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a. \(5^{4-x}+1=26\)
\(\Leftrightarrow5^{4-x}=26-1=25\)
\(\Leftrightarrow5^{4-x}=5^2\)
\(\Leftrightarrow4-x=2\)
\(\Leftrightarrow x=2\)
b. \(\left(\frac{2}{x}+1\right)^{2x}=5^{2x}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}+1=5\\\frac{2}{x}+1=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{x}=4\\\frac{2}{x}=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{3}\end{cases}}\)
c. \(\left(1-2x\right)^4-\left(1-2x\right)^6=0\)
\(\Leftrightarrow\left(1-2x\right)^4.\left[1-\left(1-2x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(1-2x\right)^4=0\\1-\left(1-2x\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}1-2x=0\\\left(1-2x\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=1\\2x=0hoac2x=-2\end{cases}}\)
\(\Leftrightarrow x=\frac{1}{2},x=0,x=-1\)
\(\left(5-x\right)^2+\left(5-x\right)^5=0\)
\(\Leftrightarrow\left(5-x\right)^2.\left[1+\left(5-x\right)^3\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(5-x\right)^2=0\\1+\left(5-x\right)^3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=0\\\left(5-x\right)^3=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\5-x=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
Vậy \(x\in\left\{5,6\right\}\)
\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)
\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)
\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)
\(a,\left(x-1,2\right)^2=4\)
\(\Rightarrow x-1,2=2\)
\(\Rightarrow x=3,2\)
\(b,\left(x+1\right)^3=-125\)
\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow x+1=-5\Rightarrow x=-6\)
\(c,\left(x-5\right)^3=2^6\)
\(\Rightarrow\left(x-5\right)^3=4^3\)
\(\Rightarrow x-5=4\Rightarrow x=9\)
\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
a) \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)
=> \(\left(7-8x\right).5=6\left(-4+2x\right)\)
=> 35 - 40x = -24 + 12x
=> 35 + 24 = 12x + 40x
=> 52x = 59
=> x = 59/52
b) \(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)
=> (1 - 3: x)2 = 82
=> \(\orbr{\begin{cases}1-3:x=8\\1-3:x=-8\end{cases}}\)
=> \(\orbr{\begin{cases}3:x=-7\\3:x=9\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{7}\\x=\frac{1}{3}\end{cases}}\)
c) \(\left(x+1\right)\left(x-2\right)\ge0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-2\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-2\le0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\le2\end{cases}}\)
=> \(-1\le x\le2\)
h) \(\left(x+1\right)\left(x-3\right)\le0\)
=> \(\hept{\begin{cases}x+1\ge0\\x-3\le0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-3\ge0\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\x\le3\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}\) (loại)
= \(-1\le x\le3\)
... câu cuối bn lm dài dòng quá r ạ -)) cái dòng sra là bỏ luôn dấu GTTĐ của VT r ạ :))
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
a: =>|x-1/4|=3/4
=>x-1/4=3/4 hoặc x-1/4=-3/4
=>x=1 hoặc x=-1/2
b: \(\left|x+\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{9}{4}=\dfrac{2-9}{4}=-\dfrac{7}{4}\)(vô lý)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+5=1-x\\2x+5=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\x=-6\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{4}{3};-6\right\}\)
e: =>|3/2-x|=0
=>3/2-x=0
hay x=3/2
|2x-1|=1,5
TH(1)2x-1=1,5
2x =1,5+1
2x =2,5
x =2,5 :2
x =1,25
TH(2) 2x-1=-1,5
2x =-1,5+1
2x =-0,5
x =-0,5:2
x =-0,25
các câu khác cứ tương tự bạn nhé
b) \(7,5-\left|5-2x\right|=-4,5\)
\(\left|5-2x\right|=7,5+4,7\)
\(\left|5-2x\right|=12\)
th1 :\(5-2x=12\)
\(2x=5-12\)
\(2x=-7\)
\(x=-7:2\)
\(x=-3,5\)
th2: \(5-2x=-12\)
\(2x=5+12\)
\(2x=17\)
\(x=17:2\)
\(x=8,5\)
c) \(-3+\left|x\right|=-1\)
\(\left|x\right|=-1+3\)
\(\left|x\right|=2\)
th1: \(x=-2\)
th2 : \(x=2\)
d)\(\left|2\dfrac{1}{3}-x\right|=\dfrac{1}{6}\)
\(\left|\dfrac{7}{3}-x\right|=\dfrac{1}{6}\)
th1 :\(\dfrac{7}{3}-x=\dfrac{1}{6}\)
\(x=\dfrac{7}{3}-\dfrac{1}{2}\)
\(x=\dfrac{11}{6}\)
th2: \(\dfrac{7}{3}-x=\dfrac{-1}{6}\)
\(x=\dfrac{7}{3}+\dfrac{1}{6}\)
\(x=\dfrac{-5}{2}\)
e) \(\dfrac{5}{7}-\left|x+1\right|=\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{5}{7}-\dfrac{1}{14}\)
\(\left|x+1\right|=\dfrac{9}{14}\)
th1 :\(x+1=\dfrac{9}{14}\)
\(x=\dfrac{9}{14}-1\)
\(x=\dfrac{-5}{14}\)
th2 : \(x+1=\dfrac{-9}{14}\)
\(x=\dfrac{-9}{14}-1\)
\(x=\dfrac{-5}{14}\)
Làm ơn có ai giúp mih câu hỏi này với
b. (2:x+1)^2x=5^2x
\(\Rightarrow\)2:x+1=5
2:x =5-1
2:x =4
x =2:4
x =1/2