Tìm X: 4³. X =192 trarb lời ngay giúp mình nha mình sẽ tick cho
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(x-128+20):192 = 0
=> x - 128 + 20 = 0
=> x - 128 = -20
=> x = -20 + 128
=> x = 108
Vậy x = 108
\(\left(x-128+20\right):192=0\)
\(x-\left(128-20\right)=0\cdot192\)
\(x-108=0\)
\(x=0+108\)
\(x=108\)
\(x-\frac{2}{8}=\frac{-1}{4}\)
\(\Leftrightarrow x=\frac{-1}{4}+\frac{2}{8}\)
\(\Leftrightarrow x=\frac{-1}{4}+\frac{1}{4}\)
\(\Leftrightarrow x=0\)
Vậy ....
(x+3)(x-2)+3x=4(x+3/4)
x(x+3)-2(x+3)+3x=4x+3
x2+3x-2x-6+3x-4x=3
x2+(3x-2x+3x-4x)-6=3
x2-6=3
x2=9
x2=32
=>x=|3|
x=3 hoặc x=-3
\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)
a
\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
b
x^3 chứ: )
\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
\(27-\left(2x+1\right)=4\)
\(2x+1=27-4\)
\(2x+1=23\)
\(2x=23-1\)
\(2x=22\)
\(x=22:2\)
\(x=11\)
Vậy x = 11
#HQX
\(\Leftrightarrow2\sqrt{x-4}=5\left(x\ge4\right)\\ \Leftrightarrow\sqrt{x-4}=\dfrac{5}{2}\\ \Leftrightarrow x-4=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)

Ta có
4³ . X = 192
64 . X = 192
X = 192 : 64
X = 3
Đáp số: X = 3.
\(4^3\cdot x=192\)
=>64x=192
=>\(x=\frac{192}{64}=3\)