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Bài 1 :
(a^2+b^2)(x^2+y^2)=(ax+by)^2
<=> a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 = a^2x^2 + 2abxy + b^2y^2
<=> a^2y^2 + b^2x^2 = 2abxy
<=> a^2y^2 + b^2x^2 - 2abxy = 0
<=> (ay - bx)^2 = 0
=> ay - bx = 0
=> ay = bx
=> a/x = b/y ( x,y khác 0)
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
a) \(\dfrac{2x-6}{x^2-x-6}\)
\(=\dfrac{2\left(x-3\right)}{x^2-3x+2x-6}\)
\(=\dfrac{2\left(x-3\right)}{x\left(x-3\right)+2\left(x-3\right)}\)
\(=\dfrac{2\left(x-3\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\dfrac{2}{x+2}\)
b) \(\dfrac{6x^2-x-2}{4x^2-1}\)
\(=\dfrac{6x^2+3x-4x-2}{\left(2x\right)^2-1^2}\)
\(=\dfrac{3x\left(2x+1\right)-2\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{\left(2x+1\right)\left(3x-2\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{3x-2}{2x-1}\)
\(c,\dfrac{x^3-x^2+3x-3}{x^3+2x^2+3x+6}\)
\(=\dfrac{x^2\left(x-1\right)+3\left(x-1\right)}{x^2\left(x+2\right)+3\left(x+2\right)}\)
\(=\dfrac{\left(x-1\right)\left(x^2+3\right)}{\left(x+2\right)\left(x^2+3\right)}=\dfrac{x-1}{x+2}\)
d,Sửa đề :
\(\dfrac{a^2-b^2+c^2+2ac}{a^2+b^2-c^2+2ab}\)
\(=\dfrac{\left(a^2+2ac+c^2\right)-b^2}{\left(a^2+2ab+b^2\right)-c^2}\)
\(=\dfrac{\left(a+c\right)^2-b^2}{\left(a+b\right)^2-c^2}\)
\(=\dfrac{\left(a-b+c\right)\left(a+b+c\right)}{\left(a+b-c\right)\left(a+b+c\right)}\)
\(=\dfrac{a-b+c}{a+b-c}\)
e,g Đề ko rõ
a) ( a + b + c ) 2 + ( a + b - c ) 2 -2 x ( a+b) 2
2a+2b+2x+2a+2b-2c-2.(2a+2b)
2a+2b+2c+2a+2b-2c-4a-4b
4a+4b-4a-4b=0
b) 2x.( 2x -1 ) 2 -3x.( x+3 )( x-3) - 4x.(x+1).2
2x.(4x-2)-3x2-9x-3x2+9x-4x(2x+2)
8x2-4x-3x2-9x-3x2+9x-8x2-8x
-12x-3x2
c) ( a-b+c).2 -(b-c).2 + 2ab - 2ac
2a-2b+2c-2b+2c+2ab-2ac
2a-4b+4c+2ab-2ac
d) (3x+1).2 - 2(3x+1)( 3x+5 )+(3x+5).2
6x+2-6x-2-6x-10+6x+10=0
Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
a) (12x-5)(4x-1)+(3x-7)(1-16x)
= (48x^2 - 12x - 20x + 5) + (3x - 48x^2 - 7 + 112x)
= 48x^2 - 12x - 20x + 5 +3x - 48x^2 -7 + 112x
= 83x-2
những phần sau bạn cứ làm tương tự theo cách nhân đa thức với đa thức và phá ngiawcj là ra nha :0))
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4
\(3x^n\left(6x^{n-3}+1\right)-2x^n\left(9x^{n-3}-1\right)=18x^{n+n-3}+3x^n-18x^{n+n-3}+2x^n=5x^n\)
\(2x\left(2x-1\right)^2-3x\left(x-3\right)\left(x+3\right)-4x\left(x+1\right)^2=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2-2x=x\left(x^2-16x-x\right)\)