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1: \(=\dfrac{x^2-1}{x\left(x^2-1\right)}=\dfrac{1}{x}\)
2: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{y\left(x-2\right)}=\dfrac{x+2}{y}\)
3: \(=\dfrac{2x^2+2xy-xy-y^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)\left(2x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2x-y}{x-y}\)
4: \(=\dfrac{x\left(x^2-1\right)}{x\left(x^2-x-2\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{x-1}{x-2}\)
a: \(=\dfrac{5x^2y^4}{-10x^2y}=-\dfrac{1}{2}y^3=-\dfrac{1}{2}\cdot8=-4\)
b: \(=\dfrac{15x^4y^2}{5x^3y}+\dfrac{20x^3y^2}{5x^2y}=3xy+4xy=7xy\)
\(=7\cdot\dfrac{1}{7}\cdot2009=2009\)
1,\(\dfrac{x^2-6x+9}{x^2-8x+15}=\dfrac{\left(x-3\right)^2}{\left(x-3\right).\left(x-5\right)}=\dfrac{x-3}{x-5}\)
2,\(\dfrac{x^2+5x}{2x+10}=\dfrac{x.\left(x+5\right)}{2.\left(x+5\right)}=\dfrac{x}{2}\)
3,\(\dfrac{25-10x+x^2}{xy-5y}=\dfrac{\left(x-5\right)^2}{y.\left(x-5\right)}=\dfrac{x-5}{y}\)
4,\(\dfrac{x^2+3x-y^2-3y}{x^2-y^2}\\ \\ =\dfrac{\left(x+y\right).\left(x-y\right)+3.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}\\ \\ =\dfrac{\left(x-y\right).\left(x+y+3\right)}{\left(x-y\right).\left(x+y\right)}\\ \\ =\dfrac{x+y+3}{x+y}\)5,\(\dfrac{x^3+2x^2-x-2}{x^3-3x+2}=\dfrac{x^2.\left(x+2\right)-\left(x+2\right)}{x.\left(x^2-1\right)-2.\left(x-1\right)}\\ \\ \dfrac{\left(x+2\right).\left(x^2-1\right)}{x.\left(x+1\right).\left(x-1\right)-2.\left(x-1\right)}\\ =\dfrac{\left(x+2\right).\left(x+1\right).\left(x-1\right)}{\left(x-1\right).\left[\left(x+1\right).x-2\right]}=\dfrac{\left(x+2\right).\left(x+1\right)}{\left(x+1\right).x-2}\)
1.x2-y2+2x+1=(x2+2x+1)-y2=(x+1)2-y2=(x+1-y)(x+1+y)
2.(x2+9)2-36x2=(x2+9)2-(6x)2=(x2+9-6x)(x2+9+6x)=(x-3)2(x+3)2
3.\(8x^3+\dfrac{1}{27}=\left(2x\right)^3+\left(\dfrac{1}{3}\right)^3\\ =\left(2x+\dfrac{1}{3}\right)\text{[}\left(2x\right)^2-2x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\text{]}\\ =\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)4.x3-8y3=x3-(2y)3=(x-2y)(x2+2xy+4y2)
a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)
\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)
\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)
\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)
\(=\left(x^2+9x+19\right)^2\)
b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)
\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)
c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)
\(=x^2-2x\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(x-y-2\right)^2\)
d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)
\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
Giải:
1) \(a^3-3a^2+3a-1\)
\(=a^3-3a^2.1+3a.1^2-1^3\)
\(=\left(a-1\right)^3\)
Vậy ...
2) \(x^3+6x^2+12x+8\)
\(=x^3+3.x^2.2+3.x.2^2+2^3\)
\(=\left(x+2\right)^3\)
Vậy ...
3) \(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)
\(=\left(2x-1\right)^3\)
Vậy ...
4) \(x^3-6x^2y+12xy^2-8y^3\)
\(=x^3-3.x^2.2y+3.x.\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(x-2y\right)^3\)
Vậy ...
1: \(\Leftrightarrow3x+4x=4\)
=>7x=4
hay x=4/7
2: \(\Leftrightarrow3x-5x-5^3:5^2=0\)
=>-2x=5
=>x=-5/2
a) A = x^3 + 6x^2y + 12xy^2 + 8y^3
=> A = ( x + 2y )^3
Thay x + 2y = -5 vào A
=> A = ( -5 )^3 = -125
Vậy khi x + 2y = -5 thì A = -125
b) B = 8x^3 - 12x^2y + 6xy^2 - y^3
=> B = ( 2x - y )^3
Thay 2x - y = 1/5 vào A
=> A = ( 1/5 )^3 = 1/125
Vậy khi 2x - y = 1/5 thì B = 1/125
c) C = x^3 + 3x^2 + 3x + 1
=> C = ( x + 1 )^3
Thay x = 99 vào C
=> C = ( 99 + 1 )^3 = 100^3 = 1000000
Vậy khi x = 99 thì C = 1000000
A=26x2+y(2x+y)-10x(x+y)
A=26x2+2xy+y2-10x2-10xy
A=16x2-8xy+y2 =(4x)2-2.4x.y+y2 =(4x-y)2
Thay x=0,25y,ta có: A=(4.0,25y - y)2=(y-y)2=0
B=x3+6x2y+12xy2+8y3
B=x3+3x22y+3x(2y)2+(2y)3 =(x+2y)3
Có x+2y=-5 ⇒ x=-5-2y
Thay x=-5-2y vào, ta có B=(-5-2y+2y)3=(-5)3=-125