Question

Answer 1

tìm x à bạn

\(a,< =>\left(x-2\right)\left(x+2\right)-8\left(x-2\right)=0< =>\left(x-2\right)\left(x+2-8\right)=0\)

\(< =>\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)\(=>\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

b,\(< =>\left(x-2\right)^2-9\left(x-2\right)=0< =>\left(x-2\right)\left(x-2-9\right)=0\)

\(=>\left[{}\begin{matrix}x-2=0\\x-11=0\end{matrix}\right.\)\(=>\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

c,\(< =>\left(2x-3\right)^2-\left(5-x\right)^2=0\)

\(< =>\left(x+2\right)\left(3x-8\right)=0< =>\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)

Answer 2

a. x² - 4 = 8(x - 2)

⇔ (x - 2)² = 8(x - 2)

⇔ x - 2 = 8 

⇔ x = 8 + 2 = 10

b. x² - 4x + 4 = 9(x - 2)

⇔ (x - 2)² = 9(x - 2)

⇔ x - 2 = 9

⇔ x = 9 + 2 = 11

c. 4x² - 12x + 9 = (5 - x)²

⇔ (2x - 3)² = (5 - x)²

⇔ 2x - 3 = 5 - x

⇔ 2x + x = 5 + 3

⇔ 3x = 8

⇔ x = \(\dfrac{8}{3}\)

Answer 3

a: Ta có: \(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-8\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

c: Ta có: \(4x^2-12x+9=\left(5-x\right)^2\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(2x-3-x+5\right)\left(2x-3+x-5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)