a:Ta có: \(16-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
a:Ta có: \(16-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Tìm x
a, 2x.(x-5)-3(5-x)=0
b, x\(^2\)-16=0
Tìm x
a, 4x\(^2\)-1-x(2x+1)=0
b, x\(^2\)-7x+12=0
c, x\(^2\)-8x+6=0
Tìm x
a, x\(^2\)-8x+6=0
b,\(\dfrac{2x-1}{3}+\dfrac{x}{5}=\dfrac{3x}{10}\)
Tìm x
a, 2x.(x-3)+3(x-3)=0
b, x(3x-1)-5(1-3x)=0
tìm x
a) 2021-1+2022x(1-2021x)=0
b)(x+2)2-x2(x-6)-5=0
Tìm x
a) 3x(4x - 3) - 2x(5 - 6x) = 0
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
d) 3x (x + 1) - 5x(3 - x) + 6(x^2 + 2x + 3) = 0
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
b) 3x (x-1) + x - 1 = 0
c) 2(x+3) - x ² - 3x = 0
d) x(x - 2) + 3x - 6 = 0
e) 4x ² - 4x +1 = 0
f) x +5x ² = 0
g) x ² 2x -3 = 0
Tim x
a, x\(^2\)-7x+12=0
b, x(x-4)-3(4-x)=0
Tìm x
a) 4x(x-2) + x-2 = 0
b) (3x-1)^2 - 9 = 0
c) x^3 - 8 + (x-2)(x+1) = 0