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a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)
b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)
c: \(=\dfrac{x}{2x+1}-\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x-1-\left(x-2\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x^2-x-1-2x^2-x+4x+2}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{1}{2x-1}\)
d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)
\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)
Bài 1:
a) \(3x\left(5x^2-2x+1\right)\)
\(=15x^3-6x^2+3x\)
b) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^2\left(x^2-1\right)+2x\left(x^2-1\right)\)
\(=x^4-x^2+2x^3-2x\)
\(=x^4+2x^3-x^2-2x\)
Bài 2:
a) \(3x^2=2x\)
\(\Leftrightarrow3x^2-2x=0\)
\(\Leftrightarrow x\left(3x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
b)\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow14x=4\Leftrightarrow x=\frac{2}{7}\)
\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)
a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y
= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )
= 3x2 - 15x + 2xy - 10y - x + 5 - 2xy + 4y
= 3x2 - 16x - 6y + 5
b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 )
= [ ( 3x )2 - 22 ] - ( 8x2 + 10x + 3 )
= 9x2 - 4 - 8x2 - 10x - 3
= x2 - 10 - 7
a, (x-1).(x-2).(x-3)
= (x2 - 2x - x + 2) . (x-3)
= (x2 - 3x + 2). (x-3)4
= x3 - 3x2 - 3x2 + 9x + 2x -6
= x3 - 6x2 + 11x -6
b) (x2 +x+1)(x2-1)(x2-x+1)
= (x4 - x2 + x3 - x+ x2 -1) . (x2 - x +1)
= (x4 + x3 -x -1) . (x2 - x +1)
= x6 - x5 + x4 + x5 - x4 + x3 - x2 + x -1
= x6 + x3 - x2 + x - 1
c) (2x-5)(4-3x)-(3x+11)(5-2x)-15(2x-5)
= (8x - 6x2 - 20 + 15x) - (15x-6x+55-22x) - 30x + 75
= 8x - 6x2 - 20 + 15x - 15x+6x-55+22x - 30x+75
= 6x-6x2 +55
d)(x2-2x+3)(3x-5)-(x2+x-1)(2x+7)
làm tương tự phần C
lưu ý trước dấu ngoặc là dấu trừ, khi phá ngoặc ra phải đổi dấu
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)
\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Câu b/ quên làm ạ :> Bù nè
b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)
\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)
\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-12x^2+11x-2\)
\(=37x-12\)