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\(a,x\left(-3x+5\right)+3x\left(x+1\right)-40=0\)
\(\left(x.-3x\right)+\left(5x\right)+3x\left(x+1\right)-40=0\)
\(-3x^2+5x+\left(3x.x\right)+\left(3x.1\right)-40=0\)
\(-3x^2+5x+3x^2+3x-40=0\)
\(\left(-3x^2+3x^2\right)+5x+3x-40=0\)
\(8x-40=0\)
\(8x=0+40=40\)
\(x=40:8=5\)
a) \(x\left(5-3x\right)+3x\left(x+1\right)-40=0\)
\(\Rightarrow5x-3x^2+3x^2+3x-40=0\)
\(\Rightarrow8x-40=0\)
\(\Rightarrow8x=40\)
\(\Rightarrow x=5\)
b) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
\(\Rightarrow83x=83\)
\(\Rightarrow x=1\)
1) \(16x\left(2-x\right)-\left(4x-5\right)^2=0\)
\(32x-16x^2-16x^2+40x-25=0\)
\(72x-16x^2-25=0\)
Đề sai ko bạn 
2) \(\left(x-7\right)^2+3=\left(x-2\right)\left(x+2\right)\)
\(\left(x^2-14x+7\right)+3-\left(x-2\right)\left(x+2\right)=0\)
\(x^2-14x+7+3-x^2+4=0\)
\(-14x+14=0\)
\(x=1\)
3) \(\left(2x-3\right)^2-\left(7x-2x\right)^2=2\)
\(\left(2x-3\right)^2-\left(5x\right)^2=2\)
\(\left(2x-3-5x\right)\left(2x-3+5x\right)=2\)
\(\left(-3x-3\right)\left(7x-3\right)=2\)
=> lập bảng tìm x
4) \(\left(5x-7\right)^2-\left(1-3x\right)^2=16x\left(x-3\right)\)
\(25x^2-70x+49-9x^2+6x-1-16x^2+48x=0\)
\(-16x+48=0\)
\(x=3\)
Bài 1 :
a, \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2-9\right)\)
\(=x^2+6x+9+x^2-6x+9+2x^2-18\)
\(=4x^2\)
b, \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)
\(=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9=8\)
1.
a) x4 + x3 + x + 1 = x3( x + 1 ) + ( x + 1 ) = ( x + 1 )( x3 + 1 ) = ( x + 1 )( x + 1 )( x2 - x + 1 ) = ( x + 1 )2( x2 - x + 1 )
b) x2y + xy2 - x - y = xy( x + y ) - ( x + y ) = ( x + y )( xy - 1 )
c) x2 - 2xy + y2 - xz + yz = ( x2 - 2xy + y2 ) - ( xz - yz ) = ( x - y )2 - z( x - y ) = ( x - y )( x - y - z )
d) ax - ab + b - x = ( ax - x ) - ( ab - b ) = x( a - 1 ) - b( a - 1 ) = ( a - 1 )( x - b )
2.
( 2x - 1 )2 - 25 = 0
<=> ( 2x - 1 )2 - 52 = 0
<=> ( 2x - 1 - 5 )( 2x - 1 + 5 ) = 0
<=> ( 2x - 6 )( 2x + 4 ) = 0
<=> \(\orbr{\begin{cases}2x-6=0\\2x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
3x( x - 1 ) + x - 1 = 0
<=> 3x( x - 1 ) + ( x - 1 ) = 0
<=> ( x - 1 )( 3x + 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)
B1:
a) \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
b) \(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(xy-1\right)\left(x+y\right)\)
c) \(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
d) \(ax-ab+b-x\)
\(=a\left(x-b\right)-\left(x-b\right)\)
\(=\left(a-1\right)\left(x-b\right)\)
\(A.x^2-16x=0\)
\(x^2-\left(4x\right)^2=0\)
\(\left(x-4x\right)\left(x+4x\right)=0\)
\(\left(-3x\right)\left(5x\right)=0\)
\(\Rightarrow\) \(-3x=0\) hoặc \(5x=0\)
\(x=\dfrac{0}{-3}\) hoặc \(x=\dfrac{0}{5}\)
Vậy \(x=0\) hoặc \(x=0.\)
B. 4x2 - 4x + 1 = 0
(2x)2 - (2x)2 + 12 = 0
(2x - 2x + 1 ) (2x + 2x +1) = 0
1 (4x + 1) =0
=> 1 (4x + 1) =0
4x + 1 = 0
4x = 0-1
Vậy x = \(\dfrac{-1}{4}.\)
C. (3x-1)2 - (2x+3)2 = 0
(3x -1 -2x +3) (3x -1 +2x +3) = 0
(x + 2)(5x + 2) = 0
=> x + 2 =0 hoặc 5x + 2 =0
x = 0 - 2 hoặc 5x = 0 - 2
Vậy x = -2 hoặc x = \(\dfrac{-2}{5}.\)
Còn về câu d thì mình hơi phân vân, tại mình dốt toán lắm
a/ \(x^2-16x=0\)
\(\Leftrightarrow x\left(x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
Vậy...
b/ \(4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy..
c/ \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1-2x-3\right)\left(3x-1+2x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy...
d/ \(2013x^2-2014x+1=0\)
\(\Leftrightarrow2013x^2-x-2013x+1=0\)
\(\Leftrightarrow x\left(2013x-1\right)-\left(2013x-1\right)=0\)
\(\Leftrightarrow\left(2013x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2013x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2013}\\x=1\end{matrix}\right.\)
Vậy..
4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)
=> (a+b)^2=(a-b)^2+4ab
- 2x – x2 + 2 – x – (3x2 + 6x + 5x +10) = – 4x2 + 2
- 2x – x2 + 2 – x – 3x2 – 6x – 5x – 10 = – 4x2 + 2 –10x = 10 x = – 1
- 2x2 – 6x + x – 3 = 0
(x – 3)(2x + 1) = 0
x = 3 hay x = -1/2
a) \(16x^2-1=0\)
\(\Rightarrow16x^2=1\)
\(\Rightarrow x^2=\frac{1}{16}\)
\(\Rightarrow x^2=\left(\pm\frac{1}{4}\right)^2\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{1}{4}\\\frac{-1}{4}\end{cases}}\)
b) \(x^2+\frac{1}{4}=0\)
Ta có: \(x^2\ge0\forall x\Rightarrow x^2+\frac{1}{4}\ge\frac{1}{4}>0\)
=> Vô nghiệm
c) \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x^2-1\right)\left(x+3\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)
Trường hợp 1: \(x-1=0\Rightarrow x=1\)
Trường hợp 2: \(x+1=0\Rightarrow x=-1\)
Trường hợp 3: \(x+3=0\Rightarrow x=-3\)