16x^3-16x=0
a) \(4x^2+16x+3=0\)
\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)
b) \(7x^2+16x+2=1+3x^2\)
\(4x^2+16x+1=0\)
\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)
c) \(4x^2+20x+4=0\)
\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)
tìm x :
16x3-16x4+4x-8x2-1=0
\(16x^3-16x^4+4x-8x^2-1=0\)
<=> \(-16x^4-4x^2+16x^3+4x-4x^2-1=0\)
<=> \(-4x^2\left(4x+1\right)+4x\left(4x^2+1\right)-\left(4x^2+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(4x^2-4x+1\right)=0\)
<=> \(-\left(4x^2+1\right)\left(2x-1\right)^2=0\)
<=> \(2x-1=0\) (do 4x2 + 1 > 0 )
<=> \(x=\frac{1}{2}\)
x^3-16x=0
16x^4 -12x^3=0
Lời giải:
$16x^4-12x^3=0$
$\Leftrightarrow 4x^3(4x-3)=0$
$\Leftrightarrow x^3=0$ hoặc $4x-3=0$
$\Leftrightarrow x=0$ hoặc $x=\frac{3}{4}$
Tính f(15) biết: f(x) = x6 - 16x5 + 16x4 - 16x3 + 16x2 - 16x - 100
Giải phương trình
a, x4 + 5x3 + 12x2 +20x + 16 = 0
b, 16x4 - 24x3 + 16x2 - 6x + 1 = 0
Giải phương trình
4x^3-16x=0
\(4x^3-16x=0\)
\(\Leftrightarrow4x\cdot\left(x^2-4\right)=0\)
\(\Leftrightarrow4x\cdot\left(x-2\right)\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
\(\Leftrightarrow\)\(4x(x^{2}-4)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} 4x=0\\ x^{2}-4=0 \end{array} \right.\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=0\\ x=2,x=-2 \end{array} \right.\)
Ta có: \(4x^3-16x=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
x^3 - 1/16x = 0
x(x^2 - 1/16) =0
x(x+1/4)(x-1/4)=0
...............(tự làm tiếp)
Tìm x biết: 3-16x^2=0
`3-16x^2=0`
`<=>(\sqrt3)^2-(4x)^2=0`
`<=>(\sqrt3+4x)(\sqrt3-4x)=0`
`<=> [(\sqrt3=-4x),(\sqrt3=4x):}`
`<=> [(x=-\sqrt3/4),(x=\sqrt3/4):}`
Vậy `S={\pm \sqrt3/4}`.
Ta có: \(3-16x^2=0\)
\(\Leftrightarrow16x^2=3\)
\(\Leftrightarrow x^2=\dfrac{3}{16}\)
hay \(x\in\left\{\dfrac{\sqrt{3}}{4};-\dfrac{\sqrt{3}}{4}\right\}\)