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\(\dfrac{1}{27}+\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{3}\\ \Leftrightarrow\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{3}-\dfrac{1}{27}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\\ \Leftrightarrow2x-\dfrac{1}{3}=\dfrac{2}{3}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
\(\dfrac{1}{27}\)+\(\left(2x-\dfrac{1}{3}\right)^3\)=\(\dfrac{1}{3}\)
⇔\(\left(2x-\dfrac{1}{3}\right)^3\)=\(\dfrac{1}{3}\)−\(\dfrac{1}{27}\)=\(\dfrac{8}{27}\)=\(\left(\dfrac{2}{3}\right)^3\)
⇔\(2x-\dfrac{1}{3}\)=\(\dfrac{2}{3}\)
⇔2\(x\)=1⇔\(x\)=\(\dfrac{1}{2}\)
Vậy \(x\)=\(\dfrac{1}{2}\)
\(\frac{x-1}{3}=\frac{2x-5}{7}\)
\(\Rightarrow\)7(x-1)=3(2x-5)
7x-7=6x-15
7x-6x=-15+7
x=-8
Vậy x=-8
\(\frac{3}{x}=\frac{4x}{27}\)
\(\Rightarrow3.27=4x.x\)
81=4x2
x2=81:4
bạn tự tính
\(a)x+\left(-5\right)=-14\)
\(\Leftrightarrow x=-14-\left(-5\right)\)
\(\Leftrightarrow x=-14+5\)
\(\Leftrightarrow x=-9\)
\(b)-x+7=-23\)
\(\Leftrightarrow-x=-23+ \left(-7\right)\)
\(\Leftrightarrow-x=-30\)
\(\Leftrightarrow x=30\)
\(c)112-x=\left(-3\right).\left(-15\right)\)
\(\Leftrightarrow112-x=45\)
\(\Leftrightarrow x=112-45\)
\(\Leftrightarrow x=67\)
\(d)\left(x-15\right)-27=5^5:5^3\)
\(\Leftrightarrow\left(x-15\right)-27=5^2\)
\(\Leftrightarrow\left(x-15\right)-27=25\)
\(\Leftrightarrow x-15=52\)
\(\Leftrightarrow x=67\)
\(e)\left(2x+1\right)^2=81\)
\(\Leftrightarrow\left(2x+1\right)^2=9^2\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
\(f)(x-5^3)=-27\)
\(f)(x-5^3)=-9^3\)
\(\Leftrightarrow x-5=-9\)
\(\Leftrightarrow x=-4\)
P/s: Bạn tự kết luận.
\(\left(2x-1\right)^2+\left(y-3\right)^8+\left(z-5\right)^{20}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\\z-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\\z=5\end{matrix}\right.\)
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
\(\left(2x-1\right)^3-27=0\)
\(\Rightarrow\left(2x-1\right)^3=27\)
\(\Rightarrow2x-1=3\Rightarrow2x=4\Rightarrow x=2\)
(2x-1)^3-3^3=0
2x-1=3
2x=3+1
2x=4
x=4:2
x=2