Question

a, (x + 2)(x + 4) - x^2 = 24

b, (x + 5)(x -5) = x^2 + x

c, (2x + 3)(2x -3) = 4x(x -1)

d, (x + 2)(x^2 -2x + 4) = 9

Answer 1

a, \(\left(x+2\right)\left(x+4\right)-x^2=24\\ \Rightarrow x^2+6x+8-x^2=24\\ \Rightarrow6x+8=24\\ \Rightarrow6x=16\\ \Rightarrow x=\dfrac{8}{3}\)

b, \(\left(x+5\right)\left(x-5\right)=x^2+x\)

\(\Rightarrow x^2+x-\left(x+5\right)\left(x-5\right)=0\)

\(\Rightarrow x^2+x-x^2+25=0\\ \Rightarrow x+25=0\\ \Rightarrow x=-25\)

Answer 2

\(a,< =>x^2+4x+2x+8-x^2=24< =>6x+8=24< =>x=\dfrac{24-8}{6}=\dfrac{8}{3}\)

b,\(< =>x^2-25-x^2-x=0< =>-25-x=0< =>x=-25\)

c,\(< =>4x^2-9-4x^2+4x=0< =>4x-9=0< =>x=\dfrac{9}{4}\)

d,\(< =>x^3+2^3=9< =>x^3=1=>x=1\)

Answer 3

c,(2x+3)(2x-3)=4x(x-1)

⇔ 4x²-9=4x²-4x

⇔ 4x=9

\(\Leftrightarrow x=\dfrac{9}{4}\)

d, (x+2)(x²-2x+4)=9

⇔ x³+8=9

⇔ x³=1

⇔ x=1

Answer 4

a: Ta có: \(\left(x+2\right)\left(x+4\right)-x^2=24\)

\(\Leftrightarrow x^2+6x+8-x^2=24\)

\(\Leftrightarrow6x=16\)

hay \(x=\dfrac{8}{3}\)

b: Ta có: \(\left(x+5\right)\left(x-5\right)=x^2+x\)

\(\Leftrightarrow x^2+x-x^2+25=0\)

\(\Leftrightarrow x+25=0\)

hay x=-25

c: Ta có: \(\left(2x+3\right)\left(2x-3\right)=4x\left(x-1\right)\)

\(\Leftrightarrow4x^2-4x-4x^2+9=0\)

\(\Leftrightarrow-4x=-9\)

hay \(x=\dfrac{9}{4}\)

d: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)=9\)

\(\Leftrightarrow x^3+8=9\)

\(\Leftrightarrow x^3=1\)

hay x=1