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\(\text{a) }x^2-9x+20\)
\(=x^2-4x-5x+20\)
\(=\left(x^2-4x\right)-\left(5x-20\right)\)
\(=x\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(x-5\right)\)
\(\text{b) }x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=\left(x^2+4x\right)+\left(5x+20\right)\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(x+5\right)\)
\(\text{c) }x^2+x-20\)
\(=x^2+5x-4x-20\)
\(=\left(x^2+5x\right)-\left(4x+20\right)\)
\(=x\left(x+5\right)-4\left(x+5\right)\)
\(=\left(x+5\right)\left(x-4\right)\)
\(\text{d) }x^2-x-20\)
\(=x^2+4x-5x-20\)
\(=\left(x^2+4x\right)-\left(5x+20\right)\)
\(=x\left(x+4\right)-5\left(x+4\right)\)
\(=\left(x+4\right)\left(x-5\right)\)
a) Ta có: \(3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)
b) Ta có: \(5x-2=x+4\)
\(\Leftrightarrow5x-x=4+2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)
a) x2 + x - 12 = x2 - 3x + 4x - 12 = x(x - 3) + 4(x - 3) = (x - 3)(x + 4)
b) x2 - x - 12 = x2 + 3x - 4x - 12 = x(x + 3) - 4(x + 3) = (x + 3)(x - 4)
c) x2 - 9x + 20 = x2 - 4x - 5x + 20 = x(x - 4) - 5(x - 4) = (x - 4)(x - 5)
d) x2 + 9x + 20 = x2 + 4x + 5x + 20 = x(x + 4) + 5(x + 4) = (x + 4)(x + 5)
a,x^2+x-12=x^2-3x+4x-12
=x(x-3)+4(x-3)
=(x-3)*(x+4)
b) x2 - x - 12 = x2 + 3x - 4x - 12 = x(x + 3) - 4(x + 3) = (x + 3)(x - 4)
x^2 - 9x + 20 = 0
x^2 - 4x - 5x + 20 = 0
x(x - 4) - 5(x-4) = 0
(x - 5) . (x - 4) = 0
=> x = 5 hoặc x = 4.
x^2-9x+20=0
x^2-2*9/2*x+81/4-1/4=0
x^2-2*9/2*x+(9/2)^2=1/4
(x-9/2)^2=1/4
x-9/2=1/2
x=5
\(x^2-9x+20=x^2-4x-5x+20=x\left(x-4\right)-5\left(x-4\right)=\left(x-4\right)\left(x-5\right)\)