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\(1,\left(2-x\right)^2-9=0\)
\(\Leftrightarrow\left(2-x-9\right)\left(2-x+9\right)=0\)
\(\Leftrightarrow\left(-7-x\right)\left(11-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\11-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
\(b,\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)=15-9\left(x+1\right)^2\)\(\Leftrightarrow x^3-9x^2+27x-27-x^3-27=15-9x^2-18x-9\)\(\Leftrightarrow x^3-9x^2+27x-x^3+9x^2+18x=15+27+27\)\(\Leftrightarrow45x=69\Rightarrow x=\dfrac{23}{15}\)
1. \(\left(2-x\right)^2-9=0\)
\(\left(2-x\right)^2=9\)
\(\left(2-x\right)^2=3^2\)
\(2-x=3\)
\(-x=-1\Rightarrow x=1\)
a) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c)\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
d) \(x^3+x=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
e)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)
\(\Leftrightarrow3x^2-6x+3+9x-21=0\)
\(\Leftrightarrow3x^2+3x-18=0\)
\(\Leftrightarrow3\left(x^2+x-6\right)=0\)
\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy \(S=\left\{2;-3\right\}\)
=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84
=> (72 - 84 ) - (20x + 36x ) = (30x - 6x ) - 240 + 84
=> -12 - 56x = 24x - 156
=> -12 + 156 = 24x + 56x
=> 144 = 80x
=> x = 144 : 80
=> x = 9/5
Chúng ta sẽ sử dụng hằng đẳng thức em nhé :)
a. \(x^3+1-x^3+3x=15\Leftrightarrow3x=14\Leftrightarrow x=\frac{14}{3}\)
b. \(x^2+x-6+3x=4x+3\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
c. \(x^3+2x^2-5x-10+5x=2x^2+17\Leftrightarrow x^3=27\Leftrightarrow x=3\)
\(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+9=0\\x-3=0\\x+21=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-9\\x=3\\x=-21\end{cases}}\)
Vậy ...................
vì 15(x+9)(x-3)(x+21)=0 nên một trong ba số x+9;x-3,x+21 sẽ bằng 0 CÓ 3TH:TH1: x+9=0 x=0-9=-9 TH2: x-3=0 x=0+3=3 TH3:x+21=0 x=0-21=-21 Vậy x =-9;3;-21. Nếu bạn ko tin thì cứ bấm máy tính nhé