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a) 3x^3-12x=0
3x(x^2-4)=0
3x(x-2)(x+2)=0
suy ra 3x=0 suy ra x=0
x-2=0 x=2
x+2=0 x= -2
b) (x-3)^2-(x-3)(3-x)^2=0
(x-3)^2-(x-3)(x-3)^2=0
(x-3)^2(1-x+3)=0
(x-3)^2(4-x)=0
suy ra x-3=0 suy ra x=3
4-x=0 x=4
a) và b) đã nhé bạn
a)
pt <=> \(x^2+4x+4+x^2-6x+9=2x^2+14x\)
<=> \(2x^2-2x+13=2x^2+14x\)
<=> \(16x=13\)
<=> \(x=\frac{13}{16}\)
b)
pt <=> \(x^3+3x^2+3x+1+x^3-3x^2+3x-1=2x^3\)
<=> \(2x^3+6x=2x^3\)
<=> \(6x=0\)
<=> \(x=0\)
c)
pt <=> \(\left(x^3-3x^2+3x-1\right)-125=0\)
<=> \(\left(x-1\right)^3=125\)
<=> \(x-1=5\)
<=> \(x=6\)
d)
pt <=> \(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
<=> \(\left(x-1\right)^2+\left(y+2\right)^2=0\) (1)
CÓ: \(\left(x-1\right)^2;\left(y+2\right)^2\ge0\forall x;y\)
=> \(\left(x-1\right)^2+\left(y+2\right)^2\ge0\) (2)
TỪ (1) VÀ (2) => DÁU "=" XẢY RA <=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
e)
pt <=> \(2x^2+8x+8+y^2-2y+1=0\)
<=> \(2\left(x+2\right)^2+\left(y-1\right)^2=0\)
TA LUÔN CÓ: \(2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x;y\)
=> DẤU "=" XẢY RA <=> \(\hept{\begin{cases}2\left(x+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)
a) ( x + 2 )2 + ( x - 3 )2 = 2x( x + 7 )
<=> x2 + 4x + 4 + x2 - 6x + 9 = 2x2 + 14x
<=> x2 + 4x + x2 - 6x - 2x2 - 14x = -4 - 9
<=> -16x = -13
<=> x = 13/16
b) ( x + 1 )3 + ( x - 1 )3 = 2x3
<=> x3 + 3x2 + 3x + 1 + x3 - 3x2 + 3x - 1 = 2x3
<=> x3 + 3x2 + 3x + x3 - 3x2 + 3x - 2x3 = -1 + 1
<=> 6x = 0
<=> x = 0
c) x3 - 3x2 + 3x - 126 = 0
<=> ( x3 - 3x2 + 3x - 1 ) - 125 = 0
<=> ( x - 1 )3 = 125
<=> ( x - 1 )3 = 53
<=> x - 1 = 5
<=> x = 6
d) x2 + y2 - 2x + 4y + 5 = 0
<=> ( x2 - 2x + 1 ) + ( y2 + 4y + 4 ) = 0
<=> ( x - 1 )2 + ( y + 2 )2 = 0 (*)
\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
e) 2x2 + 8x + y2 - 2y + 9 = 0
<=> 2( x2 + 4x + 4 ) + ( y2 - 2y + 1 ) = 0
<=> 2( x + 2 )2 + ( y - 1 )2 = 0 (*)
\(\hept{\begin{cases}2\left(x+2\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+2\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức xảy ra (*) ) <=> \(\hept{\begin{cases}x+2=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=1\end{cases}}\)
a) \(\left(x+2\right)^2-9=0\)
\(=>\left(x+2\right)^2-3^2=0\\ =>\left(x+2-3\right).\left(x+2+3\right)=0\)
\(=>\left(x-1\right).\left(x+5\right)=0\)
\(=>\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x= 1 hoặc x= -5
b) \(x^2-2x+1=25\)
\(=>x^2-2.x.x+1^2=25\)
\(=>\left(x-1\right)^2-25=0\\ =>\left(x-1\right)^2-5^2=0\)
\(=>\left(x-1-5\right).\left(x-1+5\right)=0\)
\(=>\left(x-6\right).\left(x+4\right)=0=>\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Vậy x= 6 hoặc x= -4
c) \(4x\left(x-1\right)-\left(2x+5\right)\left(2x-5\right)=1\)
\(=>4x\left(x-1\right)-\left[\left(2x\right)^2-5^2\right]=1\)
\(=>4x\left(x-1\right)-4x^2+25-1=0\)
\(=>4x\left(x-1\right)-4x^2+24=0\)
\(=>4x\left(x-1\right)-\left(4x^2-24\right)=0\\ =>4x\left(x-1\right)-4\left(x^2-6\right)=0\)
..................... tắc ròi -.-"
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+3\right)=15\)
\(=>x^3+27-x^3-3x=15\)
\(=>27-3x-15=0=>12-3x=0=>3\left(4-x\right)=0\)
Vì \(3>0=>4-x=0=>x=4\)
Vậy x= 4
e) \(3\left(x+2\right)^2+\left(2x+1\right)^2-7\left(x+3\right)\left(x-3\right)=28\)
\(=>3\left(x^2+2.x.2+2^2\right)+4x^2+4x+1-7\left(x^2-9\right)=28\)
\(=>3\left(x^2+4x+4\right)+4x^2+4x+1-7x^2+63=28\)
\(=>3x^2+12x+12+4x^2+4x+1-7x^2+63=28\)
\(=>16x+75=28=>16x=-47=>x=\frac{-47}{16}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>'-'
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
Bài 1 :
\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)
\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)
\(\Leftrightarrow\)\(9x=16\)
\(\Leftrightarrow\)\(x=\frac{16}{9}\)
Vậy \(x=\frac{16}{9}\)
Chúc bạn học tốt ~
a: Sửa đề: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
=>4x-27=1
hay x=7
b: \(x^3-8-\left(x-2\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x^2+3\right)=0\)
=>(x-2)(2x+1)=0
=>x=-1/2 hoặc x=2
c: \(\Leftrightarrow\left(2x+1-x+2\right)\left[\left(4x^2+4x+1\right)+\left(2x+1\right)\left(x-2\right)+\left(x-2\right)^2\right]=0\)
=>x+3=0
hay x=-3