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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)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
a) |2-x|=|x-5|
\(\Rightarrow2-x=x-5\)hoặc \(2-x=-\left(x-5\right)\)
\(\Rightarrow-x-x=-5-2\)hoặc \(-x+x=5-2\)(vô lý)
\(\Rightarrow x=3,5\)
a) (x - 2)3 + (3x - 1)(3x + 1) = (x + 1)3
<=> x3 - 6x2 + 12x - 8 + 9x2 - 1 - x3 - 3x2 - 3x - 1 = 0
<=> 9x - 10 = 0
<=> 9x = 10
<=> x = 10/9
Vậy S = {10/9}
b) (x + 1)(2x - 3) = (2x - 1)(x + 5)
<=> 2x2 - x - 3 - 2x2 - 9x + 5 = 0
<=> -10x + 2 = 0
<=> -10x = -2
<=> x = 1/5
Vậy S = {1/5}
c) (x - 1)3 - x(x + 1)2 = 5x(2 - x) - 11(x + 2)
<=> x3 - 3x2 + 3x - 1 - x3 - 2x2 - x = 10x - 5x2 - 11x - 22
<=> -5x2 + 2x + 5x2 + x + 22 - 1 = 0
<=> 3x = -21
<=> x = -7
Vậy S = {-7}
d) (x - 3)(x + 4) - 2(3x - 2) = (x - 4)2
<=> x2 + x - 12 - 6x + 4 - x2 + 8x - 16 = 0
<=> 3x - 24 = 0
<=> 3x = 24
<=> x = 8
Vậy S = {8}
e) x(x + 3)2 - 3x = (x + 2)3 + 1
<=> x3 + 6x2 + 9x - 3x = x3 + 6x2 + 12x + 8 + 1
<=> x3 + 6x2 + 6x - x3 - 6x2 - 12x = 9
<=> -6x = 9
<=> x = -3/2
Vậy S = {-3/2}
f) (x + 1)(x2 - x + 1) - 2x = x(x + 1)(x- 1)
<=> x3 + 1 - 2x = x3 - x
<=> x3 - 2x - x3 + x = -1
<=> -x = -1
<=> x = 1
Vậy S = {1}
a/ (x-1)2 - (x+1)2 = 3
<=> (x-1-x-1)(x-1+x+1) = 3
<=> - 2 . 2x = 3
<=> -4x = 3 <=> x = -3/4
KL: x = -3/4
b/ (x-2)(x+2) - (x-3) ^2 = 5x - 1
<=> x2 - 4 - x2 + 6x - 9 = 5x - 1
<=> 6x - 5x = -1 + 4 + 9
<=> x = 12
KL: x= 12
c/ x(x-2) - (x+3)2 - x = 4
<=> x2 - 2x - x2 - 6x - 9 - x = 4
<=> -9x = 4 + 9
<=> x = -13/9
KL: x= -13/9
d/ 6x(x-1) - 5(x-2)2 - x(x - 3) = 4
<=> 6x2 - 6x - 5(x2 - 4x + 4) - x2 - 3x = 4
<=> 6x2 - 6x - 5x2 + 20x - 20 -x2 - 3x = 4
<=> 11x = 24
<=> x = 24/11
KL: x = 24/11
a)
\(\left(x-1\right)^2-\left(x+1\right)^2=3\\ \Leftrightarrow-2.2x=3\\ \Leftrightarrow-4x=3\\ \Leftrightarrow x=-\dfrac{4}{3}\)
Vậy...
b)
\(\left(x-2\right)\left(x+2\right)-\left(x-3\right)^2=5x-1\\ \Leftrightarrow x^2-4-\left(x^2-6x+9\right)=5x-1\\ \Leftrightarrow x^2-4-x^2+6x-9=5x-1\\ \Leftrightarrow-13+6x=5x-1\\ \Leftrightarrow6x-5x=-1+13\\ \Leftrightarrow x=12\)
Vậy...