x=\frac{1}{8}\)
\(\Rightarrow2^x=2^{-3}\)
\(\Rightarrow x=-3\)
Vậy \(x=-3\)
\(6.8^{x-1}+8^{x+1}=6.8^{19}+8^{21}\)
\(\Rightarrow\hept{\begin{cases}x-1=19\\x+1=21\end{cases}\Rightarrow\hept{\begin{cases}x=20\\x=20\end{cases}}}\)
\(5.2^x+3.2^{x+2}=5.2^5+3.2^7\)
\(\Rightarrow\hept{\begin{cases}x=5\\x+2=7\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=5\end{cases}}}\)
P/s:Kết quả thì chắc chắn đúng nhưng cách trình bày bài giải có thể sai,mong bn thông cảm =.=
a: \(\Leftrightarrow2^{2x}\cdot8+2^{2x}\cdot2+2^{2x}=176\)
\(\Leftrightarrow2^{2x}=16\)
=>2x=4
=>x=2
b: \(\Leftrightarrow2^{2x}\left(2^3+2-1\right)=144\)
\(\Leftrightarrow2^{2x}=16\)
=>2x=4
=>x=2
a: \(\Leftrightarrow2x-3=x\)
=>x=3
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)
=>2^x=1/8
=>x=-3
c: =>2x+7=-4
=>2x=-11
=>x=-11/2
d: =>(4x-3)^2*(4x-4)(4x-2)=0
hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5
a) \(81^{-2x}.27^x=9^5\)
\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=3^{10}\)
\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)
\(\Rightarrow3^{-5}=3^{10}\)
\(\Rightarrow-5x=10\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
b) \(2^x+2^{x+3}=144\)
\(\Rightarrow\left(1+2^3\right).2^x=144\)
\(\Rightarrow\left(1+8\right).2^x=144\)
\(\Rightarrow9.2^x=144\)
\(\Rightarrow2^x=16\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
c) \(2^{x-1}+5.2^{x-2}=7.32\)
\(\Rightarrow\left(2+5\right).2^{x-2}=244\)
\(\Rightarrow7.2^{x-2}=244\)
\(\Rightarrow2^{x-2}=32\)
\(\Rightarrow x-2=5\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(4^x+5.2^{2x+1}=176\)
\(4^x+5.4^x.2=176\)
\(4^x.\left(1+10\right)=176\)
\(4^x=176:11\)
\(4^x=16\)
\(4^x=4^2\)
\(=>x=2\)