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NT
1
D
12 tháng 10 2017
a) \(5.2^{x+1}.2^{-2}-2^x=384\Leftrightarrow2^x\left(5.2^{-2}.2-1\right)=384\)\(\Leftrightarrow2^x.1,5=384\Leftrightarrow2^x=384:1,5=256=2^8\)
\(\Rightarrow x=8\)
b) \(3^{x+2}.5^y=45^x\Leftrightarrow3^{x+2}.5^y=3^{2x}.5^x\Leftrightarrow\frac{3^{2x}}{3^{x+2}}=\frac{5^y}{5^x}\)\(\Leftrightarrow3^{2x-x+2}=5^{y-x}\Leftrightarrow3^{x+2}=5^{y-x}\)
\(\Rightarrow x+2=y-x=0\Rightarrow x=y=-2\)
30 tháng 10 2022
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 \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)
16 tháng 9 2018
a, đầu bài sai hay sao đó bn
b,\(\dfrac{2}{3}\).3x+1 - 7.3x = -405
3x(\(\dfrac{2}{3}\).3 - 7) = -405
3x . (-5) = -405
3x = 81 mà 81 = 34 suy ra x = 4
15 tháng 9 2019
a) \(5.2^{x+1}.2^{-2}-2^x=384\)
\(\Leftrightarrow2^x.2.\frac{5}{4}-2^x=384\)
\(\Leftrightarrow2^x.\left(\frac{5}{2}-1\right)=384\)
\(\Leftrightarrow2^x.\frac{3}{2}=384\)
\(\Leftrightarrow2^x=256\)
\(\Leftrightarrow2^x=2^8\)
\(\Leftrightarrow x=8\)
c) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\)
\(\Leftrightarrow\left(x+1\right)^{x+3}-\left(x+1\right)^{x+1}=0\)
\(\Leftrightarrow\left(x+1\right)^{x+1}\left[\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{x+1}=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{0;-2\right\}\end{cases}}}\)
Vậy \(x\in\left\{0;-1;-2\right\}\)
2 tháng 10 2017
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. )
LH
30 tháng 9 2017
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 \(x=7\)
3 tháng 2 2017
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 ....
\(\Leftrightarrow\dfrac{5}{2}\cdot2\cdot2^x-2^x=384\)
\(\Leftrightarrow4\cdot2^x=384\)
\(\Leftrightarrow2^x=96\)
hay \(x\in\varnothing\)