Thực hiện phép tính
a. \(3^2\left(2x^2-5x-4\right)\)
b. \(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x\)
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Trả lời:
Câu 1:
1, A = ( 2x - 3 )( 7x + 8 ) = 14x2 + 16x - 21x - 24 = 14x2 - 5x - 24
2, A = ( 3x + 7 )( x2 - 3x + 2 ) = 3x3 - 9x2 + 6x + 7x2 - 21x + 14 = 3x3 - 2x2 - 15x + 14
3, A = ( x2 - 2x + 5 )( x2 + x - 1 ) = x4 + x3 - x2 - 2x3 - 2x2 + 2x + 5x2 + 5x - 5 = x4 - x3 + 2x2 + 7x - 5
4, A = ( x2 - 3 )( x2 - 7x + 6 ) = x4 - 7x3 + 6x2 - 3x2 + 21x - 18 = x4 - 7x3 + 3x2 + 21x - 18
5, A = ( 2x - y )( 2x + y ) = 4x2 - y2
6, A = ( x - 2y )( x2 + 2xy + 4y2 ) = x3 - 8y3
7, A = ( x + 3y )( x2 - 3xy + 9y2 ) = x3 + 27y3
8, A = ( xy + 1 )( xy - 1 ) = x2y2 - 1
\(A=\left(6x-2\right)^2+2\left(6x-2\right)\left(2-5x\right)+\left(2-5x\right)^2\)
\(=\left(6x-2+2-5x\right)^2=x^2\)
\(B=\left(2a^2+1+2a\right)\left(2a^2+1-2a\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=-4a^2\)
\(C=\left[\left(x^2+3x+1\right)-\left(3x-1\right)\right]^2\)\(=\left(x^2+3x+1-3x+1\right)^2=\left(x^2+2\right)^2\)
1.\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3x+9}\right)=\left(\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right):\left(\frac{3x-9-x^2}{3x\left(x+3\right)}\right)=-\frac{1}{x-3}\)
2.\(\left(\frac{2\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}=\frac{4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}=\frac{x-2}{2}\)
3.\(\left(\frac{3\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right):\frac{2x\left(x+5\right)}{\left(1-3x\right)^2}=\frac{-6x^2+11x+3}{\left(1-3x\right)\left(3x+1\right)}.\frac{\left(1-3x\right)^2}{2x\left(x+5\right)}=\frac{-6x^2+11x+3}{\left(3x+1\right)}.\frac{\left(1-3x\right)}{2x\left(x+5\right)}\)
4.\(\left(\frac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}=\frac{5}{x-5}-\frac{x}{x-5}=-1\)
\(1.A=14x^2+16x-21x-24=14x^2-5x-24\)
\(2.A=3x^3-2x^2-15x+14\)
\(3.A=x^4-x^3+2x^2+7x-5\)
\(4.A=x^4-7x^3+3x^2+21x-18\)
\(5.A=4x^2-y^2\)
\(6.A=x^3-8y^3\)
\(7.A=x^3+27y^3\)
\(8.A=x^2y^2-1\)
3) \(=\frac{3x\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}:\frac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)\(=\frac{3x^2+5x}{\left(1-3x\right)\left(1+3x\right)}.\frac{\left(1-3x\right)^2}{2x\left(3x+5\right)}=\frac{1-3x}{2\left(1+3x\right)}\)
4) \(=\frac{x.x-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}.\frac{x^2+5x}{2x-5}+\frac{x}{5-x}\)\(=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}\)
\(=\frac{5}{x-5}+\frac{-x}{x-5}=\frac{5-x}{x-5}=-1\)
5) \(=\left[\frac{x^2+xy}{\left(x+y\right)\left(x^2+y^2\right)}+\frac{y}{x^2+y^2}\right]:\left[\frac{1}{x-y}-\frac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right]\)
\(=\frac{x+y}{x^2+y^2}:\frac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)\(=\frac{x+y}{x^2+y^2}.\frac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)^2}=\frac{x+y}{x-y}\)
Bài 1 a
\(x^4+3x^3-6x^2-8x=x\left(x^3-8\right)+3x^2\left(x-2\right)\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)+3x^2\left(x-2\right)\)
\(=\left(x-2\right)\left[x\left(x^2+2x+4\right)+3x^2\right]\)
\(=x\left(x-2\right)\left(x^2+5x+4\right)=x\left(x-2\right)\left(x+1\right)\left(x+4\right)\)
Bài 2 :
a, \(8x^3y^5z:2x^2y^3=4xy^2z\)
b, \(\frac{1}{2}x^3y^5:\left(-\frac{3}{4x^2y^3}\right)=-\frac{2x^5y^8}{3}\)
c, \(\left(5x^5-x^3+3x^2\right):3x^2=\frac{5}{3}x^3-3x+1\)
Trả lời:
a, \(3^2\left(2x^2-5x-4\right)=18x^2-45x-36\)
b, \(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
a, \(9\left(2x^2-5x-4\right)=18x^2-45x-36\)
b, \(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x\)
\(=x^2+2x+1+x^2+x-6-4x=2x^2-x-5\)