Tìm x biết: 3-16x^2=0
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NQ
3
12 tháng 10 2017
1)\(x^2-x=x\left(x-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
12 tháng 9 2018
\(16x^3-12x^2+3x-7=0\)
\(\Leftrightarrow16x^3-16x^2-3x^2+3x+7x^2-7=0\)
\(\Leftrightarrow16x^2\left(x-1\right)-3x\left(x-1\right)+7\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow16x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\left(7x+7\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(16x^2-3x+7x+7\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(16x^2+4x+7\right)=0\)
<=> x - 1 = 0
<=> x = 1
T
12 tháng 9 2018
\(\Leftrightarrow16x^3-16x^2+4x^2-4x+7x-7=0\)
\(\Leftrightarrow16x^2.\left(x-1\right)+4x.\left(x-1\right)+7.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(16x^2+4x+7\right)=0\)
Ta có \(16x^2+4x+7=\left(4x\right)^2+2.4x.\frac{1}{2}+\frac{1}{4}+\frac{27}{4}\)
\(=\left(4x+\frac{1}{2}\right)^2+\frac{27}{4}>0\)
nên \(\left(x-1\right).\left(16x^2+4x+7\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Rightarrow x=1\)
KC
23 tháng 7 2017
\(16x^3-12x^2+3x-7=0\)
\(16x^3-16x^2+4x^2-4x+7x-7=0\)
\(16x^2\left(x-1\right)+4x\left(x-1\right)+7\left(x-1\right)=0\)
\(\left(x-1\right)\left(16x^2+4x+7\right)=0\)
Vì \(0< 16x^2+4x+7\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
30 tháng 10 2016
= 16x3 -16x2 + 4x2 - 4x + 7x - 7
= 16x2(x-1)+4x(x-1)+7(x-1)
=(x-1)(16x2+4x+7)
19 tháng 6 2021
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
19 tháng 6 2021
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
TA
26 tháng 4 2023
a) \(x^3-16x=0\)
⇔\(x\left(x^2-16\right)=0\)
⇒\(x=0\) hoặc \(x^2-16=0\)
\(TH_1:x=0\)
\(TH_2:x^2-16=0\) ⇔ \(x^2=16\) ⇔ \(x=\pm4\)
Vậy \(x\in\left\{0;\pm4\right\}\)
b) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
⇒ \(2x+1=x-1\)
⇒ \(2x+2=x\)
⇒ \(2\left(x+1\right)=x\) ⇒ x = -2
Vậy x = -2
30 tháng 7 2021
a) Ta có: \(x^4-16x^2=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b) Ta có: \(x^8+36x^4=0\)
\(\Leftrightarrow x^4\left(x^4+36\right)=0\)
\(\Leftrightarrow x^4=0\)
hay x=0
c) Ta có: \(\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
d) Ta có: \(5\left(x-2\right)-x^2+4=0\)
\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
NT
2
NC
30 tháng 8 2020
a) Ta có: \(x^3-3x^2-16x+48=0\)
\(\Leftrightarrow x^2\left(x-3\right)-16\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm4\end{cases}}\)
b) Ta có: \(10x^2-33x-7=0\)
\(\Leftrightarrow\left(10x^2-35x\right)+\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\5x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{5}\end{cases}}\)
30 tháng 8 2020
x3 - 3x2 - 16x + 48 = 0
<=> ( x3 - 3x2 ) - ( 16x - 48 ) = 0
<=> x2( x - 3 ) - 16( x - 3 ) = 0
<=> ( x - 3 )( x2 - 16 ) = 0
<=> ( x - 3 )( x - 4 )( x + 4 ) = 0
<=> x = 3 hoặc x = 4 hoặc x = -4
10x2 - 33x - 7 = 0
<=> 10x2 + 2x - 35x - 7 = 0
<=> ( 10x2 + 2x ) - ( 35x + 7 ) = 0
<=> 2x( 5x + 1 ) - 7( 5x + 1 ) = 0
<=> ( 5x + 1 )( 2x - 7 ) = 0
<=> \(\orbr{\begin{cases}5x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{5}\\x=\frac{7}{2}\end{cases}}\)
`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 \(x\in\left\{\dfrac{\sqrt{3}}{4};-\dfrac{\sqrt{3}}{4}\right\}\)