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a) \(\left(x+5\right)^3=64\)
\(\Leftrightarrow\left(x+5\right)^3=4^3\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
Vậy x = - 1
b) \(x:\left(-\frac{3}{5}\right)^2=-\frac{3}{5}\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^2.\left(-\frac{3}{5}\right)\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^3\)
\(\Leftrightarrow x=-0,216\)
Vậy x = - 0, 216
c) \(\left(\frac{4}{7}\right)^4.x=\left(\frac{4}{7}\right)^6\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^6:\left(\frac{4}{7}\right)^4\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^2\)
\(\Leftrightarrow\text{x}=\frac{16}{49}\)
Vậy x = 16/49
d) \(\left(-\frac{1}{3}\right)^3x=\frac{1}{81}\)
\(\Leftrightarrow-\frac{1}{27}x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = - 1/3
a. x = {3;-3}
b. x thuộc rỗng
c. x2-4=0
x2 = 4
x={2;-2}
d. x2+1=82
x2 =83
x thuộc rỗng
e. (2x)2=6
x thuộc rỗng
f. (x-1)2=9
TH1: x-1=3=>x=4
TH2: x-1=-3=>x=-2
Vậy x={4;-2}
g.(2x+3)2=25
TH1: 2x+3=5=> x=1
Th2: 2x+3=-5=>x=-4
VẬY X={1;-4}
a, x^2= 9
=>\(\sqrt{9}=3\)
b,\(x^2=5=>x=\sqrt{5}\)
c, x^2-4=0
=>x^2=4
=>x=2
d, x^2+1=82
=>x^2=81 =>\(\sqrt{81}=9\)
3, 2x^2=6
=>x= \(\sqrt{6}\)
f, {x-1} ^2=9
=> x-1=3
=>x=2
g{ 2x+3}^2=25
=> 2x+3=5
=>2x=2
=>x=1
\(x+1+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow x+1+x+2+x+3+...+x+100=5750\)
\(\Rightarrow100x+1+2+3+...+100=5750\)
\(\Rightarrow100x+\left[\left(\dfrac{100-1}{1}+1\right):2\right]\left(100+1\right)=5750\)
\(\Rightarrow100x+5050=5750\)
\(\Rightarrow100x=700\Rightarrow x=7\)
\(25-\left(30+x\right)=x-\left(123-67\right)\)
\(\Rightarrow25-30+x=x-123+67\)
\(\Rightarrow-5+x=x-56\)
\(\Rightarrow x\in\varnothing\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\Rightarrow x=5\\\left(x-5\right)^2-1=0\Rightarrow\left(x-5\right)^2=1\Rightarrow x=6;4\end{matrix}\right.\)
\(\left(x^2+1\right)\left(x-3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+1>0\Rightarrow x^2>-1\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+1< 0\Rightarrow x^2< -1\\x-3>0\Rightarrow x>3\end{matrix}\right.\end{matrix}\right.\)
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