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-4x2 + 4x - 3 < 0
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8 tháng 8 2023
a: =>2^x*4-2^x*3=32
=>2^x=32
=>x=5
b: =>(4x-3)^2-(4x-3)=0
=>(4x-3)(4x-3-1)=0
=>(4x-3)(4x-4)=0
=>x=3/4 hoặc x=1
c: =>7^2x+7^2x*7^3=344
=>7^2x=1
=>2x=0
=>x=0
d: =>(7x-3)^2012-(7x-3)^2010=0
=>(7x-3)^2010*[(7x-3)^2-1]=0
=>(7x-3)^2010*(7x-4)(7x-2)=0
=>x=2/7; x=4/7; x=3/7
e: =>(4x^2-3)^3=-8
=>4x^2-3=-2
=>4x^2=1
=>x^2=1/4
=>x=1/2 hoặc x=-1/2
T
8 tháng 8 2023
a) 2x(22 - 3) = 32
2x.1=25
=> x = 5
b) (4x - 3)2 = 4x -3
=> (4x - 3)2 - (4x - 3) = 0
(4x-3)[(4x - 3) - 1] = 0
(4x-3)(4x - 4)=0
\(\Rightarrow\left[{}\begin{matrix}4x-3=0\\4x-4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=1\end{matrix}\right.\)
c) 72x + 72x+3 = 344
=> 72x(1 + 73) =344
72x . 344 = 344
=> 2x = 0 => x = 0
d) (7x - 3)2012 = (3 - 7x)2010
=> (7x - 3)2012 - (7x - 3)2010 = 0
(7x - 3)2010 [(7x - 3)2 - 1] = 0
\(\Rightarrow\left[{}\begin{matrix}7x-3=0\\\left(7x-3\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\7x=4\\7x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{7}\\x=\dfrac{4}{7}\\x=\dfrac{2}{7}\end{matrix}\right.\)
e) (4x2 - 3)3 + 8 = 0
(4x2 - 3)3 = (-2)3
=> 4x2 - 3 = -2
4x2 = 1
x2 = 1/4
=> \(x=\pm\dfrac{1}{2}\)
13 tháng 2 2022
\(a,\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ b,4x^2-1=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(c,x^2-4x+3=0\\ \Leftrightarrow x^2-3x-x+3=0\\ \Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,9x^2-6x+1=0\\ \Leftrightarrow\left(3x-1\right)^2=0\\ \Leftrightarrow3x-1=0\\ \Leftrightarrow x=\dfrac{1}{3}\)
LB
1
AH
Akai Haruma
Giáo viên
23 tháng 12 2021
Lời giải:
PT $\Leftrightarrow (4x^2-4x+1)-3|2x-1|+2=0$
$\Leftrightarrow (2x-1)^2-3|2x-1|+2=0$
$\Leftrightarrow |2x-1|^2-3|2x-1|+2=0$
$\Leftrightarrow (|2x-1|-1)(|2x-1|-2)=0$
$\Rightarrow |2x-1|=1$ hoặc $|2x-1|=2$
$\Leftrightarrow 2x-1=\pm 1$ hoặc $2x-1=\pm 2$
$\Rightarrow x\in \left\{0; 1; \frac{3}{2}; \frac{-1}{2}\right\}$
PT
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2020
a.
\(\left(4x^2+4x+1\right)-y^2=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)
b.
\(\Leftrightarrow2x^2+2x-x-1-2x^2-3x+1=0\)
\(\Leftrightarrow-2x=0\)
\(\Leftrightarrow x=0\)
19 tháng 8 2021
a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(3x+5+2x\right)\left(3x+5-2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
19 tháng 6 2021
\(4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy...
TG
19 tháng 6 2021
\(4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
<=> 2x + 1 = 0
<=> x = -1/2
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
\(x^6+2x^3+1=0\)
\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)
\(\Leftrightarrow\left(x^3+1\right)^2=0\)
\(\Leftrightarrow x^3=\left(-1\right)^3\)
\(\Leftrightarrow x=-1\)
___________
\(x\left(x-5\right)=4x-20\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
_____________
\(x^4-2x^2=8-4x^2\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+4\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
_______________
\(\left(x^3-x^2\right)-4x^2+8x-4\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
-4x2 + 4x - 3
= -4x2 + 4x - 1 - 2
= -( 4x2 - 4x + 1 ) - 2
= -( 2x - 1 )2 - 2
\(-\left(2x-1\right)^2\le0\forall x\Rightarrow-\left(2x-1\right)^2-2\le-2< 0\forall x\)( đpcm )