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c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(\left(x+3\right)^3-\left(x+1\right)^3=56\)
⇔ \(x^3+3.x^2.3+3.x.3^2+3^3-\left(x^3+3.x^2+3.x+1\right)=56\)
⇔ \(x^3+9x^2+27x+27-x^2-3x^2-3x-1=56\)
⇔ \(6x^2+24x+26=56\)
⇔ \(6x\left(x-4\right)=30\)
...
\(x^3+9x^2+27x+3-x^3-3x^2-3x-1=56\)
=>\(6x^2+24x=54\)
=>\(x^2+4x=9\)
=>\(\left(x+2\right)^2=13\)
=>x+2=\(\sqrt{13}\) hoặc x+2=\(-\sqrt{13}\)
=>x=\(\sqrt{13}-2\) hoặc x=\(-\sqrt{13}-2\)
a. 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0
<=> 3(6x2-5x+1)-(18x2-29x+3)=0
<=> 14x=0
<=> x=0
b. (x - 3)(x - 5) + 3 (x - 1) = (x - 1)(x - 3)
<=> (x-3)(x-5-x+1)+3(x-1)=0
<=> -4(x-3)+3(x-1)=0
<=> -x+9=0
<=> x=9
c. (x - 1)(x - 2) - (x + 2)(x + 1) = 8
<=> x2-3x+2-(x2+3x+2)=8
<=> -6x=8
<=> \(x=\frac{-4}{3}\)
(x+1)(x2-x+1)-x(x-3)(x+3)=8
x3+1-x(x2-9)=8
x3+1-x3-9x=8
(x3-x3)+(1-8)-9x=0
-7-9x=0
-9x=-7
x=7/9