(4x - 1)^3 = 27
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Sửa đề: \(\dfrac{16}{15}\rightarrow\dfrac{16}{25}\)
Giải:
\(\left(x-\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{4}{5}\right)^2\\\left(x-\dfrac{3}{5}\right)^2=\left(\dfrac{-4}{5}\right)^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{5}=\dfrac{4}{5}\\x-\dfrac{3}{5}=\dfrac{-4}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{-1}{5}\end{matrix}\right.\)
\(\left(\dfrac{5}{3}-x\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(\dfrac{5}{3}-x\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\dfrac{5}{3}-x=\dfrac{1}{3}\)
\(x=\dfrac{5}{3}-\dfrac{1}{3}\)
\(x=\dfrac{4}{3}\)
\(\left(4x-1\right)^3=27\)
⇔ \(\left(4x-1\right)^3=3^3\)
⇒ \(4x-1=3\)
⇔ \(x=1\)
(4x-3)(2x-5) +(3-4x)(x-1)=0
(4x-3)(2x-5)-(4x-3)(x-1)=0
(4x-3)(2x-5-x+1)=0
(4x-3)(x-4)=0
4x-3=0 hoặc x-4=0
x=\(\frac{3}{4}\)hoặc x=4
\(2x-1-x^2=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\\ \left(1-3\right)^3-1=\left(-2\right)^3-1=-2-1=-3\\ \left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\\ \left(x+2\right)^3+1=\left(x+2+1\right)\left[\left(x+2\right)^2+\left(x+2\right)+1\right]\\ =\left(x+3\right)\left(x^2+4x+4+x+2+1\right)\\ =\left(x+3\right)\left(x^2+5x+7\right)\)
`4x(x-5)-(x-1) (4x-3)-5=0`
`=> 4x*x - 4x*5 - ( x*4x-3*x-1*4x+ 1*3) -5=0`
`=> 4x^2 - 20x-(4x^2 -3x-4x+3)-5=0`
`=> 4x^2 - 20x-4x^2+3x+4x-3-5=0`
`=>-13x-8=0`
`=> -13x=8`
`=> x=-8/13`
Vậy `x=-8/13`
`4x(x-5)-(x-1)(4x-3)-5 = 0`
`=> 4x^2 - 20x - (4x^2 -3x-4x+3)= 5`
`=> 4x^2 - 20x - 4x^2 + 3x + 4x -3 = 5`
`=> (4x^2 - 4x^2) - (20x - 3x - 4x) = 8`
`=> -13x = 8`
`=> x = -8/13`
\(\left(4x-1\right)^3=27\Leftrightarrow4x-1=3\Leftrightarrow4x=4\Leftrightarrow x=1\)
\(\left(4x-1\right)^3=27\\ \Leftrightarrow\left(4x-1\right)^3=3^3\\ \Leftrightarrow4x-1=3\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\)