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H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
\(x^6+2x^3+1=0\)
\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)
\(\Leftrightarrow\left(x^3+1\right)^2=0\)
\(\Leftrightarrow x^3=\left(-1\right)^3\)
\(\Leftrightarrow x=-1\)
___________
\(x\left(x-5\right)=4x-20\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
_____________
\(x^4-2x^2=8-4x^2\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+4\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
_______________
\(\left(x^3-x^2\right)-4x^2+8x-4\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
TN
19 tháng 5 2016
x2(x+2)2+4x2=12(x+2)2
=>x2(x+2)2+4x2-12(x+2)2=0
VT=(x2-2x-4)(x2+6x+12)
pt trở thành (x2-2x-4)(x2+6x+12)=0
=>x2-2x-4=0 hoặc x2+6x+12=0
Th1:x2-2x-4=0
denta:(-2)2-(-4(1.4))=20
x1:(2+\(\sqrt{20}\)):2=1+\(\sqrt{5}\)
x2:(2-\(\sqrt{20}\)):2=\(\sqrt{5}\)+1
Th2:x2+6x+12=0
denta:62-4(1.12)=-12
=>\(\Delta< 0\)
=>vô nghiệm
vậy pt có nghiệm là 1-\(\sqrt{5}\)và \(\sqrt{5}\)+1
T
1
18 tháng 8 2023
a: =>2x^2+9x-6x-27=0
=>x(2x+9)-3(2x+9)=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
b: =>-10x^2+6x-5x+3=0
=>-2x(5x-3)-(5x-3)=0
=>(5x-3)(-2x-1)=0
=>x=-1/2 hoặc x=5/3
c: =>-x^3+2x^2-x^2+4=0
=>-x^2(x-2)-(x-2)(x+2)=0
=>(x-2)(-x^2-x-2)=0
=>x-2=0
=>x=2
d: =>(x^3+8)-4x(x+2)=0
=>(x+2)(x^2-2x+4)-4x(x+2)=0
=>(x+2)(x^2-6x+4)=0
=>x=-2 hoặc \(x=3\pm\sqrt{5}\)
CM
0
CM
0
4 tháng 3 2019
Với dạng bài này ta chỉ việc chia hoocne là ra nhé!
\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)
\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
TA
21 tháng 1 2018
a) [x(x+1].[(x-1)(x+2)]=24
(x2+x)(x2+x+2)=24
Dat x2+x=a , ta dc: a(a+2)=24
=> a2+2a-24=0
=> (a-4)(a+6)=0
=> a=4 hoac a=-6
Thay vao roi tu tim x nha
b)
23 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Lời giải:
\((x^3-x^2)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2(x-1)-4(x^2-2x+1)=0\)
\(\Leftrightarrow x^2(x-1)-4(x-1)^2=0\)
\(\Leftrightarrow (x-1)[x^2-4(x-1)]=0\)
\(\Leftrightarrow (x-1)(x^2-4x+4)=0\)
\(\Leftrightarrow (x-1)(x-2)^2=0\)
\(\Rightarrow \left[\begin{matrix} x=1\\ x=2\end{matrix}\right.\)