(4x-3)^3=-125
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\(\left(4x-15\right)^3=125\)
\(\left(4x-15\right)^3=5^3\)
\(4x-15=5\)
\(4x=20\)
\(x=5\)
\(\left(4x-15\right)^3=125\)
\(\left(4x-15\right)^3=5^3\)
\(=>4x-15=5\)
\(4x=5+15\)
\(4x=20\)
\(x=20:4\)
\(x=5\)
(4x - 3)x =-125
x(4x−3)=−125 4x2−3x=−125 4x2−3x+125=0 x=\(\dfrac{3+\sqrt{1991\iota}}{8}\);\(\dfrac{3-\sqrt{1991\iota}}{8}\)\(\left(3.4^x-3\right).\left(x^3-125\right)=0\\ \rightarrow\left[{}\begin{matrix}3.4^x-3=0\\x^3-125=0\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}3.4^x=3\\x^3=125\end{matrix}\right.\)
\(\rightarrow\left[{}\begin{matrix}4^x=1=4^0\\x^3=5^3\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(\left(3\cdot4^x-3\right)\left(x^3-125\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3\cdot4^x-3=0\\x^3-125=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3\left(4^x-1\right)=0\\x^3-5^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4^x=1\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
(4x-3)x=-125
4x.x-3x=-125
5x-3x=-125
2x=-125
x=-125:2
x=-62,5
\(\left(3-4x\right)^3=-125\)
\(\Rightarrow\left(3-4x\right)^3=\left(-5\right)^3\)
\(\Rightarrow3-4x=-5\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
Vậy x = 2
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\(420-\left(4x+3\right)=125+52=177.\)
\(4x+3=420-177=243\)
\(4x=243-3=240\)
\(x=60\)
\(4x^3+15=47\)
\(\Leftrightarrow4x^3=32\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x=2\)
\(4\cdot2^x-3=125\)
\(\Leftrightarrow4\cdot2^x=128\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
4x3 + 15 = 47
=> 4x3 = 47 - 15 = 32
=> x3 = 32 : 4 = 8 = 23
=> x = 2
4.2x - 3 = 125
=> 4.2x = 125 + 3 = 128
=> 2x = 128 : 4 = 32 = 25
=>x = 5
(4x-3)^3= -125
(4x-3)^3=(-5)^3
4x-3=-5
4x=-2
x=-1/2
\(\left(4x-3\right)^3=-125\)
\(\left(4x-3\right)^3=\left(-5\right)^3\)
\(4x-3=-5\)
\(x=-\frac{1}{2}\)