Question

Tìm x 

a,(x+4)^2-1=0

b,(2x-3)^2=100

c,x^2+8x+16=0

d,4x^2-12x=-9

Answer 1

a.x = -3

b.\(\frac{13}{2}\)

c. (x+4)²

d. không bik

Answer 2

Bài làm:

a) \(\left(x+4\right)^2-1=0\)

\(\Leftrightarrow\left(x+4\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

b) \(\left(2x-3\right)^2=100\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=10\\2x-3=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=13\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{2}\\x=-\frac{7}{2}\end{cases}}\)

c) \(x^2+8x+16=0\)

\(\Leftrightarrow\left(x+4\right)^2=0\)

\(\Rightarrow x+4=0\)

\(\Rightarrow x=-4\)

d) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Rightarrow2x-3=0\)

\(\Rightarrow x=\frac{3}{2}\)

Answer 3

a)\(\left(x+4\right)^2-1=0\Leftrightarrow\left(x+4\right)^2=1\)

\(\Leftrightarrow\left(x+4\right)^2=1^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)

b) \(\left(2x-3\right)^2=100\)

\(\Leftrightarrow\left(2x-3\right)^2=10^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=10\\2x-3=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=13\\2x=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{2}\\x=-\frac{7}{2}\end{cases}}\)

c) \(x^2+8x+16=0\)

\(\Leftrightarrow\left(x+4\right)^2=0\)

\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)

d) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)