x ^{ 4 } +3x ^{ 3 } -12x ^{ 2 } -3x+1 = 0
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NH
23 tháng 9 2018
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
4 tháng 9 2022
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
23 tháng 10 2017
a)
\(12x^2-3x=6\\ \Leftrightarrow x^2-\dfrac{1}{4}x=\dfrac{1}{2}\\ \Leftrightarrow x^2-2.\dfrac{1}{8}x+\left(\dfrac{1}{8}\right)^2=\dfrac{1}{2}+\left(\dfrac{1}{8}\right)^2=\dfrac{33}{64}\\ \Leftrightarrow\left(x-\dfrac{1}{8}\right)^2=\dfrac{33}{64}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{8}=\dfrac{\sqrt{33}}{8}\\x-\dfrac{1}{8}=-\dfrac{\sqrt{33}}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{33}}{8}\\x=\dfrac{1-\sqrt{33}}{8}\end{matrix}\right.\)
b)
\(x^2-4x+3=0\\ \Leftrightarrow x^2-4x+4=-3+4=1\\ \Leftrightarrow\left(x-2\right)^2=1\\ \Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
23 tháng 10 2017
c)
\(3x^2-12x=0\\ \Leftrightarrow x^2-4x=0\\ \Leftrightarrow x^2-4x+4=4\\ \Leftrightarrow\left(x-2\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
d) TH1:
\(x^2+3x+4=0\\ \Leftrightarrow x^2+2.1,5x+\left(1,5\right)^2=\left(1,5\right)^2-4=-\dfrac{7}{4}\\ \Leftrightarrow\left(x+1,5\right)^2=-\dfrac{7}{4}\left(vô\:lí\right)\)
do đó pt trên vô nghiệm
TH2:
\(x^2+3x-4=0\\ \Leftrightarrow x^2+2.\dfrac{3}{2}x+\dfrac{3}{2}=4+\dfrac{3}{2}=\dfrac{25}{4}\\ \Leftrightarrow\left(x+\dfrac{3}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{2}=\dfrac{5}{2}\\x+\dfrac{3}{2}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{2}=1\\x=-\dfrac{8}{2}=-4\end{matrix}\right.\)
25 tháng 1 2021
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
L
1
AH
Akai Haruma
Giáo viên
28 tháng 1
Lời giải:
Xét hiệu \(\frac{(x+1)^2}{3x^2-2x+1}-\frac{12x+4}{3}=\frac{-(3x-1)^2(4x+1)}{3(3x^2-2x+1)}\)
Với $x>0$ thì $-(3x-1)^2(4x+1)<0$
$3(3x^2-2x+1)=3[2x^2+(x-1)^2]>0$ với mọi $x>0$
$\Rightarrow \frac{(x+1)^2}{3x^2-2x+1}-\frac{12x+4}{3}=\frac{-(3x-1)^2(4x+1)}{3(3x^2-2x+1)}<0$
$\Rightarrow \frac{(x+1)^2}{3x^2-2x+1}<\frac{12x+4}{3}$
Ta có đpcm.
tính x