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a) \(\frac{1}{2}+\frac{2}{3}:x=\frac{3}{4}\)
=> \(\frac{2}{3}:x=\frac{3}{4}-\frac{1}{2}\)
=> \(\frac{2}{3}:x=\frac{1}{4}\)
=> \(x=\frac{2}{3}:\frac{1}{4}=\frac{8}{3}\)
b) \(5,4-3\left|x-\frac{21}{10}\right|=0\)
=> \(3\left|x-\frac{21}{10}\right|=\frac{27}{5}\)
=> \(\left|x-\frac{21}{10}\right|=\frac{27}{5}:3=\frac{9}{5}\)
=> \(\left|x-\frac{21}{10}\right|=\frac{9}{5}\)
Trường hợp 1 : \(x-\frac{21}{10}=\frac{9}{5}\)
=> \(x=\frac{9}{5}+\frac{21}{10}=\frac{39}{10}\)
Trường hợp 2 : \(x-\frac{21}{10}=-\frac{9}{5}\)
=> \(x=-\frac{9}{5}+\frac{21}{10}=\frac{3}{10}\)
Vậy : ...
c) \(10\sqrt{x-5}=25\)
=> \(\sqrt{x-5}=\frac{5}{2}\)
=> \(\left(x-5\right)^2=\frac{25}{4}\)
Trường hợp 1 :
\(x-5=\frac{25}{4}\)=> \(x=\frac{25}{4}+5=\frac{45}{4}\)
Trường hợp 2 :
\(x-5=-\frac{25}{4}\)=> \(x=-\frac{25}{4}+5=-\frac{5}{4}\)(loại)
Vậy \(x=\frac{45}{4}\)
a) \(\frac{x-2}{5}=\frac{3}{8}\)
(x-2).8=5.3
(x-2).8=15
x-2=15:8
x-2=\(\frac{15}{8}\)
x=\(\frac{15}{8}+2\)
x=\(\frac{31}{8}\)
b)\(\frac{x-1}{x+5}=\frac{6}{7}\)
(x-1).7=(x+5).6
7x-7=6x+30
7x=6x+30+7
7x=6x+37
7x-6x=37
x=37
c)\(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2.25=6.24\)
\(x^2.25=144\)
\(x^2=144:25\)
\(x^2=\frac{144}{25}\)
\(x^2=\left(\frac{12}{5}\right)^2\)
\(x=\frac{12}{5}\)
a) \(\frac{16}{2^x}=1\Leftrightarrow2^x=16\Leftrightarrow2^x=2^4\Leftrightarrow x=4\)
b)\(5^{x+2}=625\Leftrightarrow5^{x+2}=5^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c)\(\frac{x+3}{8}=\frac{2}{x-3}\left(đk:x\ne3\right)\Leftrightarrow\left(x+3\right).\left(x-3\right)=2.8\Leftrightarrow x^2-9=16\Leftrightarrow x^2=25\Leftrightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
d)\(\frac{x^2}{6}=\frac{24}{25}\Leftrightarrow25x^2=24.6\Leftrightarrow\left(5x\right)^2=144\Leftrightarrow\orbr{\begin{cases}5x=12\\5x=-12\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{5}\\x=-\frac{12}{5}\end{cases}}\)
a) 16=2^x \(\Leftrightarrow\)x=4
b)5^x+2=5^4\(\Leftrightarrow\)x+2=4\(\Leftrightarrow\)x=2
k đi, mk làm tiếp cho
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
e)
\(\left(x+3\right)^3=\left(x+3\right)^5\)
\(\Rightarrow\)\(x+3=1;0\)
TH1: TH2
\(x+3=0\) \(x+3=1\)
\(x=-3\) \(x=-2\)
\(x\in\left\{-3;-2\right\}\)
a) \(x^2=1\)
\(\Rightarrow\orbr{\begin{cases}x^2=1^2\\x^2=\left(-1\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Vậy x = 1 hoặc x = -1
b) \(x^2=x\)
\(\Rightarrow x^2-x=0\)
\(\Rightarrow x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x = 0 hoặc x = -1
c) \(x^{10}=25.x^8\)
\(\Rightarrow x^{10}:x^8=25\)
\(\Rightarrow x^2=25\)
\(\Rightarrow\orbr{\begin{cases}x^2=5^2\\x^2=\left(-5^2\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
Vậy x = 5 hoặc x = -5
_Chúc bạn học tốt_
\(1^2=1\) = > x=1 \(\left(-1\right)^2=1\)=> x= -1
\(1^2=1\)=> x=1
\(x^{10}=25.x^8\Rightarrow x^{10}=5^2.x^8\Rightarrow x=5\)
Học tốt^^