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21 tháng 7 2021
a, \(\left(x^2-9\right)^2-\left(x-3\right)\left(x+3\right)\left(x^2+9\right)=\left(x^2-9\right)^2-\left(x^2-9\right)\left(x^2+9\right)\)
\(=x^4-18x^2+81-x^4+81=-18x^2+162\)
b, \(\left(x^2+x-3\right)\left(x^2-x+3\right)=\left[x^4-\left(x-3\right)^2\right]\)
\(=x^4-x^2+6x-9\)
BB
4
AD
31 tháng 7 2021
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)=27\)
\(\Leftrightarrow x^3+27-x\left(x^2-1\right)=27\)
\(\Leftrightarrow x^3+27-x^3+x=27\)
\(\Leftrightarrow27+x=27\)
\(\Leftrightarrow x=0\)
#H
19 tháng 7 2016
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)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
DN
0
17 tháng 7 2017
Ta có : 3(2x - 1)2 \(\ge0\forall x\)
7(3y + 5)2 \(\ge0\forall x\)
Mà : 3(2x - 1)2 + 7(3y + 5)2 = 0
Nên : 3(2x - 1)2 = 7(3y + 5)2 = 0
\(\Leftrightarrow\hept{\begin{cases}3\left(2x-1\right)^2=0\\7\left(3y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^2=0\\\left(3y+1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)=0\\\left(3y+1\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x=1\\3y=-1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{1}{3}\end{cases}}\)
10 tháng 7 2023
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)
\(< =>\left(x^2-9\right)^2-\left[3\left(x-3\right)\right]^2=0\)
\(< =>\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)
\(< =>\left(x^2-9+3x-9\right)\left(x^2-9-3x+9\right)=0\)
\(< =>\left(x^2+3x-18\right)\left(x^2-3x\right)=0\)
\(=>\left[{}\begin{matrix}x^2+3x-18=0\\x^2-3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}\left(x+6\right)\left(x-3\right)=0\\x\left(x-3\right)=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=-6\\x=3\\x=0\end{matrix}\right.\)
Nhanh đấy tốc độ đấy =))