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4 tháng 10 2021
\(7x\left(16x^2-1\right)=0\)
\(\Leftrightarrow7x\left(4x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
11 tháng 1 2023
=>2x(x^2-11x+18)=0
=>x(x-2)(x-9)=0
=>\(x\in\left\{0;2;9\right\}\)
14 tháng 10 2021
\(a,\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ b,\Leftrightarrow\left(3x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{3}\)
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) \(D=\left(4x-3\right)\left(16x^2+12x+9\right)-\left(4x+3\right)\left(16x^2-12x+9\right)\)
\(=\left(4x\right)^3-3^3-\left[\left(4x\right)^3+3^3\right]\)
\(=64x^3-27-\left(64x^3+27\right)\)
\(=64x^3-27-64x^3-27\)
\(=-27-27\)
\(=-54\)
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
\(4-x=2\left(x-4\right)^2\)
\(\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)
\(\Leftrightarrow4-x=2x^2-16x+32\)
\(\Leftrightarrow2x^2-15x+28=0\)
\(\Leftrightarrow2x^2-7x-8x+28=0\)
\(\Leftrightarrow x\left(2x-7\right)-4\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
___________
\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Leftrightarrow x^3-2x^2+x-2+2x=4\)
\(\Leftrightarrow x^3-2x^2+3x-2-4=0\)
\(\Leftrightarrow x^3-2x^2+3x-6=0\)
\(\Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-3\left(\text{vô lý}\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
________________
\(x^4-16x^2=0\)
\(\Leftrightarrow\left(x^2\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)x\left(x+4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
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
CC
1
10 tháng 8 2023
\(x^4-6x^3+16x^2-22x+16=0\)
\(\Rightarrow x^4-2x^3+3x^2-4x^3+8x^2-12x+5x^2-10x+15+1=0\)
\(\Rightarrow x^2\left(x^2-2x+3\right)-4x\left(x^2-2x+3\right)+5\left(x^2-2x+3\right)x^2+1=0\)
\(\Rightarrow\left(x^2-2x+3\right)\left(x^2-4x+5\right)=-1\)
\(\Rightarrow\left(x^2-2x+1+2\right)\left(x^2-4x+4+1\right)=-1\)
\(\Rightarrow\left[\left(x-1\right)^2+2\right]\left[\left(x-2\right)^2+1\right]=-1\left(1\right)\)
mà \(\left\{{}\begin{matrix}\left(x-1\right)^2+2>0,\forall x\\\left(x-2\right)^2+1>0,\forall x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[\left(x-1\right)^2+2\right]\left[\left(x-2\right)^2+1\right]>0,\forall x\\\left[\left(x-1\right)^2+2\right]\left[\left(x-2\right)^2+1\right]=-1\end{matrix}\right.\) (vô lí)
Vậy phương trình trên vô nghiệm (dpcm)
C
2
7 tháng 6 2021
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
7 tháng 6 2021
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
pt <=> (16x^2-24x)+(2x-3) = 0
<=> (2x-3).(8x+1) = 0
<=> 2x-3=0 hoặc 8x+1=0
<=> x=3/2 hoặc x=-1/8
Vậy ..............
Theo bài ra ta có:\(16.x^2+22x-3=0\)
\(\Rightarrow\left(16x+22\right)x=3=1.3=3.1=\left(-1\right).\left(-3\right)=\left(-3\right).\left(-1\right)\)
Tự kẻ bẳng nha