2x(x-3)-16x2(3-x)=0
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4T
2
NH
5 tháng 3 2023
1)
x^3 -16x=0`
`<=>x(x^2 -16)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b)
`x^4 -2x^3=0`
`<=>x^3 (x-2)=0`
\(< =>\left[{}\begin{matrix}x^3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
3)
`(2x-11)(x^2 -1)=0`
\(< =>\left[{}\begin{matrix}2x-11=0\\x^2-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=11\\x^2=1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=1\\x=-1\end{matrix}\right.\)
4)
`x^3 -36x=0`
`<=>x(x^2 -36)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
5)
`2x+19=0`
`<=>2x=-19`
`<=>x=-19/2`
2 tháng 10 2021
Bài 2:
a: \(x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b: \(x^8+36x^4=0\)
\(\Leftrightarrow x^4=0\)
hay x=0
2 tháng 10 2021
a(b+3)-b(3+b)
=(3+b)(a-b)
Thay số, có: (3+1997).(2003-1997)
= 2000.6 =12000
xy(x+y)-2x-2y
xy(x+y)- 2(x+y)
(x+y).(xy-2)
Thay số, co: 7. (8-2)
7.4=28
TT
2
H
8 tháng 8 2023
`4-x=2(x-4)^2`
`<=>4-x=2(x^2-8x+16)`
`<=> 4-x=2x^2 - 16x+32`
`<=> 4-x-2x^2+16x-32=0`
`<=> -2x^2 +15x-28=0`
`<=> -(2x^2-15x+28)=0`
`<=>-(2x^2-7x-8x+28)=0`
`<=> - [x(2x-7) - 4(2x-7)]=0`
`<=> -(2x-7)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}-2x+7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-7\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
__
`(x^2 +1) (x-2)+2x=4`
`<=> x^3 -2x^2 +x-2+2x-4=0`
`<=> x^3 -2x^2 +3x-6=0`
`<=> (x^3+3x)-(2x^2+6)=0`
`<=> x(x^2 +3) -2(x^2+3)=0`
`<=>(x^2+3)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)
__
`x^4 -16x^2=0`
`<=> x^2 (x^2 -16)=0`
`<=>x^2(x-4)(x+4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
\(4-x=2\left(x-4\right)^2\)
\(\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)
\(\Leftrightarrow4-x=2x^2-16x+32\)
\(\Leftrightarrow2x^2-15x+28=0\)
\(\Leftrightarrow2x^2-7x-8x+28=0\)
\(\Leftrightarrow x\left(2x-7\right)-4\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
___________
\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Leftrightarrow x^3-2x^2+x-2+2x=4\)
\(\Leftrightarrow x^3-2x^2+3x-2-4=0\)
\(\Leftrightarrow x^3-2x^2+3x-6=0\)
\(\Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-3\left(\text{vô lý}\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
________________
\(x^4-16x^2=0\)
\(\Leftrightarrow\left(x^2\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)x\left(x+4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
7 tháng 10 2021
a: \(8x\left(x-2017\right)-2x+4034=0\)
\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
18 tháng 11 2021
\(a,\Leftrightarrow x-28=-45\\ \Leftrightarrow x=-27\\ b,\Leftrightarrow3+x=0\\ \Leftrightarrow x=-3\\ c,\Leftrightarrow\left[{}\begin{matrix}7-x=0\\-x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow16\left(x^2-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
22 tháng 12 2022
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
3 tháng 9 2023
c: C=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2(x^2-4)
=250x^3+120x-2x^2+8
=250x^3-2x^2+120x+8
d: D=(4x)^3-3^3-(4x)^3-3^3
=64x^3-27-64x^3-27
=-54
H9
HT.Phong (9A5)
CTVHS
3 tháng 9 2023
c) \(C=\left(5x+2\right)^3+\left(5x-2\right)^3-2\left(x-2\right)\left(x+2\right)\)
\(=\left[\left(5x\right)^3+3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2+2^3\right]+\left[\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2-2^3\right]-2\left(x^2-4\right)\)
\(=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2x^2+8\)
\(=\left(125x^3+125x^3\right)+\left(150x^2-150x^2-2x^2\right)+\left(60x+60x\right)+\left(8-8+8\right)\)
\(=250x^3-2x^2+120x+8\)
d) \(D=\left(4x-3\right)\left(16x^2+12x+9\right)-\left(4x+3\right)\left(16x^2-12x+9\right)\)
\(=\left(4x\right)^3-3^3-\left[\left(4x\right)^3+3^3\right]\)
\(=64x^3-27-\left(64x^3+27\right)\)
\(=64x^3-27-64x^3-27\)
\(=-27-27\)
\(=-54\)
\(2x\left(x-3\right)-16x^2\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(1+8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-\frac{1}{8}\end{matrix}\right.\)