3 nhân ( 4+x) -2xmũ2-8x=0
K
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17 tháng 7 2021
a) \(x^2-9=\left(x-3\right)\left(x+3\right)\)
b) \(x^2+1-\dfrac{41}{25}=x^2-\dfrac{16}{25}=\left(x-\dfrac{4}{5}\right)\left(x+\dfrac{4}{5}\right)\)
LN
1
16 tháng 8 2018
a
4x2--25=0
=> (2x)22 --52 =0
=> (2x-5)(2x+5)=0
\(\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}X=\frac{5}{2}\\X=\frac{-5\:\:. \:\:\:\:\:\:\:\:\:\:TT}{2}\end{cases}Mình\:}\)
16 tháng 8 2018
\(4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow x=\sqrt{\frac{25}{4}}\) \(=\frac{5}{2}\)
\(\left(x^3-x^2\right)^2-\left(4x^2-8x+4\right)=0\)
= \(\left(x^3-x^2\right)^2-\left(2x-2\right)^2=0\)
=(\(\left(x^3-x^2-2x+2\right)\left(x^3-x^2+2x-2\right)=0\)
=\(\left[x^2\left(x-1\right)-2\left(x-1\right)\right]\) \(\left[x^2\left(x-1\right)+2\left(x-1\right)\right]\)=0
=\(\left(x-1\right)\left(x^2-2\right)\left(x-1\right)\left(x^2+2\right)\) = 0
= \(\left(x-1\right)\left(x^2-2\right)\left(x^2+2\right)=0\)
=\(\left(x-1\right)\left(x^4-4\right)\) = 0
=> \(x-1=0\) hoặc \(x^4-4=0\)
=> \(x=1\) hoặc \(x=\pm\sqrt{2}\)
câu 2
a)\(\left(3x^2\right)^3-\left(2x\right)^3\)
= \(\left(3x^2-2x\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)
= \(x\left(3x-2\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)
may be wrong , but chawsc k nhiều , chỗ nào k hiểu ib hỏi mk sai nha <3
14 tháng 8 2018
1,
a, \(\left(2x-5\right)\cdot\left(2x+5\right)=0\)
\(x=\frac{5}{2}\)
x\(=-\frac{5}{2}\)
b \(\left(x^3-x^2\right)^2-\left(2x-2\right)^2\)=0
(x-2x+2)(x+2x-2)=0
x=2
x=2/3
2,
a (3x^2)^3-(2x)^3
(3x^2-2x)(9x^4+6x^3+4x^2)
14 tháng 8 2018
\(4x^2-25=0\)
\(\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{5}{2}\end{cases}}\)
\(27x^6-8x^3=\left(3x^2\right)^3-\left(2x\right)^3=\left(3x^2-2x\right)\left[\left(3x^2\right)^2+3x^2.2x+\left(2x\right)^2\right]=x^3.\left(3x-2\right).\left(3x^2+6x+4\right)\)
RR
14 tháng 8 2018
1a) 4x2 - 25 = 0 => 4x2 = 25 => x2 = \(\frac{25}{4}\)= \(\left(\frac{5}{2}\right)^2\)=> x = \(\frac{5}{2}\)
KK
1
MH
17 tháng 2 2021
1/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
2/ \(x^3+3x^2+6x+4=0\)
\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))
\(\Leftrightarrow x=-1\).
3/ \(x^3-6x^2+8x=0\)
\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)
4/ \(x^4-8x^3-9x^2=0\)
\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)
4 tháng 3 2019
Với dạng bài này ta chỉ việc chia hoocne là ra nhé!
\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)
\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
KN
27 tháng 12 2019
\(M=3x^4-8x^3-6x^2+8x+3\)
\(=3x^4-12x^3+4x^3+9x^2+x^2-16x^2+12x-4x+3\)
\(=\left(3x^4-12x^3+9x^2\right)+\left(4x^3-16x^2+12x\right)+\left(x^2-4x+3\right)\)
\(=3x^2\left(x^2-4x+3\right)+4x\left(x^2-4x+3\right)+\left(x^2-4x+3\right)\)
\(=\left(3x^2+4x+1\right)\left(x^2-4x+3\right)\)
\(=\left(3x^2+3x+x+1\right)\left(x^2-3x-x+3\right)\)
\(=\left[3x\left(x+1\right)+\left(x+1\right)\right]\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(3x+1\right)\left(x+1\right)\left(x-1\right)\left(x-3\right)\)
M = 0\(\Leftrightarrow\left(3x+1\right)\left(x+1\right)\left(x-1\right)\left(x-3\right)=0\)
\(\Rightarrow x\in\left\{\frac{-1}{3};-1;1;3\right\}\)
\(3\left(4+x\right)-2x^2-8x=0\)
\(\Leftrightarrow2x^2+5x-12=0\)
Giải PT bậc 2 \(x_1=-4;x_2=\frac{3}{2}\)