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a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
Bài 1.
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left[x^2+2x+7+2\left(x+2\right)-5\right]=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\)
Ta có:
\(x^2+4x+6\)
\(=x^2+2.x.2+4+2\)
\(=\left(x+2\right)^2+2\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+2\right)^2+2\ge2\) với mọi x
\(\Rightarrow x^2+4x+6\) vô nghiệm
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
b) \(3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(2\left(x+3\right)x^2-3x=0\)
\(\Rightarrow x\left[2\left(x+3\right)x-3\right]=0\)
\(\Rightarrow x\left(2x^2+6x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x^2+6x-3=0\end{matrix}\right.\)
Ta có:
\(2x^2+6x-3\)
\(=2\left(x^2+3x-\dfrac{3}{2}\right)\)
\(=2\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}-\dfrac{3}{2}\right)\)
\(=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\)
Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\ge-\dfrac{15}{2}\) với mọi x
\(\Rightarrow2x^2+6x-3\) vô nghiệm
\(\Rightarrow x=0\)
a,\(3x\left(x-1\right)+x-1=0\)
\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(3x+1\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
c,\(\left(2x-1\right)^2-25=0\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.