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\(5^{x+2}=125\Rightarrow5^x.5^2=125\Rightarrow5^x=125:5^2\Rightarrow5^x=5\Rightarrow5^x=5^1\Rightarrow x=1\)
\(\left|x+3\right|-\frac{1}{3}=1\)
\(\Rightarrow\left|x+3\right|=\frac{1}{3}\)
\(\Rightarrow\begin{cases}x+3=\frac{1}{3}\\x+3=-\frac{1}{3}\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\)
a) 5x+2 = 125
=> 5x+2 = 52
=> x + 2 = 2
=> x = 2 - 2
=> x = 0
b) \(\left|x+3\right|-\frac{2}{3}=1\)
=> \(\left|x+3\right|=1+\frac{2}{3}\)
=> \(\left|x+3\right|=\frac{5}{3}\)
=> \(\left[\begin{array}{nghiempt}x+3=\frac{5}{3}\\x+3=-\frac{5}{3}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{4}{3}\\x=-\frac{14}{3}\end{array}\right.\)
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\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)
\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)
\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)
\(\Rightarrow x+3=376\)
\(\Rightarrow x=373\)
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b)
(x-1/5)3=8/125
(x-1/5)3=(2/5)3
=>x-1/5=2/5
x=2/5+1/5
x=3/5
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a) 7/2 - x = 0,75
x = 7/2 - 0,75
x = 11/4
b) 2 |x - 1/3 | + 1/ 5 = 1
2 | x - 1/3 | = 1 - 1,5
2| x - 1/3 | = - 0,5
x - 1/3 = -0,5 : 2 => x = -0,25 = -1/4
x - 1/3 = 0,5 : 2 => x = 0,25 = 1/4
Mình nghĩ câu c bạn nên suy nghĩ nhé , chúc bạn may mắn
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A)
\(x^3=-125\)
\(x=-5\)
Vậy x = -5
B)
| x | = x + 1
\(\Leftrightarrow\orbr{\begin{cases}x=x+1\\x=-\left(x+1\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-x=1\\x=-x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0=1\left(VN\right)\\2x=-1\end{cases}}\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy x = -1/2
C) \(x^2=x\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x1 = 0 ; x2 = 1
A) x3 = -125
x = -125 : 3
x \(\approx\)-41,5
B) | x | = x + 1
\(\orbr{\begin{cases}x=x\\x=-x\end{cases}}\)
=>\(\orbr{\begin{cases}x>0\\x< 0\end{cases}}\)
C) x2 = x
x \(\varepsilon\)N
~ Hok tốt ~
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a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
\(\left(x+1\right)^3=-125\)
\(\left(x+1\right)^3=-5^3\)
\(\Leftrightarrow x+1=-5\)
\(\Rightarrow x=-6\)
\(\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=-5^3\)
\(\Leftrightarrow\left(x+1\right)=-5\)
\(\Leftrightarrow x=-6\)
Chúc em học tốt ~~~