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2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
Bài 1.
A = x2 + 2xy + y2 = ( x + y )2 = ( -1 )2 = 1
B = x2 + y2 = ( x2 + 2xy + y2 ) - 2xy = ( x + y )2 - 2xy = (-1)2 - 2.(-12) = 1 + 24 = 25
C = x3 + 3xy( x + y ) + y3 = ( x3 + y3 ) + 3xy( x + y ) = ( x + y )( x2 - xy + y2 ) + 3xy( x + y )
= -1( 25 + 12 ) + 3.(-12).(-1)
= -37 + 36
= -1
D = x3 + y3 = ( x3 + 3x2y + 3xy2 + y3 ) - 3x2y - 3xy2 = ( x + y )3 - 3xy( x + y ) = (-1)3 - 3.(-12).(-1) = -1 - 36 = -37
Bài 2.
M = 3( x2 + y2 ) - 2( x3 + y3 )
= 3( x2 + y2 ) - 2( x + y )( x2 - xy + y2 )
= 3( x2 + y2 ) - 2( x2 - xy + y2 )
= 3x2 + 3y2 - 2x2 + 2xy - 2y2
= x2 + 2xy + y2
= ( x + y )2 = 12 = 1
1/Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=81\)
\(\Rightarrow M=ab+bc+ca=\frac{\left(81-141\right)}{2}\)
a: \(=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x+1\right)\left(2x-1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x\cdot5}{4x\left(2x+1\right)}=\dfrac{10}{2x+1}\)
b: \(=\left(\dfrac{1}{x^2+1}+\dfrac{x-2}{x+1}\right):\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{x+1+x^3+x-2x^2-2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x^3-2x^2+2x-1}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x\left(x^2-x+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
\(1,3x-24y=3\left(x-8y\right)\)
\(2,6x^3y^2-12x^2y^2-3x^2y=3x^2y\left(2xy-4y-1\right)\)
\(3,7x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(7x-8\right)\)
...(tương tự)
\(10,5x-5y+x^2-xy=5\left(x-y\right)+x\left(x-y\right)=\left(x-y\right)\left(x+5\right)\)
\(11,x^2+2xy+y^2-16=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)
x+y=9=>x=9-y thay x=9-y vào xy=14
=> (9-y).y=14
=> 9y-y^2 -14=0
=> 7y+2y-y^2-14=0
=> y(7-y)-2(7-y)=0
=>(7-y)(y-2)=0
=>y=2 hoặc 7
=> x=7 hoặc 2
=> (x;y)=(2,7);(7,2)
a.
x-y = 7-2 = 5
hoặc 2-7=-5
b. x^2+y^2= 7^2+2^2 =53
hoặc 2^2+7^2=53
c. x^3+y^3 = 7^3 +2^3 = 351
hoặc 2^3 +7^3 =351
d.x^4+y^4= 7^4 +2^4= 2417
hoặc 2^4+7^4=2417
Câu 2:
a: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2-2x+1=\left(x-1\right)^2\)
b: \(=\dfrac{x^3-3x^2+2x^2-6x-x+3}{x-3}=x^2+2x-1\)
13.
M \(=\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)\)\(+16\)
\(=\)\(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)\) \(+16\)
\(=\left(x^2+10x+20\right)^2-16+16\)
\(=\left(x^2+10x+20\right)^2\) là một số chính phương
Nhiều quá, nhìn đã thấy ớn lạnh :(
Bạn nên chia nhỏ ra , post 1 hoặc 2 bài 1 lần thôi, đăng 1 lần 1 nùi thế này không ai dám làm đâu, bội thực chữ viết.
Bài 1:
a: \(x\left(x+y\right)+5y-x^2\)
\(=x^2+xy+5y-x^2\)
=xy+5y
b: \(\left(x-2\right)\left(y+1\right)-xy+4\)
\(=xy+x-2y-2-xy+4\)
=-2y+x+2
c: \(\dfrac{\left(4x^2y+12xy^2-8xy\right)}{2xy}\)
\(=\dfrac{2xy\cdot2x+2xy\cdot6y-2xy\cdot4}{2xy}\)
=2x+6y-4
d: \(\left(x-4\right)^2+8x-7\)
\(=x^2-8x+16+8x-7\)
\(=x^2+9\)