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\(\left(2x-2\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-2\right)-\left(2x-1\right)\right]\cdot\left[\left(2x-2\right)+\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-2-2x+1\right)\cdot\left(2x-2+2x-1\right)=0\)
\(\Leftrightarrow\left(2x-2x-2+1\right)\cdot\left(2x+2x-2-1\right)=0\)
\(\Leftrightarrow\left(-1\right)\cdot\left(4x-3\right)=0\)
\(\Leftrightarrow4x-3=0\div\left(-1\right)\)
\(\Leftrightarrow4x-3=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
\(\left(2x-2\right)^2-\left(2x-1\right)^2=0\)
\(\left[2x-2-\left(2x-1\right)\right]\left[2x-2+\left(2x-1\right)\right]=0\)
\(\left(2x-2-2x+1\right)\left(2x-2+2x-1\right)=0\)
\(-1\left(4x-3\right)=0\)
\(-4x+3=0\)
\(-4x=-3\)
\(x=\frac{3}{4}\)
a) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
b) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
b)x^3 - 6x^2 +11x-6=0
<=>x^3 - x^2 - 5x^2 +5x + 6x - 6=0
<=>x^2(x - 1) - 5x(x - 1) +6(x - 1)=0
<=>(x-1).(x^2 - 5x + 6)=0
<=>(x - 1).(x^2 - 2x - 3x + 6)=0
<=>(x - 1).[(x(x-2)-3(x-2)]=0
<=>(x-1)(x-2)(x-3)=0
<=>x-1=0hoac x-2=0 hoac x-3=0
<=>x=1hoac x=2 hoac x=3
a) (2x + 1)2 + 2(2x + 1)(5x - 1) + (5x - 1)2 = (2x + 1 + 5x - 1)2 = (7x)2 = 49x2
b) (x2 - 1)(x + 2) - (x - 1)(x2 + x + 1) = x3 + 2x2 - x - 2 - x3 + 1 = 2x2 - x - 1
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
Ta có : (x + 1)2 - (x + 2)(x - 2) = 0
<=> (x + 1)2 - (x2 - 22) = 0
<=> x2 + 2x + 1 - x2 + 4 = 0
<=> 2x + 5 = 0
=> 2x = -5
=> x = \(-\frac{5}{2}\)