a) (x - 2)2 - (x - 3)(x + 3) = 17
⇔ (x2 - 4x + 4) - (x2 - 9) = 17
⇔ x2 - 4x + 4 - x2 + 9 = 17
⇔ 13 - 4x = 17
⇔ - 4x = -4
⇔ x = 1
b) 4(x - 3)2 - (2x - 1)(2x + 1) = 10
⇔ [2(x - 3)]2 - (4x2 - 1) = 10
⇔ (2x - 6)2 - 4x2 + 1 = 10
⇔ 4x2 - 24x + 36 - 4x2 + 1 = 10
⇔ - 24x = -27
⇔ x = \(\dfrac{9}{8}\)
c) (x - 4)2 - (x - 2)(x + 2) = 36
⇔ x2 - 8x + 16 - x2 + 4 = 36
⇔ -8x = 16
⇔ x = -2
d) (2x + 3)2 - (2x - 1)(2x + 1) = 10
⇔ 4x2 + 12x + 9 - 4x2 + 1 = 10
⇔ 12x = 0
⇔ x = 0
Tìm x ,biết :
a, \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=17\)
\(\Rightarrow x^2-4x+4-x^2+9=17\)
\(\Rightarrow-4x+13=17\)
\(\Rightarrow-4x=4\)
\(\Rightarrow x=-1\)
b,\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+1=10\)
\(\Rightarrow4x^2-24x+36-4x^2+1=10\)
⇒ \(-24x+37=10\)
\(\Rightarrow-24x=-27\)
\(\Rightarrow x=\dfrac{-27}{-24}=\dfrac{9}{8}\)
c,\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=36\)
⇒ \(x^2-8x+16-x^2+4=36\)
⇒ \(-8x+20=36\)
⇒ \(-8x=16\Rightarrow x=-2\)
d,\(\left(2x+3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Rightarrow4x^2+12x+9-4x^2+1=10\)
\(\Rightarrow12x+10=10\)
\(\Rightarrow12x=0\Rightarrow x=0\)
a.
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=17\\ \Leftrightarrow\left(x^2-4x+4\right)-\left(x^2-9\right)=17\\ \Leftrightarrow-4x+13=17\\ \Leftrightarrow x=-1\)
b.
\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\\ \Leftrightarrow\left(4x^2-24x+36\right)-\left(4x^2-1\right)=10\\ \Leftrightarrow-24x+37=10\\ \Leftrightarrow x=\dfrac{9}{8}\)
c.
\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=36\\ \Leftrightarrow\left(x^2-8x+16\right)-\left(x^2-4\right)=36\\ \Leftrightarrow-8x+20=36\\ \Leftrightarrow x=-2\)
d.
\(\left(2x+3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\\ \Leftrightarrow\left(4x^2+12x+9\right)-\left(4x^2-1\right)=10\\ \Leftrightarrow12x+10=10\\ \Leftrightarrow x=0\)
