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a, \(x^3-2x=0\Leftrightarrow x\left(x^2-2\right)=0\Leftrightarrow x=;x=\pm\sqrt{2}\)
b, \(x^2\left(x-3\right)+12-4x=0\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=\pm2;x=3\)
c, \(\left(x-2\right)^2=x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-x^2-2x-4\right)=0\Leftrightarrow\left(x-2\right)\left(-x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x+6>0\right)=0\Leftrightarrow x=2\)
d, \(x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow x=2;x=3\)
e, \(x^3-4x^2+2x-1=0\Leftrightarrow x=3,5...\)
A) x2 - 4x + 4 = (x - 2)2 (hằng đẳng thức số 2)
Cm : x2 - 4x + 4 = x2 - 2x - 2x + 4 = x(x - 2) - 2(x - 2) = (x - 2)(x - 2) = (x - 2)2
b tương tự
a, 3x3 . 5x2 = 15x5
b, 2x . ( 3x2 + 2x ) = 6x3 + 4x2
c, -3xy . ( 2x + 5y ) = -6x2y +-15xy2
d, 3x2. ( 6 - x2 + 2x ) = 18x2 - 3x3 + 6x3
e, ( x + 2 ) . ( x + 3 ) = x2 + 5x + 6
i, ( x - 4 ) . ( x + 4 ) = x2 - 16
h, ( 1 - 2x ) . ( 3x + 2 ) = 2 - 6x2 - x
k, ( x - y ) . ( x + y ) = x2 - y2
t, ( 2x + 1 ) . ( 4x2 - 2x + 1 ) = 8x3 - 1
a, 3x3 . 5x2 = 15x5
b, 2x . ( 3x2 + 2x ) = 6x3 + 4x2
c, -3xy . ( 2x + 5y ) = -6x2y +-15xy2
d, 3x2. ( 6 - x2 + 2x ) = 18x2 - 3x3 + 6x3
e, ( x + 2 ) . ( x + 3 ) = x2 + 5x + 6
i, ( x - 4 ) . ( x + 4 ) = x2 - 16
h, ( 1 - 2x ) . ( 3x + 2 ) = 2 - 6x2 - x
k, ( x - y ) . ( x + y ) = x2 - y2
t, ( 2x + 1 ) . ( 4x2 - 2x + 1 ) = 8x3 - 1
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{3;-\dfrac{5}{2}\right\}\)
\(b,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow-\left(3x-2\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-11-2+5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{\dfrac{2}{3};\dfrac{13}{4}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-\dfrac{1}{2};3\right\}\)
\(d,\left(x-1\right)\left(2x-1\right)=x\left(1-x\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1+x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;\dfrac{1}{3}\right\}\)
\(e,0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;3\right\}\)
\(f,\left(x+2\right)\left(3-4x\right)=x^2+4x=4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-x^2-4x-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-2;\dfrac{1}{5}\right\}\)
\(g,\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\forall x\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\\x=-3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-3\right\}\)
\(h,2x\left(x-1\right)=x^2-1\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(S=\left\{1\right\}\)
b.
2x(x-1) = (x2-1)
2x(x-1)-(x-1)(x+1)=0
(X-1)(2x-x-1)=0
X-1=0 hoặc x-1=0
Vậy x=1
a) \(\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
Đến đây ok chưa?