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Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x-2}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{x^2-2x+1}{x\left(x+1\right)}:\dfrac{x^2-2x+1}{x}\)
\(=\dfrac{1}{x+1}\)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
Trả lời:
Bài 4:
b, B = ( x + 1 ) ( x7 - x6 + x5 - x4 + x3 - x2 + x - 1 )
= x8 - x7 + x6 - x5 + x4 - x3 + x2 - x + x7 - x6 + x5 - x4 + x3 - x2 + x - 1
= x8 - 1
Thay x = 2 vào biểu thức B, ta có:
28 - 1 = 255
c, C = ( x + 1 ) ( x6 - x5 + x4 - x3 + x2 - x + 1 )
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7 + 1
Thay x = 2 vào biểu thức C, ta có:
27 + 1 = 129
d, D = 2x ( 10x2 - 5x - 2 ) - 5x ( 4x2 - 2x - 1 )
= 20x3 - 10x2 - 4x - 20x3 + 10x2 + 5x
= x
Thay x = - 5 vào biểu thức D, ta có:
D = - 5
Bài 5:
a, A = ( x3 - x2y + xy2 - y3 ) ( x + y )
= x4 + x3y - x3y - x2y2 + x2y2 + xy3 - xy3 - y4
= x4 - y4
Thay x = 2; y = - 1/2 vào biểu thức A, ta có:
A = 24 - ( - 1/2 )4 = 16 - 1/16 = 255/16
b, B = ( a - b ) ( a4 + a3b + a2b2 + ab3 + b4 )
= a5 + a4b + a3b2 + a2b3 + ab4 - ab4 - a3b2 - a2b3 - ab4 - b5
= a5 + a4b - ab4 - b5
Thay a = 3; b = - 2 vào biểu thức B, ta có:
B = 35 + 34.( - 2 ) - 3.( - 2 )4 - ( - 2 )5 = 243 - 162 - 48 + 32 = 65
c, ( x2 - 2xy + 2y2 ) ( x2 + y2 ) + 2x3y - 3x2y2 + 2xy3
= x4 + x2y2 - 2x3y - 2xy3 + 2x2y2 + 2y4 + 2x3y - 3x2y2 + 2xy3
= x4 + 2y4
Thay x = - 1/2; y = - 1/2 vào biểu thức trên, ta có:
( - 1/2 )4 + 2.( - 1/2 )4 = 1/16 + 2. 1/16 = 1/16 + 1/8 = 3/16
b: \(\dfrac{xy}{2x-y}-\dfrac{2x^2}{y-2x}=\dfrac{xy}{2x-y}+\dfrac{2x^2}{2x-y}=\dfrac{xy+2x^2}{2x-y}\)
b: \(\dfrac{3x^2-x}{x-1}+\dfrac{x+2}{1-x}+\dfrac{3-2x^2}{x-1}\)
\(=\dfrac{3x^2-x-x-2+3-2x^2}{x-1}\)
\(=\dfrac{x^2-2x+1}{x-1}=x-1\)
\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\\ =\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\\ =\dfrac{6.2}{2x\left(x+4\right)}+\dfrac{3x}{2x\left(x+4\right)}\\ =\dfrac{12+3x}{2x\left(x+4\right)}\\ =\dfrac{3\left(4+x\right)}{2x\left(x+4\right)}\\ =\dfrac{3}{2x}\)
________
\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\\ \left(\text{đ}k\text{x}\text{đ}:x\ne\pm2\right)\\ =\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}+\dfrac{x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x+x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2x^2-8}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{2\left(x^2-4\right)}{x^2-4}\\ =2\)
a: \(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)
\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)
b: \(=\dfrac{\left(x+1\right)\left(x+2\right)+\left(x-2\right)^2+x-14}{x^2-4}\)
\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{x^2-4}=\dfrac{2x^2-8}{x^2-4}=2\)
Ta có: