\(36x^2-49=0\Leftrightarrow x^2=\frac{49}{36}\Leftrightarrow x=+-\frac{7}{6}\)
Vậy...
\(x^3-16x=0\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\Leftrightarrow x=+-4\end{cases}}\)
Vậy...
Tìm x:
a) x2+9x=0
b) (x+4)2-16=0
c) x3-16x=0
d) x2-10x+25=0
bài 7 tìm x
3,(x+1)=(x+1) 4,x(2x-3)-2(3-2x)=0
6,
Tìm x biết.
a) 4x^2 - 49 = 0 b) x^2 + 36 = 12x
c) 1/16x^2 - x + 4 = 0 d) x^3 -3√3x2 + 9x - 3√3 = 0
e) (x - 2)^2 - 16 = 0 f) x^2 - 5x - 14 = 0
g) 8x(x - 3) + x - 3 = 0
Tìm x
a, 3/4x*(x2-9)=0
b, x3-16x=0
c, (x-1)(x+2)-x-2=0
d, 3x3-27x=0
e, x2(x+1)+2x(x+1)=0
f, x(2x-3)-2(3-2x)=0
Bài 1: Phân tích đa thức thành nhân tử: a) 4y3 + 16y2 + 16y b) 8x2-48x+6xy-36y c) 8x2-48x-6xy+36y d) a2 –2ab+b2 –4 e) 4–x2 –4xy–4y2 f) 8a2 –16a+8ax–16x g) 16–4x2 +8xy–4y2 h) –4x2 –16xy–16y2 Bài 2: Tìm x, biết: a) x3 – 6x2 + 9x = 0 b) 5x(x–6)+3x–18=0 c) 5x(x – 6) – 18 + 3x = 0 d) 5x(x – 6) – 3x + 18 = 0 e) (2x – 3)2 = (5 – x)2 f) (2x + 1)2 = (3x – 2)2 g) 16(2x–3)=-25x2 (3–2x)
tìm x
a)16x2 - (4x -5)2 =15
b)(3x +4)2 - (3x -1) (3x+1)= 49
c)(3x -1)2 -(3x -2)2=0
d)(2x +1)2 - (x-1)2 =0
Giải hộ mình với.
Tìm x, biết:
a) 16x2 - (4x - 5)2 = 15
b) (2x + 3)2 - 4 (x - 1) (x +1) =49
c) (3x - 1)2 - (3x - 2)2 = 0
Tìm x:(5x-4)²-16x²=0