1 Thực hiện phép tính
a) 2 . ( x^2 - x ) - 2x^2 = 3
b) 2 . ( x^2 + x ) - 2x = 8
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a) \(=\dfrac{x+15}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{x+3}=\dfrac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
b) \(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{y^2+x^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)^2-\left(x-y\right)^2+2\left(x^2+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2\left(x^2+y^2+2xy\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}\)
a) 2(x-1)2 - 4(x+3)2 + 2x(x-5)
= 2(x2 -2x +1)- 4(x2 + 6x +9) + 2x2 -10x
= 2x2 - 4x + 2 -4x2 - 24x - 36 + 2x2 - 10x
= (2x2 + 2x2 - 4x2) - (4x + 24x+10x) +(2-36)
= -38x-34
b) 2(2x+5)2 -3(4x+1)(1-4x)
= 2(4x2 + 20x + 25) + 3(4x+1)(4x-1)
= 8x2 +40x + 50 + 3(16x2 -1)
= 8x2 + 40x + 50 + 48x2 - 3
=56x2 +40x + 47
a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)
\(=2\left(x^2-2x+1\right)-4\left(x^2+6x+9\right)+2x\left(x-5\right)\)
\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)
b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)
\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)
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
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
a) \(2\left(x^2-x\right)-2x^2=3\)
\(\Leftrightarrow2x^2-2x-2x^2=3\)
\(\Leftrightarrow-2x=3\Leftrightarrow x=-\frac{3}{2}\)
b) \(2\left(x^2+x\right)-2x=8\)
\(\Leftrightarrow2x^2+2x-2x=8\)
\(\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
a) 2 . ( x^2 - x ) - 2x^2 = 3
2x2-2x-2x2=0
-2x=0
x=0
Vậy x=0
b) 2 . ( x^2 + x ) - 2x = 8
2x2+2x-2x=8
2x2=8
x2=4=22=(-2)2
Vậy x=2;-2